Galvanic and Electrolytic Cells Question Bank Set 5

CHEM 121 - GENERAL CHEMISTRY I - Galvanic and Electrolytic Cells Question Bank - Set 5 Liberty University Question 1 Question ◦ Consider a galvanic cell that consists of a standard hydrogen electrode (EH + /H = 2 ◦ 0 V) and a copper electrode in a 1 M copper(II) sulfate solution (ECu2+ /Cu = 0.34 V). Calculate the cell potential at 25°C. Solution Step 1: Write the overall cell reaction by combining the half-reactions of the two electrodes: Cu2+ (aq) + 2e− → Cu(s) (Reduction half-reaction) 2H+ (aq) + 2e− → H2 (g) (Oxidation half-reaction) 2+ Cu + (aq) + 2H (aq) → Cu(s) + H2 (g) (Overall cell reaction) Step 2: Calculate the standard cell potential using the standard reduction potentials for the half-reactions: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EH+ /H 2 ◦ Ecell = 0.34 V − 0 V = 0.34 V Step 3: Calculate the actual cell potential at 25°C using the Nernst equation: ◦ Ecell = Ecell − RT ln(Q) nF where R = 8.314 J · K−1 · mol−1 (gas constant), T = 298 K (temperature), n = 2 (number of moles of electrons exchanged), F = 96485 C/mol (Faraday 2+ + ] = 11 = 1 (since the concentration of Cu2+ and constant), and Q = [Cu [H2][H ] + H are both 1 M). Ecell = 0.34 V − (8.314 J · K−1 · mol−1 )(298 K) × ln(1) 2(96485 C/mol) Ecell = 0.34 V − (8.314 J · K−1 · mol−1 )(298 K) ×0 2(96485 C/mol) Ecell = 0.34 V Therefore, the cell potential for the galvanic cell at 25°C is 0.34 V. Question 2 Question ◦ Consider a galvanic cell with a standard cell potential of Ecell = 1.10 V. If 2+ the cell operates until the concentration of Cu ions in the copper half-cell is reduced to 0.025 M, calculate the cell potential at this point assuming ideal behavior under standard conditions (T = 25◦ C). Solution Step 1: Write the cell reaction and Nernst equation for the given scenario. The cell reaction for the given galvanic cell is: Cu2+ (aq) + 2e− → Cu(s) The Nernst equation is: ◦ Ecell = Ecell − 0.0592 log n  [Cu2+ ] [Cu2+ ]◦  ◦ where: - Ecell is the cell potential at any given point, - Ecell is the standard cell potential, - n is the number of moles of electrons transferred in the balanced cell reaction, - [Cu2+ ] is the concentration of Cu2+ ions at any given point, [Cu2+ ]◦ is the standard concentration of Cu2+ ions (usually 1 M), - 0.0592 is RT the value of at 25◦ C, and - log denotes the logarithm to the base 10. F Step 2: Calculate the cell potential at the given point. Given that the initial 2 concentration of Cu2+ ions is 1 M and the final concentration is 0.025 M:   0.0592 0.025 Ecell = 1.10 V − log 2 1 = 1.10 V − 0.0296 × log(0.025) = 1.10 V − 0.0296 × (−1.6) = 1.10 V − (−0.0474) = 1.1474 V Therefore, the cell potential at this point is 1.1474 V. Question 3 Question A galvanic cell is constructed with a copper electrode immersed in a 1.0 M Cu2+ solution and a silver electrode immersed in a 1.0 M Ag + solution. If the ◦ ◦ standard reduction potentials are ECu 2+ /Cu = +0.34 V and EAg + /Ag = +0.80 V, determine the cell potential at 25°C. Solution To determine the cell potential, we can use the formula: ◦ ◦ Ecell = Ecathode − Eanode ◦ where Ecathode is the standard reduction potential of the half-reaction at the ◦ cathode and Eanode is the standard reduction potential of the half-reaction at the anode. Step 1: Write the overall cell reaction, with the cathode and anode halfreactions: At the cathode (Cu2+ reduction): Cu2+ (aq) + 2e− → Cu(s) At the anode (Ag + reduction): Ag + (aq) + e− → Ag(s) Overall cell reaction: 2Ag + (aq) + Cu(s) → 2Ag(s) + Cu2+ (aq) Step 2: Calculate the cell potential by using the standard reduction potentials: ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ Ecell = EAg + /Ag − ECu2+ /Cu 3 Ecell = (+0.80 V) − (+0.34 V) Ecell = +0.46 V Therefore, the cell potential at 25°C is +0.46 V. Question 4 Question Consider the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V 2H+ (aq) + 2e− → H2 (g) E ◦ = 0.00 V (a) Write the overall cell reaction for a galvanic cell constructed using these half-reactions. Indicate the anode and cathode. (b) Calculate the standard cell potential, E°cell, for the galvanic cell. Solution (a) The overall cell reaction for a galvanic cell constructed using these halfreactions can be obtained by adding the two half-reactions. The half-reaction with the more positive standard reduction potential will be reduced and the other oxidized. The overall cell reaction is: Zn2+ (aq) + 2H+ (aq) → Zn(s) + H2 (g) In this cell reaction, Zn2+ (aq)isreducedtoZn(s)atthecathode, while2H + (aq)isoxidizedtoH2(g)attheanode. (b) The standard cell potential, E°cell, can be determined by adding the standard reduction potentials for the cathode and anode half-reactions. The ◦ standard cell potential, Ecell , is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode Substitute the given standard reduction potentials: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.00 V − (−0.76 V) ◦ Ecell = 0.76 V Therefore, the standard cell potential for the galvanic cell is 0.76 V. 4 Question 5 Question Consider a galvanic cell with the following half-reactions: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) ◦ Calculate the cell potential (Ecell ) at 25°C for this galvanic cell using standard reduction potentials. Solution To calculate the cell potential, we will use the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode Step 1: Find the standard reduction potentials for each half-reaction. The ◦ standard reduction potentials (Ered ) for the given half-reactions are as follows: ◦ Ered,Zn 2+ +2 e− − −→Zn = −0.76 V ◦ Ered,Cu 2+ +2 e− − −→Cu = 0.34 V ◦ Step 2: Plug the values into the formula to find Ecell . ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.34 V − (−0.76 V) ◦ Ecell = 1.10 V ◦ Therefore, the cell potential (Ecell ) at 25°C for this galvanic cell is 1.10 V. Question 6 Question Consider a galvanic cell with the following half-reactions: Anode: Cathode: Zn(s) → Zn2+ (aq) + 2e− Cu2+ (aq) + 2e− → Cu(s) ◦ ◦ If the standard reduction potentials are EZn 2+ /Zn = −0.76 V and ECu2+ /Cu = ◦ ◦ 0.34 V, calculate the cell potential (Ecell ) at 25 C and comment on the spontaneity of the redox reaction. 5 Solution ◦ 1. Calculate Ecell : ◦ The cell potential (Ecell ) can be calculated using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ where Ecathode is the standard reduction potential of the cathode and ◦ Eanode is the standard reduction potential of the anode. ◦ ◦ Given data: EZn 2+ /Zn = −0.76 V ECu2+ /Cu = 0.34 V ◦ ◦ ◦ Therefore, Ecell = ECu 2+ /Cu − EZn2+ /Zn ◦ Ecell = 0.34 V − (−0.76 V) ◦ Ecell = 1.10 V ◦ Hence, the cell potential (Ecell ) is 1.10 V. 2. Comment on spontaneity: ◦ ◦ The redox reaction is spontaneous when Ecell > 0. In this case, Ecell = 1.10 V, which is greater than 0. Therefore, the redox reaction is spontaneous at 25◦ C. The positive cell potential indicates that the reaction will proceed spontaneously from the anode (Zn) to the cathode (Cu). Question 7 Question Consider a galvanic cell with the following half-reactions: Cathode: Cu2+ (aq) + 2e− → Cu(s) Anode: Zn(s) → Zn2+ (aq) + 2e− Calculate the cell potential at standard conditions (T = 298K, P = 1 atm, and [Cu2+ ] = 1.0M, [Zn2+ ] = 1.0M). Solution Step 1: Write the overall cell reaction by combining the two half-reactions: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Step 2: Write the cell potential equation using the standard reduction potentials of the half-reactions. The standard reduction potentials (E ◦ ) for the given half-reactions are: ◦ ECu 2+ /Cu = 0.34 V ◦ EZn 2+ /Zn = −0.76 V 6 ◦ The cell potential (Ecell ) can be calculated using the equation: ◦ ◦ ◦ Ecell = Ecathode − Eanode Plug in the values: ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EZn2+ /Zn ◦ Ecell = 0.34 V − (−0.76 V) ◦ Ecell = 1.10 V Step 3: Calculate the cell potential under the given standard conditions. The Nernst equation relates the standard cell potential to the reaction quotient (Q), temperature, and standard cell potential: ◦ E = Ecell − 0.0592 log(Q) n where n is the number of electrons transferred. In this case, n = 2 because 2 electrons are transferred in both half-reactions. Since the reaction is at standard conditions, Q = 1 and the Nernst equation simplifies to: ◦ E = Ecell ◦ Substitute the value of Ecell into the equation: E = 1.10 V Therefore, the cell potential at standard conditions is 1.10 V. Question 8 Question Consider a Galvanic Cell with a standard cell potential of 1.10 V that operates under standard conditions. If the cell potential drops to 0.95 V due to a concentration change in one of the half-cells, calculate the change in concentration of the reactant. Assume the temperature and pressure remain constant. Given: ˆ Standard reduction potential of the half-cell reaction: Cu2+ (aq) + 2e− → Cu(s): E ◦ = 0.34 V ˆ Concentration of Cu2+ ions in the original cell: 1.0 M Solution Step 1: Determine the half-cell reactions and their standard reduction potentials. The two half-cell reactions involved are: Zn(s) → Zn2+ (aq) + 2e− 7 ◦ Standard reduction potential for Zn half-cell: EZn = −0.76 V Cu2+ (aq) + 2e− → Cu(s) ◦ Given standard reduction potential for Cu half-cell: ECu = 0.34 V Step 2: Calculate the standard cell potential based on the standard reduction ◦ potentials of the two half-cell reactions. The standard cell potential (Ecell ) is given by: ◦ ◦ ◦ Ecell = ECu − EZn ◦ Ecell = 0.34 − (−0.76) = 1.10 V Step 3: Calculate the reaction quotient Q at the lower cell potential of 0.95 V. At equilibrium, the cell potential is equal to zero. Therefore, at 0.95 V, the reaction quotient Q is given by: ◦ 0.95 = Ecell − 0.0592 log Q 2 Substitute known values into the equation and solve for Q: 0.95 = 1.10 − 0.0296 log Q 0.15 = 0.0296 log Q log Q = 5.0676 Q = 105.0676 = 112260 Step 4: Calculate the new concentration of Cu2+ ions in the lower potential cell. The reaction quotient Q is related to the concentrations of the products and reactants in the cell. For the copper half-cell: Q= [Cu](solid) [Cu2+ (aq)]2 Since the concentration of Cu solid does not change significantly, we can approximate the concentration of Cu2+ in the lower potential cell to be: r r 1.0 1.0 2+ [Cu ] = = Q 112260 [Cu2+ ] ≈ 0.01 M Step 5: Calculate the change in concentration of the reactant. The change in concentration of the reactant Cu2+ is given by: ∆[Cu2+ ] = [Cu2+ ]initial − [Cu2+ ]final ∆[Cu2+ ] = 1.0 − 0.01 ∆[Cu2+ ] = 0.99 M Therefore, the change in concentration of the Cu2+ reactant is 0.99 M. 8 Question 9 Question Consider a galvanic cell with the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V 2H+ (aq) + 2e− → H2 (g) E ◦ = 0.00 V Calculate the cell potential at 25◦ C when the concentrations are [Zn + M and [H ] = 0.10 M. 2+ ] = 1.0 Solution Step 1: Write the overall cell reaction based on the given half-reactions. Identify the oxidizing and reducing agents. The overall cell reaction is the sum of the two half-reactions: Zn2+ (aq) + 2H+ (aq) → Zn(s) + H2 (g) Zinc is the reducing agent (it is oxidized) and hydrogen ions are the oxidizing agent (they are reduced). Step 2: Calculate the standard cell potential (E ◦ ) using the standard reduction potentials given. The standard cell potential is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode Plugging in the standard reduction potentials: ◦ Ecell = 0.00 V − (−0.76 V) = 0.76 V Step 3: Calculate the Nernst equation to find the cell potential under nonstandard conditions. The Nernst equation is given by:   Q 0.0592 ◦ log Ecell = Ecell − n K where n is the number of moles of electrons transferred and Q is the reaction quotient. Step 4: Calculate the cell potential with the given concentrations. First, calculate the reaction quotient: Q= [products] PZn · PH2 = [reactants] PZn2+ · PH2 + Substitute the concentrations and partial pressures: Q= 1 · 1 atm = 10 atm 1 · (0.1)2 9 Step 5: Calculate the cell potential using the Nernst equation. Substitute the values into the Nernst equation: Ecell = 0.76 V − 0.0592 log(10) ≈ 0.716 V 2 Therefore, the cell potential at 25◦ C when the concentrations are [Zn + 1.0 M and [H ] = 0.10 M is approximately 0.716 V. 2+ ]= Question 10 Question Consider a galvanic cell with a standard voltage of 1.23 V. If the cell potential drops to 1.10 V when the cell is operating, calculate the reaction quotient Q for the cell and determine if the reaction at the anode or cathode is favored. Solution Step 1: Write the Nernst equation to relate the cell potential (E), standard cell potential (E ◦ ), the reaction quotient (Q), the gas constant (R), absolute temperature (T ), and the Faraday constant (F ). E = E◦ − RT ln(Q) nF Step 2: As the cell potential drops to 1.10 V, we have: 1.10 V = 1.23 V − (0.0257 V) · T ln(Q) n Step 3: Rearrange the equation to solve for ln(Q). ln(Q) = (1.23 − 1.10) V · n 0.0257 V ln(Q) = 0.13 V · n 0.0257 V ln(Q) = 5.05n Step 4: Exponentiate both sides of the equation to solve for Q. Q = e5.05n Step 5: Since Q is based on the reaction quotient, we can determine that if Q > 1, the reaction is product-favored (occurs spontaneously), and if Q < 1, the reaction is reactant-favored (does not occur spontaneously). Therefore, if Q > 1, the reaction at the anode is favored; if Q < 1, the reaction at the cathode is favored. 10 Question 11 Question Consider a galvanic cell with a standard cell potential of 0.98 V. If the concentration of Cu2+ ions is 0.10 M and the concentration of Zn2+ ions is 2.0 M, determine the standard cell potential when the cell operates under non-standard conditions with Cu2+ at 0.025 M and Zn2+ at 1.5 M. Solution Let’s label the half-reactions for the galvanic cell: Zn(s) → Zn2+ (aq) + 2e− E ◦ = −0.76 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V ◦ The standard cell potential, Ecell , is calculated as follows: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.34 V − (−0.76 V) = 1.10 V Next, we will determine the reaction quotient Q under the new concentrations of Cu2+ and Zn2+ : Q= 0.025 [Cu2+ ] 2+ = 1.5 = 0.0167 [Zn ] The Nernst equation is used to find the cell potential, Ecell , under nonstandard conditions: 0.0592 ◦ Ecell = Ecell − log(Q) 2 Ecell = 1.10 V − (0.0296) log(0.0167) Ecell = 1.10 V − (0.0296)(−1.78) Ecell = 1.10 V + 0.0529 Ecell = 1.15 V Therefore, the standard cell potential under non-standard conditions is 1.15 V. Question 12 Question A student is given a cell with a nickel electrode and a copper electrode. The ◦ ◦ standard reduction potentials are ENi 2+ /Ni = −0.25 V and ECu2+ /Cu = 0.34 V. The student is asked to determine if a spontaneous redox reaction will occur between these two electrodes. 11 Solution Step 1: Write the overall cell reaction and calculate the standard cell potential, ◦ Ecell . The overall cell reaction is: Ni(s) + Cu2+ (aq) → Ni2+ (aq) + Cu(s) ◦ ◦ ◦ Using the standard reduction potentials given: Ecell = Ecathode − Eanode ◦ ◦ ◦ ◦ ◦ Ecell = ECu2+ /Cu − ENi2+ /Ni Ecell = 0.34 V − (−0.25 V) Ecell = 0.59 V ◦ Step 2: Determine the spontaneity of the redox reaction. Since Ecell is positive (0.59 V), the cell reaction is spontaneous as written. This means that a spontaneous redox reaction will occur between the nickel and copper electrodes. Question 13 Question ◦ Consider a galvanic cell where the standard cell potential (Ecell ) is 1.02 V. If the 2+ + cell reaction is 2H (aq) + Pb(s) → Pb (aq) + H2 (g), determine the standard reduction potential of the Pb half-reaction. Additionally, explain the difference between a galvanic cell and an electrolytic cell. Solution Step 1: Calculate the standard reduction potential of the Pb half-reaction. ◦ = 1.02 V and the half-reaction: Given the standard cell potential Ecell 2H+ (aq) + Pb(s) → Pb2+ (aq) + H2 (g) We know that the overall standard cell potential is the difference in the standard reduction potentials of the two half-reactions: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ where Ecathode is the reduction potential of the cathode (Pb in this case) and ◦ Eanode is the reduction potential of the anode (H+ in this case). ◦ Given Ecell = 1.02 V, the reduction potential of the Pb half-reaction can be calculated as: ◦ ◦ ◦ EPb = Ecell − EH + ◦ EPb = 1.02 V − 0 V ◦ EPb = 1.02 V Therefore, the standard reduction potential of the Pb half-reaction is 1.02 V. Step 2: Explain the difference between a galvanic cell and an electrolytic cell. A galvanic cell is an electrochemical cell that produces electrical energy from a spontaneous chemical reaction. It involves the conversion of chemical energy into electrical energy. The cell reaction is driven by a negative Gibbs free energy change (∆G < 0). Electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). 12 On the other hand, an electrolytic cell is an electrochemical cell that consumes electrical energy to drive a non-spontaneous chemical reaction. It involves the conversion of electrical energy into chemical energy. The cell reaction is driven by a positive Gibbs free energy change (∆G > 0). Electrons are forced to flow from the external electrical source to the cathode (where reduction occurs), and from the anode (where oxidation occurs) back to the electrical source. In summary, the key difference between galvanic and electrolytic cells lies in their energy source and the spontaneity of the cell reaction. Galvanic cells operate spontaneously, producing electrical energy, while electrolytic cells require external electrical energy input to drive non-spontaneous reactions. Question 14 Question Consider a galvanic cell with the following half-reactions: Zn → Zn2+ + 2e− E ◦ = −0.76 V Cu2+ + 2e− → Cu E ◦ = 0.34 V If the concentrations of Zn2+ and Cu2+ are both 1.0 M, calculate the cell potential at 25°C. Determine whether the cell is functioning as a galvanic cell or as an electrolytic cell. Solution Step 1: Write the equation for the overall cell reaction. The overall cell reaction is the combination of the two half-reactions. To do this, we reverse the first half-reaction and add the two reactions together: Cu2+ + Zn → Zn2+ + Cu ◦ Step 2: Calculate the standard cell potential (Ecell ). The standard cell potential is the difference in standard reduction potentials for the two halfreactions: ◦ ◦ ◦ ◦ Ecell = Ecathode − Eanode Ecell = 0.34 V − (−0.76 V) = 1.10 V Step 3: Calculate the reaction quotient (Q). The reaction quotient is the ratio of the concentrations of the products to the concentrations of the reactants raised to the power of their stoichiometric coefficients: [Zn2+ ][Cu] = 1.0 Q= [Cu2+ ][Zn] Step 4: Calculate the cell potential (Ecell ) at 25°C using the Nernst equation. The Nernst equation relates the cell potential to the standard cell potential and the reaction quotient: 0.0592 log(Q) n where n is the number of electrons transferred in the overall balanced equation. For the given reaction, n = 2, so: ◦ Ecell = Ecell − 13 0.0592 log(1.0) 2 = 1.10 V Ecell = 1.10 V − Ecell Step 5: Conclusion. The cell potential is 1.10 V, which is positive. Therefore, the cell is functioning as a galvanic cell since a positive cell potential indicates a spontaneous redox reaction occurring in the cell. Question 15 Question Consider a galvanic cell with the following half-reactions: Anode: Zn(s) → Zn2+ (aq) + 2e− Cathode: 2H+ (aq) + 2e− → H2 (g) Determine the cell potential at standard conditions for this galvanic cell. Solution Step 1: Write the overall cell reaction. Zn(s) + 2H+ (aq) → Zn2+ (aq) + H2 (g) Step 2: Find the standard reduction potentials for each half-reaction from a table. ◦ EZn 2+ /Zn = −0.76 V ◦ EH + /H = 0 V 2 Step 3: Calculate the cell potential at standard conditions. ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0 V − (−0.76 V) ◦ Ecell = 0.76 V Therefore, the cell potential at standard conditions for this galvanic cell is 0.76 V. Question 16 Question A voltaic cell is constructed using a standard hydrogen electrode (SHE) as the anode and an unknown metal electrode as the cathode. The measured cell potential is 0.67 V. Determine the standard reduction potential for the unknown metal electrode. 14 Solution Step 1: Write the cell reaction. The overall cell reaction for the voltaic cell is: − H+ (aq) + e → 1 H2 (g) 2 Let the unknown metal electrode be represented as M: − Mn+ (aq) + ne → M(s) The overall cell reaction can be written as: Mn+ (aq) + n H2 (g) → M(s) + nH+ (aq) 2 Step 2: Determine the standard cell potential. Given: Measured cell potential, Ec ell = 0.67V T hestandardcellpotential, E c ell, canbecalculatedusingtheN ernstequation : E cell = Ecell − RT nF ln Q Since the cell is at standard conditions, Q = 1 and the equation simplifies to: E cell = Ecell Therefore, the standard cell potential is 0.67 V. Step 3: Apply the standard cell potential. By comparing with the standard reduction potential table, we know that the standard reduction potential for the standard hydrogen electrode (SHE) is 0 V. The standard reduction potential for the unknown metal electrode can be calculated as: E cell = E cathode − E anode 0.67 = E cathode − 0 E cathode = 0.67 V Therefore, the standard reduction potential for the unknown metal electrode is 0.67 V. Question 17 Question Consider a galvanic cell with a standard cell potential of E ◦ = 0.87 V. The cell reaction is given by: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Determine the standard cell potential for the corresponding electrolytic cell involving the same half-reactions. 15 Solution Step 1: Write the cell reaction for the electrolytic cell by reversing the reaction for the galvanic cell. Zn2+ (aq) + Cu(s) → Zn(s) + Cu2+ (aq) Step 2: Determine the standard cell potential for the electrolytic cell. Since the cell reaction has been reversed, the standard cell potential for the electrolytic cell is the negative of the standard cell potential for the galvanic cell. ◦ Eelectrolytic = −E ◦ = −0.87 V Therefore, the standard cell potential for the corresponding electrolytic cell ◦ involving the same half-reactions is Eelectrolytic = −0.87 V. Question 18 Question Consider a galvanic cell with a standard cell potential of E ◦ = 1.10 V. The half-reactions in the cell are: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) Calculate the equilibrium constant K for the cell reaction at 25°C. Solution Step 1: Write the cell reaction equation. The cell reaction equation is the sum of the half-reactions at the anode and cathode: Zn(s) + Cu2+ (aq) −−→ Zn2+ (aq) + Cu(s) Step 2: Write expression for the equilibrium constant K. The equilibrium constant K for the cell reaction is related to the standard cell potential E ◦ by the Nernst equation: RT ln(K) E = E◦ − nF where n is the number of electrons transferred in the cell reaction, R is the gas constant, T is the temperature in kelvin, and F is the Faraday constant. Step 3: Determine the number of electrons transferred n. From the balanced cell reaction, we see that n = 2, as 2 electrons are transferred in the reaction. Step 4: Plug in the known values and solve for K. Given: E ◦ = 1.10 V R = 8.314 J/mol · K T = 298 K F = 96485 C/mol 16 Substitute the values into the Nernst equation to solve for K: 1.10 V = 1.10 V − (8.314 J/mol · K)(298 K) ln(K) (2)(96485 C/mol) Solving for ln(K): ln(K) = − (8.314 J/mol · K)(298 K) (2)(96485 C/mol) ln(K) = −0.0109 K = e−0.0109 K ≈ 0.989 Therefore, the equilibrium constant K for the cell reaction at 25°C is approximately 0.989. Question 19 Question Consider a galvanic cell with the following half-cell reactions: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) ◦ ◦ If the standard reduction potentials are EZn 2+ /Zn = −0.76 V and ECu2+ /Cu = ◦ 0.34 V, calculate the standard cell potential (Ecell ) for this galvanic cell. Solution Step 1: Write the overall cell reaction. Zn(s) + Cu2+ (aq) −−→ Zn2+ (aq) + Cu(s) ◦ Step 2: Calculate the standard cell potential (Ecell ) using the standard reduction potentials. ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EZn2+ /Zn ◦ Ecell = 0.34 V − (−0.76 V) ◦ Ecell = 1.10 V Therefore, the standard cell potential for this galvanic cell is 1.10 V. 17 Question 20 Question Describe the differences between galvanic cells and electrolytic cells. Provide an example of each and explain how they operate. Solution A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy through a spontaneous redox reaction, while an electrolytic cell is an electrochemical cell that requires an external electrical energy source to drive a non-spontaneous redox reaction. Differences: ˆ Energy Conversion: Galvanic cells convert chemical energy into electrical energy, while electrolytic cells use electrical energy to drive a nonspontaneous reaction. ˆ Spontaneity: Galvanic cells have spontaneous redox reactions, while electrolytic cells have non-spontaneous reactions. ˆ Electrodes: In galvanic cells, the anode is negative and the cathode is positive; in electrolytic cells, the anode is positive and the cathode is negative. Example of a Galvanic Cell (Voltaic Cell): The Daniell cell is a common example of a galvanic cell. In this cell, a zinc electrode is placed in a solution of zinc sulfate, connected by a salt bridge to a copper electrode in a solution of copper sulfate. The zinc undergoes oxidation at the anode, releasing electrons, which flow through the external circuit to the copper cathode, where reduction occurs. Operation of a Galvanic Cell: ˆ Step 1: Zinc metal atoms undergo oxidation at the anode, releasing electrons: Zn(s) → Zn2+ (aq) + 2e− ˆ Step 2: Electrons flow through the external circuit to the copper cathode. ˆ Step 3: Copper ions in solution gain electrons and are reduced at the cathode: Cu2+ (aq) + 2e− → Cu(s) ˆ Overall Cell Reaction: 2+ Zn(s) + Cu (aq) → Zn2+ (aq) + Cu(s) 18 Example of an Electrolytic Cell: An example of an electrolytic cell is the electrolysis of water. In this cell, an electric current is passed through water to split it into hydrogen and oxygen gases. Operation of an Electrolytic Cell: ˆ Step 1: Water molecules undergo electrolysis to form oxygen gas at the anode and hydrogen gas at the cathode: 2H2 O(l) → 2H2 (g) + O2 (g) Question 21 Question Consider the following electrochemical cell reaction: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) ◦ Given that the standard reduction potentials are EZn 2+ /Zn = −0.76 V and ◦ ECu2+ /Cu = 0.34 V, determine if the reaction will be spontaneous under standard conditions. If not, calculate the minimum potential at which the reaction will be spontaneous. Solution ◦ Step 1: Calculate the cell potential (Ecell ) under standard conditions. ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.34 V − (−0.76) V ◦ Ecell = 1.10 V Step 2: Determine if the reaction is spontaneous under standard conditions. ◦ Since Ecell is positive, the reaction is spontaneous under standard conditions. Step 3: Calculate the minimum potential to make the reaction spontaneous. For a reaction to be spontaneous, the cell potential (Ecell ) must be positive. ◦ Ecell = Ecell − 0.0592 log Q n where Q is the reaction quotient and n is the number of electrons transferred. Step 4: Calculate the reaction quotient (Q) at equilibrium. Since the reaction is at equilibrium when Ecell = 0, we have: 0.0592 log Q 2 0 = 1.10 − log Q = 1.10 × 2 0.0592 19 log Q = 37.1622 Q = 1037.1622 Step 5: Calculate the minimum potential required for the reaction to be spontaneous. 0.0592 log Q′ Ecell = 0 = 1.10 − 2 Substitute Q with Q′ : 0 = 1.10 − 0.0592 log 1037.1622 2 0.0592 × 37.1622 2 0 = 1.10 − 1.0968 0 = 1.10 − 1.0968 = 1.10 1.0968 > 1.10 Therefore, the minimum potential required for the reaction to be spontaneous is greater than 1.10 V. Question 22 Question Consider a galvanic cell with a standard cell potential of 0.78 V. If the cell potential drops to 0.70 V at 25°C, calculate the change in standard Gibbs free energy (∆G◦ ) for the reaction occurring in the cell. Solution Step 1: Calculate the change in cell potential (∆E ◦ ): Given: Standard cell potential, E ◦ = 0.78 V Cell potential at 25°C, E = 0.70 V The change in cell potential can be calculated using the equation: ∆E ◦ = E ◦ − E ∆E ◦ = 0.78 − 0.70 = 0.08 V Step 2: Calculate the Faraday constant (F ): The Faraday constant, F , is given by: F = 96485 C/mol Step 3: Using the equation relating ∆G◦ , ∆E ◦ , and n (the number of moles of electrons transferred in the reaction), we have: ∆G◦ = −nF ∆E ◦ 20 Step 4: Determine the number of moles of electrons transferred, n: Since the cell potential dropped, we know the reaction is running in the reverse direction. Therefore, the number of moles of electrons transferred, n, is: n = −2 Step 5: Calculate the change in standard Gibbs free energy (∆G◦ ): Substitute the values of n, F , and ∆E ◦ into the equation: ∆G◦ = −(−2)(96485)(0.08) ∆G◦ = 15435.76 J/mol Therefore, the change in standard Gibbs free energy for the reaction occurring in the cell is 15435.76 J/mol. Question 23 Question Consider a galvanic cell with a standard cell potential of 1.13 V that involves the reaction: Zn(s) + 2Fe3+ (aq) → Zn2+ (aq) + 2Fe2+ (aq) (a) Write the cell notation for this galvanic cell. (b) Calculate the standard cell potential of the galvanic cell if [Fe3+ ] = 0.20 M and [Fe2+ ] = 2.00 M. Solution (a) The cell notation for the galvanic cell can be written as: Zn(s) | Zn2+ (aq) || Fe3+ (aq), Fe2+ (aq) | Pt(s) (b) The standard cell potential can be calculated using the Nernst equation: RT ln Q nF where E is the cell potential, E ◦ is the standard cell potential, R is the gas constant (8.314 J/(mol K)), T is the temperature in Kelvin, n is the number of moles of electrons transferred in the balanced equation, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient. In this case, n = 2 because two moles of electrons are transferred in the balanced equation. The reaction quotient Q can be expressed as: E = E◦ − Q= [Zn2+ ] 1 3+ 2+ = (0.20)(2.00)2 = 1.25 [Fe ][Fe ] 21 Substitute the given values into the Nernst equation: E = 1.13 V − (8.314)(298) ln 1.25 ≈ 1.08 V 2(96485) Question 24 Question ◦ Consider a voltaic cell with a standard cell potential of Ecell = 1.20 V. The cell reaction is: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Determine the standard reduction potentials for the Zn and Cu half-reactions and write the half-reactions. Also, calculate the standard cell potential for the reverse reaction. Solution Step 1: Write the half-reactions and determine their standard reduction potentials. Zn2+ + 2e− → Zn ◦ EZn = −0.76 V Cu2+ + 2e− → Cu ◦ ECu = 0.34 V Step 2: Write the balanced overall cell reaction for the voltaic cell. Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Step 3: Calculate the standard cell potential for the reverse reaction. Given ◦ the standard cell potential for the forward reaction is Ecell = 1.20 V, the stan◦ dard cell potential for the reverse reaction is −Ecell . ◦ Ecell, reverse = −1.20 V Question 25 Question Consider the following reaction in an electrochemical cell: Pb2+ (aq) + 2Ag(s) → Pb(s) + 2Ag+ (aq) Determine which electrode acts as the anode and which as the cathode. Explain your reasoning. 22 Solution Step 1: Write the standard reduction potentials for the half-reactions involved. The standard reduction potentials for the half-reactions involved are: Ag+ (aq) + e− → Ag(s) E ◦ = +0.80 V Pb2+ (aq) + 2e− → Pb(s) E ◦ = −0.13 V Step 2: Identify the half-reaction with the more positive standard reduction potential as the reduction half-reaction (cathode). Since the reduction potential for the reduction of Ag+ to Ag is higher (+0.80 V), this is the reduction halfreaction. Therefore, the Ag electrode acts as the cathode. Step 3: Deduce the anode based on the cathode. The other electrode, the Pb electrode, must therefore act as the anode. Therefore, in the given cell setup: - The Ag electrode acts as the cathode. The Pb electrode acts as the anode. Question 26 Question Consider a galvanic cell and an electrolytic cell, both involving the same redox reaction. The galvanic cell produces an electric current when the reaction occurs spontaneously, while the electrolytic cell requires an external electric current to drive the non-spontaneous reaction. Explain the key differences between these two types of cells, focusing on their operating principles and applications. Solution Step 1: Galvanic Cell A galvanic cell, also known as a voltaic cell, converts chemical energy into electrical energy. It consists of two half-cells connected by a salt bridge or porous barrier. In the spontaneous redox reaction, electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs), generating an electric current. This current can be used to power external devices. Step 2: Electrolytic Cell An electrolytic cell uses an external electric current to drive a non-spontaneous redox reaction. It consists of two electrodes (anode and cathode) immersed in an electrolyte solution. When an electric current is applied, electrons are forced to move against their natural direction. This process allows for the reduction of cations at the cathode and the oxidation of anions at the anode. Step 3: Key Differences 1. Energy Conversion: Galvanic cells convert chemical energy into electrical energy, while electrolytic cells require electrical energy to drive a non-spontaneous reaction. 2. Spontaneity: Galvanic cells run spontaneously, while electrolytic cells need an external power source to operate. 3. 23 Electrode Reactions: In galvanic cells, oxidation occurs at the anode and reduction at the cathode. In electrolytic cells, the reactions are reversed due to the external electric current. 4. Applications: Galvanic cells are used in batteries to power devices, while electrolytic cells are used in processes like electroplating, metal refining, and electrolysis of water. In summary, galvanic cells produce electrical energy from chemical reactions, while electrolytic cells use electrical energy to drive non-spontaneous reactions. They have different applications and operate based on opposite principles of redox reactions. Question 27 Question ◦ Consider a galvanic cell that consists of a standard hydrogen electrode (Ecell = 0.00 V) and a silver electrode in a 1.0 M solution of silver nitrate. Part a: Write the overall cell reaction for this galvanic cell. ◦ Part b: If the standard reduction potential of silver is EAg + /Ag = 0.80 V, calculate the cell potential at standard conditions. Part c: Will the cell potential increase or decrease if the concentration of silver nitrate is increased? Justify your answer. Solution Part a: The overall cell reaction for the galvanic cell involving a standard hydrogen electrode and a silver electrode is given by: 2H+ (aq) + 2e− → H2 (g) E ◦ = 0.00 V Ag+ (aq) + e− → Ag(s) E ◦ = 0.80 V Adding these two half-reactions together, we get the overall cell reaction: 2H+ (aq) + 2Ag+ (aq) → H2 (g) + 2Ag(s) Part b: ◦ The cell potential Ecell at standard conditions can be calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.80 V − 0.00 V = 0.80 V Therefore, the cell potential at standard conditions is 0.80 V. 24 Part c: If the concentration of silver nitrate is increased, the concentration of Ag+ in the cathode compartment increases. According to the Nernst equation, an increase in the concentration of a reactant shifts the equilibrium towards products resulting in a higher cell potential. Therefore, increasing the concentration of silver nitrate will increase the cell potential. Question 28 Question Consider a galvanic cell with the half-reactions: Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V Fe2+ (aq) + 2e− → Fe(s) E ◦ = −0.44 V Suppose the standard cell potential is measured to be 0.78 V. Determine the concentrations of Cu2+ and Fe2+ if the cell operates under standard conditions. Solution Step 1: Write the cell reaction. The cell reaction is obtained by adding the half-reactions together: Cu2+ (aq) + Fe(s) → Cu(s) + Fe2+ (aq) Step 2: Calculate the standard cell potential. The standard cell potential, E◦ , is the difference between the standard reduction potentials of the two halfreactions: E◦ = E ◦ (cathode) − E ◦ (anode) E ◦ = 0.34 V − (−0.44 V) E ◦ = 0.78 V Step 3: Determine the reaction quotient. Since the reaction is at equilibrium, the reaction quotient, Q, is equal to 1. Step 4: Use the Nernst equation to find the ratio of concentrations. The Nernst equation relates the standard cell potential to the reaction quotient and the concentrations of species involved in the cell reaction: E = E◦ − 0.0592 log Q n where n is the number of moles of electrons transferred in the cell reaction. Step 5: Solve for the concentrations of Cu2+ and Fe2+ . Given that Q = 1 in this case and n = 2, we can write: 0.78 V = 0.34 V − 0.0592 log 1 2 0.44 = 0.0592 log 1 25 0 = log 1 Thus, at standard conditions, the concentrations of Cu2+ and Fe2+ do not change. Question 29 Question A student is studying galvanic and electrolytic cells in their electrochemistry course. They are given a cell diagram but are unsure whether the cell is galvanic or electrolytic. The diagram shows a zinc metal electrode dipping into a solution of zinc sulfate and a silver metal electrode dipping into a solution of silver nitrate. The two electrodes are connected by a wire. The student must determine whether the cell is galvanic or electrolytic, and explain their reasoning. Solution To determine whether the given cell is galvanic or electrolytic, we need to consider the standard electrode potentials of the half-reactions involved. Step 1: Write the half-reactions The half-reaction at the zinc electrode is the oxidation of zinc metal: Zn(s) → Zn2+ (aq) + 2e− The half-reaction at the silver electrode is the reduction of silver ions: Ag+ (aq) + e− → Ag(s) Step 2: Determine the standard electrode potentials Given standard reduction potentials: ◦ EZn 2+ /Zn = −0.76 V ◦ EAg + /Ag = 0.80 V Step 3: Determine if the cell is galvanic or electrolytic To determine if the cell is galvanic or electrolytic, we can calculate the cell potential using the standard reduction potentials. The overall cell reaction is the sum of the two half-reactions: Zn(s) + Ag+ (aq) → Zn2+ (aq) + Ag(s) ◦ The overall standard cell potential (Ecell ) can be calculated as follows: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = EAg + /Ag − EZn2+ /Zn 26 ◦ Ecell = 0.80 V − (−0.76 V) = 1.56 V Since the cell potential is positive (1.56 V), the cell is galvanic. In a galvanic cell, the cell potential is positive, indicating that the spontaneous electrochemical reaction occurs. Question 30 Question Consider a galvanic cell involving the reaction: Cr3+ (aq) + Zn(s) → Zn2+ (aq) + Cr2+ (aq) The standard reduction potentials are: ◦ ECr 3+ /Cr2+ = −0.74 V ◦ EZn 2+ /Zn = −0.76 V (a) Calculate the standard cell potential for this galvanic cell. (b) Will the reaction in this galvanic cell proceed spontaneously at standard conditions? (c) If a voltage of 1.5 V is applied to the cell in the reverse direction, what reactions will take place at the electrodes? Solution (a) To calculate the standard cell potential, we use the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode Given that reduction occurs at the cathode and oxidation occurs at the anode, we have: ◦ ◦ ◦ Ecell = ECr 2+ /Cr3+ − EZn2+ /Zn ◦ Ecell = (−0.74 V) − (−0.76 V) ◦ Ecell = 0.02 V Therefore, the standard cell potential for this galvanic cell is 0.02 V. (b) For a reaction to proceed spontaneously, the standard cell potential must be positive. Since the standard cell potential for this galvanic cell is 0.02 V (which is positive), the reaction will proceed spontaneously at standard conditions. (c) When a voltage of 1.5 V is applied to the cell in the reverse direction, the reactions at the electrodes will also be reversed. This means that Zn will be plated at the cathode and Cr will be oxidized at the anode. The reactions will be: 27 At the cathode (Zn): Zn2+ (aq) + 2e− → Zn(s) At the anode (Cr): Cr2+ (aq) → Cr3+ (aq) + e− Question 31 Question ◦ Consider a galvanic cell with a standard cell potential of Ecell = 1.10 V. If the 3+ concentration of Fe in one of the half-cells is 0.10 M and the concentration of Fe2+ in the other half-cell is 1.0 M at 25◦ C, determine the cell potential when the concentration of Fe3+ in the first half-cell is reduced to 0.050 M and the concentration of Fe2+ in the second half-cell is increased to 4.0 M. Solution Step 1: Write the overall cell reaction for the galvanic cell: The half-reactions for the cell are: Fe3+ + 3e− → Fe2+ E ◦ = −0.77 V Fe2+ → Fe + 2e− E ◦ = −0.44 V The overall cell reaction is the sum of the two half-reactions: Fe3+ + Fe2+ → Fe + Fe2+ Step 2: Calculate the initial cell potential using the Nernst equation: The standard cell potential is E ◦ = 1.10 V. The initial concentrations of Fe3+ and Fe2+ are 0.10 M and 1.0 M, respectively. The Nernst equation is:  2+  0.0592 [Fe ] ◦ Ecell = Ecell − log n [Fe3+ ] Substitute the values and calculate the initial cell potential:   1.0 0.0592 log Ecell = 1.10 − 2 0.10 Ecell = 1.10 − 0.0296 × 1 ≈ 1.0704 V Step 3: Calculate the final cell potential after changing the concentrations: The new concentrations are 0.050 M for Fe3+ and 4.0 M for Fe2+ . Use the Nernst equation again to calculate the new cell potential:   0.0592 4.0 Ecell = 1.10 − log 2 0.050 Ecell = 1.10 − 0.0296 × 3.322 ≈ 1.0068 V Therefore, the cell potential after changing the concentrations is approximately 1.0068 V. 28 Question 32 Question ◦ Consider a galvanic cell with a standard cell potential of Ecell = 0.80 V. If the reaction in the cell is spontaneous, what can you conclude about the standard Gibbs free energy change for the cell reaction? Explain your answer. Solution Step 1: Recall the relationship between standard Gibbs free energy change ◦ (∆G◦ ) and standard cell potential (Ecell ): ◦ ∆G◦ = −nF Ecell where n is the number of electrons transferred in the cell reaction and F is the Faraday constant (96,485 C/mol). Step 2: Since the cell reaction is spontaneous, it means that the standard Gibbs free energy change for the reaction is negative: ∆G◦ < 0 Step 3: Substituting the given standard cell potential into the equation, we have: ∆G◦ = −nF (0.80 V) Step 4: To determine the sign of ∆G◦ , we need to consider the signs of n ◦ and F . Since Ecell is positive, n must also be positive to ensure a negative value ◦ for ∆G . Step 5: Therefore, we can conclude that the standard Gibbs free energy change for the cell reaction is negative, indicating that the reaction is spontaneous and capable of performing work. Question 33 Question Consider a galvanic cell constructed with a copper electrode in a 1.0 M Cu2+ solution and a silver electrode in a 1.0 M Ag+ solution. The standard reduction ◦ ◦ potentials are ECu 2+ /Cu = 0.34 V and EAg+ /Ag = 0.80 V. Determine the cell potential when the concentration of Cu2+ is 0.10 M and Ag+ is 0.50 M. Solution Step 1: Write the half-reactions and standard reduction potentials: The halfreactions that occur in this cell are: Cathode (Reduction at Ag+ ): Ag+ (aq) + 29 ◦ 2+ e− → Ag(s) (EAg ): Cu(s) → Cu2+ (aq)+ + /Ag = 0.80 V) Anode (Oxidation at Cu ◦ 2e− (ECu 2+ /Cu = 0.34 V) Step 2: Calculate the cell potential under standard conditions: The cell ◦ potential under standard conditions (Ecell ) is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.80 V − 0.34 V = 0.46 V Step 3: Use the Nernst equation to calculate the cell potential when the concentrations are not standard: The Nernst equation is: ◦ Ecell = Ecell − 0.0592 [Ag+ ] log n [Cu2+ ] where n is the number of electrons transferred in the balanced chemical equation. In this case, n = 1. Plugging in the values: Ecell = 0.46 V − 0.50 0.0592 log 1 0.10 Ecell = 0.46 V − 0.0592 log 5 Ecell = 0.46 V − 0.11 V = 0.35 V Therefore, the cell potential when the concentration of Cu2+ is 0.10 M and Ag is 0.50 M is 0.35 V. + Question 34 Question Consider a galvanic cell with a standard cell potential of E ◦ = 1.20 V at 25◦ C that consists of a standard hydrogen electrode (Pt—H2 (1 atm)—H+ ) and a silver electrode (Ag). Calculate the cell potential when the concentration of Ag+ is 0.10 M and the pressure of hydrogen gas is 0.5 atm at 25◦ C. Solution Step 1: Write the cell reaction for the galvanic cell: Pt(s) + 2H+ (aq) + 2e− → H2 (g) Ag(s) → Ag+ (aq) + e− Step 2: Write the balanced overall cell reaction: Pt(s) + 2Ag+ (aq) → 2Ag(s) + 2H+ (aq) 30 Step 3: Use the Nernst equation to calculate the cell potential: E = E◦ − Q 0.0592 log n K where E is the cell potential, E ◦ is the standard cell potential, n is the number of electrons transferred in the balanced cell reaction, Q is the reaction quotient, and K is the equilibrium constant. Step 4: Calculate the reaction quotient Q and the number of electrons transferred n based on the balanced cell reaction: Q= [Ag+ ]2 (0.10)2 + 2 = (1)2 [H ] n=2 Step 5: Substitute the values into the Nernst equation: E = 1.20 − 0.0592 0.01 log 2 1 E = 1.20 − 0.0296 log(0.01) E = 1.20 − 0.0296(−2) E = 1.20 + 0.0592 E = 1.2592 V Therefore, the cell potential when the concentration of Ag+ is 0.10 M and the pressure of hydrogen gas is 0.5 atm at 25◦ C is 1.2592 V. Question 35 Question Consider a galvanic cell in which the following reaction occurs: 2+ Zn(s) — Zn (aq, 1.00 M)||Cu2+ (aq, 1.00 M) — Cu(s) ◦ Calculate the cell potential at 25°C. Given: Ecell (Cu2+ /Cu) = +0.34 V and 2+ ◦ Ecell (Zn /Zn) = −0.76 V. Solution Step 1: Write the overall cell reaction. The overall cell reaction can be obtained by summing the half-reactions for each electrode. The standard cell potential, ◦ Ecell , is the difference between the standard reduction potentials of the halfreactions. The two half-reactions and their standard reduction potentials are: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V 31 Question 5 Question Consider a galvanic cell with the following half-reactions: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) ◦ Calculate the cell potential (Ecell ) at 25°C for this galvanic cell using standard reduction potentials. Solution To calculate the cell potential, we will use the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode Step 1: Find the standard reduction potentials for each half-reaction. The ◦ standard reduction potentials (Ered ) for the given half-reactions are as follows: ◦ Ered,Zn 2+ +2 e− − −→Zn = −0.76 V ◦ Ered,Cu 2+ +2 e− − −→Cu = 0.34 V ◦ Step 2: Plug the values into the formula to find Ecell . ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.34 V − (−0.76 V) ◦ Ecell = 1.10 V ◦ Therefore, the cell potential (Ecell ) at 25°C for this galvanic cell is 1.10 V. Question 6 Question Consider a galvanic cell with the following half-reactions: Anode: Cathode: Zn(s) → Zn2+ (aq) + 2e− Cu2+ (aq) + 2e− → Cu(s) ◦ ◦ If the standard reduction potentials are EZn 2+ /Zn = −0.76 V and ECu2+ /Cu = ◦ ◦ 0.34 V, calculate the cell potential (Ecell ) at 25 C and comment on the spontaneity of the redox reaction. 5 Solution ◦ 1. Calculate Ecell : ◦ The cell potential (Ecell ) can be calculated using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ where Ecathode is the standard reduction potential of the cathode and ◦ Eanode is the standard reduction potential of the anode. ◦ ◦ Given data: EZn 2+ /Zn = −0.76 V ECu2+ /Cu = 0.34 V ◦ ◦ ◦ Therefore, Ecell = ECu 2+ /Cu − EZn2+ /Zn ◦ Ecell = 0.34 V − (−0.76 V) ◦ Ecell = 1.10 V ◦ Hence, the cell potential (Ecell ) is 1.10 V. 2. Comment on spontaneity: ◦ ◦ The redox reaction is spontaneous when Ecell > 0. In this case, Ecell = 1.10 V, which is greater than 0. Therefore, the redox reaction is spontaneous at 25◦ C. The positive cell potential indicates that the reaction will proceed spontaneously from the anode (Zn) to the cathode (Cu). Question 7 Question Consider a galvanic cell with the following half-reactions: Cathode: Cu2+ (aq) + 2e− → Cu(s) Anode: Zn(s) → Zn2+ (aq) + 2e− Calculate the cell potential at standard conditions (T = 298K, P = 1 atm, and [Cu2+ ] = 1.0M, [Zn2+ ] = 1.0M). Solution Step 1: Write the overall cell reaction by combining the two half-reactions: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Step 2: Write the cell potential equation using the standard reduction potentials of the half-reactions. The standard reduction potentials (E ◦ ) for the given half-reactions are: ◦ ECu 2+ /Cu = 0.34 V ◦ EZn 2+ /Zn = −0.76 V 6 ◦ The cell potential (Ecell ) can be calculated using the equation: ◦ ◦ ◦ Ecell = Ecathode − Eanode Plug in the values: ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EZn2+ /Zn ◦ Ecell = 0.34 V − (−0.76 V) ◦ Ecell = 1.10 V Step 3: Calculate the cell potential under the given standard conditions. The Nernst equation relates the standard cell potential to the reaction quotient (Q), temperature, and standard cell potential: ◦ E = Ecell − 0.0592 log(Q) n where n is the number of electrons transferred. In this case, n = 2 because 2 electrons are transferred in both half-reactions. Since the reaction is at standard conditions, Q = 1 and the Nernst equation simplifies to: ◦ E = Ecell ◦ Substitute the value of Ecell into the equation: E = 1.10 V Therefore, the cell potential at standard conditions is 1.10 V. Question 8 Question Consider a Galvanic Cell with a standard cell potential of 1.10 V that operates under standard conditions. If the cell potential drops to 0.95 V due to a concentration change in one of the half-cells, calculate the change in concentration of the reactant. Assume the temperature and pressure remain constant. Given: ˆ Standard reduction potential of the half-cell reaction: Cu2+ (aq) + 2e− → Cu(s): E ◦ = 0.34 V ˆ Concentration of Cu2+ ions in the original cell: 1.0 M Solution Step 1: Determine the half-cell reactions and their standard reduction potentials. The two half-cell reactions involved are: Zn(s) → Zn2+ (aq) + 2e− 7 ◦ Standard reduction potential for Zn half-cell: EZn = −0.76 V Cu2+ (aq) + 2e− → Cu(s) ◦ Given standard reduction potential for Cu half-cell: ECu = 0.34 V Step 2: Calculate the standard cell potential based on the standard reduction ◦ potentials of the two half-cell reactions. The standard cell potential (Ecell ) is given by: ◦ ◦ ◦ Ecell = ECu − EZn ◦ Ecell = 0.34 − (−0.76) = 1.10 V Step 3: Calculate the reaction quotient Q at the lower cell potential of 0.95 V. At equilibrium, the cell potential is equal to zero. Therefore, at 0.95 V, the reaction quotient Q is given by: ◦ 0.95 = Ecell − 0.0592 log Q 2 Substitute known values into the equation and solve for Q: 0.95 = 1.10 − 0.0296 log Q 0.15 = 0.0296 log Q log Q = 5.0676 Q = 105.0676 = 112260 Step 4: Calculate the new concentration of Cu2+ ions in the lower potential cell. The reaction quotient Q is related to the concentrations of the products and reactants in the cell. For the copper half-cell: Q= [Cu](solid) [Cu2+ (aq)]2 Since the concentration of Cu solid does not change significantly, we can approximate the concentration of Cu2+ in the lower potential cell to be: r r 1.0 1.0 2+ [Cu ] = = Q 112260 [Cu2+ ] ≈ 0.01 M Step 5: Calculate the change in concentration of the reactant. The change in concentration of the reactant Cu2+ is given by: ∆[Cu2+ ] = [Cu2+ ]initial − [Cu2+ ]final ∆[Cu2+ ] = 1.0 − 0.01 ∆[Cu2+ ] = 0.99 M Therefore, the change in concentration of the Cu2+ reactant is 0.99 M. 8 Question 9 Question Consider a galvanic cell with the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V 2H+ (aq) + 2e− → H2 (g) E ◦ = 0.00 V Calculate the cell potential at 25◦ C when the concentrations are [Zn + M and [H ] = 0.10 M. 2+ ] = 1.0 Solution Step 1: Write the overall cell reaction based on the given half-reactions. Identify the oxidizing and reducing agents. The overall cell reaction is the sum of the two half-reactions: Zn2+ (aq) + 2H+ (aq) → Zn(s) + H2 (g) Zinc is the reducing agent (it is oxidized) and hydrogen ions are the oxidizing agent (they are reduced). Step 2: Calculate the standard cell potential (E ◦ ) using the standard reduction potentials given. The standard cell potential is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode Plugging in the standard reduction potentials: ◦ Ecell = 0.00 V − (−0.76 V) = 0.76 V Step 3: Calculate the Nernst equation to find the cell potential under nonstandard conditions. The Nernst equation is given by:   Q 0.0592 ◦ log Ecell = Ecell − n K where n is the number of moles of electrons transferred and Q is the reaction quotient. Step 4: Calculate the cell potential with the given concentrations. First, calculate the reaction quotient: Q= [products] PZn · PH2 = [reactants] PZn2+ · PH2 + Substitute the concentrations and partial pressures: Q= 1 · 1 atm = 10 atm 1 · (0.1)2 9 Step 5: Calculate the cell potential using the Nernst equation. Substitute the values into the Nernst equation: Ecell = 0.76 V − 0.0592 log(10) ≈ 0.716 V 2 Therefore, the cell potential at 25◦ C when the concentrations are [Zn + 1.0 M and [H ] = 0.10 M is approximately 0.716 V. 2+ ]= Question 10 Question Consider a galvanic cell with a standard voltage of 1.23 V. If the cell potential drops to 1.10 V when the cell is operating, calculate the reaction quotient Q for the cell and determine if the reaction at the anode or cathode is favored. Solution Step 1: Write the Nernst equation to relate the cell potential (E), standard cell potential (E ◦ ), the reaction quotient (Q), the gas constant (R), absolute temperature (T ), and the Faraday constant (F ). E = E◦ − RT ln(Q) nF Step 2: As the cell potential drops to 1.10 V, we have: 1.10 V = 1.23 V − (0.0257 V) · T ln(Q) n Step 3: Rearrange the equation to solve for ln(Q). ln(Q) = (1.23 − 1.10) V · n 0.0257 V ln(Q) = 0.13 V · n 0.0257 V ln(Q) = 5.05n Step 4: Exponentiate both sides of the equation to solve for Q. Q = e5.05n Step 5: Since Q is based on the reaction quotient, we can determine that if Q > 1, the reaction is product-favored (occurs spontaneously), and if Q < 1, the reaction is reactant-favored (does not occur spontaneously). Therefore, if Q > 1, the reaction at the anode is favored; if Q < 1, the reaction at the cathode is favored. 10 Question 11 Question Consider a galvanic cell with a standard cell potential of 0.98 V. If the concentration of Cu2+ ions is 0.10 M and the concentration of Zn2+ ions is 2.0 M, determine the standard cell potential when the cell operates under non-standard conditions with Cu2+ at 0.025 M and Zn2+ at 1.5 M. Solution Let’s label the half-reactions for the galvanic cell: Zn(s) → Zn2+ (aq) + 2e− E ◦ = −0.76 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V ◦ The standard cell potential, Ecell , is calculated as follows: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.34 V − (−0.76 V) = 1.10 V Next, we will determine the reaction quotient Q under the new concentrations of Cu2+ and Zn2+ : Q= 0.025 [Cu2+ ] 2+ = 1.5 = 0.0167 [Zn ] The Nernst equation is used to find the cell potential, Ecell , under nonstandard conditions: 0.0592 ◦ Ecell = Ecell − log(Q) 2 Ecell = 1.10 V − (0.0296) log(0.0167) Ecell = 1.10 V − (0.0296)(−1.78) Ecell = 1.10 V + 0.0529 Ecell = 1.15 V Therefore, the standard cell potential under non-standard conditions is 1.15 V. Question 12 Question A student is given a cell with a nickel electrode and a copper electrode. The ◦ ◦ standard reduction potentials are ENi 2+ /Ni = −0.25 V and ECu2+ /Cu = 0.34 V. The student is asked to determine if a spontaneous redox reaction will occur between these two electrodes. 11 Solution Step 1: Write the overall cell reaction and calculate the standard cell potential, ◦ Ecell . The overall cell reaction is: Ni(s) + Cu2+ (aq) → Ni2+ (aq) + Cu(s) ◦ ◦ ◦ Using the standard reduction potentials given: Ecell = Ecathode − Eanode ◦ ◦ ◦ ◦ ◦ Ecell = ECu2+ /Cu − ENi2+ /Ni Ecell = 0.34 V − (−0.25 V) Ecell = 0.59 V ◦ Step 2: Determine the spontaneity of the redox reaction. Since Ecell is positive (0.59 V), the cell reaction is spontaneous as written. This means that a spontaneous redox reaction will occur between the nickel and copper electrodes. Question 13 Question ◦ Consider a galvanic cell where the standard cell potential (Ecell ) is 1.02 V. If the 2+ + cell reaction is 2H (aq) + Pb(s) → Pb (aq) + H2 (g), determine the standard reduction potential of the Pb half-reaction. Additionally, explain the difference between a galvanic cell and an electrolytic cell. Solution Step 1: Calculate the standard reduction potential of the Pb half-reaction. ◦ = 1.02 V and the half-reaction: Given the standard cell potential Ecell 2H+ (aq) + Pb(s) → Pb2+ (aq) + H2 (g) We know that the overall standard cell potential is the difference in the standard reduction potentials of the two half-reactions: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ where Ecathode is the reduction potential of the cathode (Pb in this case) and ◦ Eanode is the reduction potential of the anode (H+ in this case). ◦ Given Ecell = 1.02 V, the reduction potential of the Pb half-reaction can be calculated as: ◦ ◦ ◦ EPb = Ecell − EH + ◦ EPb = 1.02 V − 0 V ◦ EPb = 1.02 V Therefore, the standard reduction potential of the Pb half-reaction is 1.02 V. Step 2: Explain the difference between a galvanic cell and an electrolytic cell. A galvanic cell is an electrochemical cell that produces electrical energy from a spontaneous chemical reaction. It involves the conversion of chemical energy into electrical energy. The cell reaction is driven by a negative Gibbs free energy change (∆G < 0). Electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs). 12 On the other hand, an electrolytic cell is an electrochemical cell that consumes electrical energy to drive a non-spontaneous chemical reaction. It involves the conversion of electrical energy into chemical energy. The cell reaction is driven by a positive Gibbs free energy change (∆G > 0). Electrons are forced to flow from the external electrical source to the cathode (where reduction occurs), and from the anode (where oxidation occurs) back to the electrical source. In summary, the key difference between galvanic and electrolytic cells lies in their energy source and the spontaneity of the cell reaction. Galvanic cells operate spontaneously, producing electrical energy, while electrolytic cells require external electrical energy input to drive non-spontaneous reactions. Question 14 Question Consider a galvanic cell with the following half-reactions: Zn → Zn2+ + 2e− E ◦ = −0.76 V Cu2+ + 2e− → Cu E ◦ = 0.34 V If the concentrations of Zn2+ and Cu2+ are both 1.0 M, calculate the cell potential at 25°C. Determine whether the cell is functioning as a galvanic cell or as an electrolytic cell. Solution Step 1: Write the equation for the overall cell reaction. The overall cell reaction is the combination of the two half-reactions. To do this, we reverse the first half-reaction and add the two reactions together: Cu2+ + Zn → Zn2+ + Cu ◦ Step 2: Calculate the standard cell potential (Ecell ). The standard cell potential is the difference in standard reduction potentials for the two halfreactions: ◦ ◦ ◦ ◦ Ecell = Ecathode − Eanode Ecell = 0.34 V − (−0.76 V) = 1.10 V Step 3: Calculate the reaction quotient (Q). The reaction quotient is the ratio of the concentrations of the products to the concentrations of the reactants raised to the power of their stoichiometric coefficients: [Zn2+ ][Cu] = 1.0 Q= [Cu2+ ][Zn] Step 4: Calculate the cell potential (Ecell ) at 25°C using the Nernst equation. The Nernst equation relates the cell potential to the standard cell potential and the reaction quotient: 0.0592 log(Q) n where n is the number of electrons transferred in the overall balanced equation. For the given reaction, n = 2, so: ◦ Ecell = Ecell − 13 0.0592 log(1.0) 2 = 1.10 V Ecell = 1.10 V − Ecell Step 5: Conclusion. The cell potential is 1.10 V, which is positive. Therefore, the cell is functioning as a galvanic cell since a positive cell potential indicates a spontaneous redox reaction occurring in the cell. Question 15 Question Consider a galvanic cell with the following half-reactions: Anode: Zn(s) → Zn2+ (aq) + 2e− Cathode: 2H+ (aq) + 2e− → H2 (g) Determine the cell potential at standard conditions for this galvanic cell. Solution Step 1: Write the overall cell reaction. Zn(s) + 2H+ (aq) → Zn2+ (aq) + H2 (g) Step 2: Find the standard reduction potentials for each half-reaction from a table. ◦ EZn 2+ /Zn = −0.76 V ◦ EH + /H = 0 V 2 Step 3: Calculate the cell potential at standard conditions. ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0 V − (−0.76 V) ◦ Ecell = 0.76 V Therefore, the cell potential at standard conditions for this galvanic cell is 0.76 V. Question 16 Question A voltaic cell is constructed using a standard hydrogen electrode (SHE) as the anode and an unknown metal electrode as the cathode. The measured cell potential is 0.67 V. Determine the standard reduction potential for the unknown metal electrode. 14 Solution Step 1: Write the cell reaction. The overall cell reaction for the voltaic cell is: − H+ (aq) + e → 1 H2 (g) 2 Let the unknown metal electrode be represented as M: − Mn+ (aq) + ne → M(s) The overall cell reaction can be written as: Mn+ (aq) + n H2 (g) → M(s) + nH+ (aq) 2 Step 2: Determine the standard cell potential. Given: Measured cell potential, Ec ell = 0.67V T hestandardcellpotential, E c ell, canbecalculatedusingtheN ernstequation : E cell = Ecell − RT nF ln Q Since the cell is at standard conditions, Q = 1 and the equation simplifies to: E cell = Ecell Therefore, the standard cell potential is 0.67 V. Step 3: Apply the standard cell potential. By comparing with the standard reduction potential table, we know that the standard reduction potential for the standard hydrogen electrode (SHE) is 0 V. The standard reduction potential for the unknown metal electrode can be calculated as: E cell = E cathode − E anode 0.67 = E cathode − 0 E cathode = 0.67 V Therefore, the standard reduction potential for the unknown metal electrode is 0.67 V. Question 17 Question Consider a galvanic cell with a standard cell potential of E ◦ = 0.87 V. The cell reaction is given by: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Determine the standard cell potential for the corresponding electrolytic cell involving the same half-reactions. 15 Solution Step 1: Write the cell reaction for the electrolytic cell by reversing the reaction for the galvanic cell. Zn2+ (aq) + Cu(s) → Zn(s) + Cu2+ (aq) Step 2: Determine the standard cell potential for the electrolytic cell. Since the cell reaction has been reversed, the standard cell potential for the electrolytic cell is the negative of the standard cell potential for the galvanic cell. ◦ Eelectrolytic = −E ◦ = −0.87 V Therefore, the standard cell potential for the corresponding electrolytic cell ◦ involving the same half-reactions is Eelectrolytic = −0.87 V. Question 18 Question Consider a galvanic cell with a standard cell potential of E ◦ = 1.10 V. The half-reactions in the cell are: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) Calculate the equilibrium constant K for the cell reaction at 25°C. Solution Step 1: Write the cell reaction equation. The cell reaction equation is the sum of the half-reactions at the anode and cathode: Zn(s) + Cu2+ (aq) −−→ Zn2+ (aq) + Cu(s) Step 2: Write expression for the equilibrium constant K. The equilibrium constant K for the cell reaction is related to the standard cell potential E ◦ by the Nernst equation: RT ln(K) E = E◦ − nF where n is the number of electrons transferred in the cell reaction, R is the gas constant, T is the temperature in kelvin, and F is the Faraday constant. Step 3: Determine the number of electrons transferred n. From the balanced cell reaction, we see that n = 2, as 2 electrons are transferred in the reaction. Step 4: Plug in the known values and solve for K. Given: E ◦ = 1.10 V R = 8.314 J/mol · K T = 298 K F = 96485 C/mol 16 Substitute the values into the Nernst equation to solve for K: 1.10 V = 1.10 V − (8.314 J/mol · K)(298 K) ln(K) (2)(96485 C/mol) Solving for ln(K): ln(K) = − (8.314 J/mol · K)(298 K) (2)(96485 C/mol) ln(K) = −0.0109 K = e−0.0109 K ≈ 0.989 Therefore, the equilibrium constant K for the cell reaction at 25°C is approximately 0.989. Question 19 Question Consider a galvanic cell with the following half-cell reactions: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) ◦ ◦ If the standard reduction potentials are EZn 2+ /Zn = −0.76 V and ECu2+ /Cu = ◦ 0.34 V, calculate the standard cell potential (Ecell ) for this galvanic cell. Solution Step 1: Write the overall cell reaction. Zn(s) + Cu2+ (aq) −−→ Zn2+ (aq) + Cu(s) ◦ Step 2: Calculate the standard cell potential (Ecell ) using the standard reduction potentials. ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EZn2+ /Zn ◦ Ecell = 0.34 V − (−0.76 V) ◦ Ecell = 1.10 V Therefore, the standard cell potential for this galvanic cell is 1.10 V. 17 Question 20 Question Describe the differences between galvanic cells and electrolytic cells. Provide an example of each and explain how they operate. Solution A galvanic cell is an electrochemical cell that converts chemical energy into electrical energy through a spontaneous redox reaction, while an electrolytic cell is an electrochemical cell that requires an external electrical energy source to drive a non-spontaneous redox reaction. Differences: ˆ Energy Conversion: Galvanic cells convert chemical energy into electrical energy, while electrolytic cells use electrical energy to drive a nonspontaneous reaction. ˆ Spontaneity: Galvanic cells have spontaneous redox reactions, while electrolytic cells have non-spontaneous reactions. ˆ Electrodes: In galvanic cells, the anode is negative and the cathode is positive; in electrolytic cells, the anode is positive and the cathode is negative. Example of a Galvanic Cell (Voltaic Cell): The Daniell cell is a common example of a galvanic cell. In this cell, a zinc electrode is placed in a solution of zinc sulfate, connected by a salt bridge to a copper electrode in a solution of copper sulfate. The zinc undergoes oxidation at the anode, releasing electrons, which flow through the external circuit to the copper cathode, where reduction occurs. Operation of a Galvanic Cell: ˆ Step 1: Zinc metal atoms undergo oxidation at the anode, releasing electrons: Zn(s) → Zn2+ (aq) + 2e− ˆ Step 2: Electrons flow through the external circuit to the copper cathode. ˆ Step 3: Copper ions in solution gain electrons and are reduced at the cathode: Cu2+ (aq) + 2e− → Cu(s) ˆ Overall Cell Reaction: 2+ Zn(s) + Cu (aq) → Zn2+ (aq) + Cu(s) 18 Example of an Electrolytic Cell: An example of an electrolytic cell is the electrolysis of water. In this cell, an electric current is passed through water to split it into hydrogen and oxygen gases. Operation of an Electrolytic Cell: ˆ Step 1: Water molecules undergo electrolysis to form oxygen gas at the anode and hydrogen gas at the cathode: 2H2 O(l) → 2H2 (g) + O2 (g) Question 21 Question Consider the following electrochemical cell reaction: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) ◦ Given that the standard reduction potentials are EZn 2+ /Zn = −0.76 V and ◦ ECu2+ /Cu = 0.34 V, determine if the reaction will be spontaneous under standard conditions. If not, calculate the minimum potential at which the reaction will be spontaneous. Solution ◦ Step 1: Calculate the cell potential (Ecell ) under standard conditions. ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.34 V − (−0.76) V ◦ Ecell = 1.10 V Step 2: Determine if the reaction is spontaneous under standard conditions. ◦ Since Ecell is positive, the reaction is spontaneous under standard conditions. Step 3: Calculate the minimum potential to make the reaction spontaneous. For a reaction to be spontaneous, the cell potential (Ecell ) must be positive. ◦ Ecell = Ecell − 0.0592 log Q n where Q is the reaction quotient and n is the number of electrons transferred. Step 4: Calculate the reaction quotient (Q) at equilibrium. Since the reaction is at equilibrium when Ecell = 0, we have: 0.0592 log Q 2 0 = 1.10 − log Q = 1.10 × 2 0.0592 19 log Q = 37.1622 Q = 1037.1622 Step 5: Calculate the minimum potential required for the reaction to be spontaneous. 0.0592 log Q′ Ecell = 0 = 1.10 − 2 Substitute Q with Q′ : 0 = 1.10 − 0.0592 log 1037.1622 2 0.0592 × 37.1622 2 0 = 1.10 − 1.0968 0 = 1.10 − 1.0968 = 1.10 1.0968 > 1.10 Therefore, the minimum potential required for the reaction to be spontaneous is greater than 1.10 V. Question 22 Question Consider a galvanic cell with a standard cell potential of 0.78 V. If the cell potential drops to 0.70 V at 25°C, calculate the change in standard Gibbs free energy (∆G◦ ) for the reaction occurring in the cell. Solution Step 1: Calculate the change in cell potential (∆E ◦ ): Given: Standard cell potential, E ◦ = 0.78 V Cell potential at 25°C, E = 0.70 V The change in cell potential can be calculated using the equation: ∆E ◦ = E ◦ − E ∆E ◦ = 0.78 − 0.70 = 0.08 V Step 2: Calculate the Faraday constant (F ): The Faraday constant, F , is given by: F = 96485 C/mol Step 3: Using the equation relating ∆G◦ , ∆E ◦ , and n (the number of moles of electrons transferred in the reaction), we have: ∆G◦ = −nF ∆E ◦ 20 Step 4: Determine the number of moles of electrons transferred, n: Since the cell potential dropped, we know the reaction is running in the reverse direction. Therefore, the number of moles of electrons transferred, n, is: n = −2 Step 5: Calculate the change in standard Gibbs free energy (∆G◦ ): Substitute the values of n, F , and ∆E ◦ into the equation: ∆G◦ = −(−2)(96485)(0.08) ∆G◦ = 15435.76 J/mol Therefore, the change in standard Gibbs free energy for the reaction occurring in the cell is 15435.76 J/mol. Question 23 Question Consider a galvanic cell with a standard cell potential of 1.13 V that involves the reaction: Zn(s) + 2Fe3+ (aq) → Zn2+ (aq) + 2Fe2+ (aq) (a) Write the cell notation for this galvanic cell. (b) Calculate the standard cell potential of the galvanic cell if [Fe3+ ] = 0.20 M and [Fe2+ ] = 2.00 M. Solution (a) The cell notation for the galvanic cell can be written as: Zn(s) | Zn2+ (aq) || Fe3+ (aq), Fe2+ (aq) | Pt(s) (b) The standard cell potential can be calculated using the Nernst equation: RT ln Q nF where E is the cell potential, E ◦ is the standard cell potential, R is the gas constant (8.314 J/(mol K)), T is the temperature in Kelvin, n is the number of moles of electrons transferred in the balanced equation, F is the Faraday constant (96485 C/mol), and Q is the reaction quotient. In this case, n = 2 because two moles of electrons are transferred in the balanced equation. The reaction quotient Q can be expressed as: E = E◦ − Q= [Zn2+ ] 1 3+ 2+ = (0.20)(2.00)2 = 1.25 [Fe ][Fe ] 21 Substitute the given values into the Nernst equation: E = 1.13 V − (8.314)(298) ln 1.25 ≈ 1.08 V 2(96485) Question 24 Question ◦ Consider a voltaic cell with a standard cell potential of Ecell = 1.20 V. The cell reaction is: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Determine the standard reduction potentials for the Zn and Cu half-reactions and write the half-reactions. Also, calculate the standard cell potential for the reverse reaction. Solution Step 1: Write the half-reactions and determine their standard reduction potentials. Zn2+ + 2e− → Zn ◦ EZn = −0.76 V Cu2+ + 2e− → Cu ◦ ECu = 0.34 V Step 2: Write the balanced overall cell reaction for the voltaic cell. Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Step 3: Calculate the standard cell potential for the reverse reaction. Given ◦ the standard cell potential for the forward reaction is Ecell = 1.20 V, the stan◦ dard cell potential for the reverse reaction is −Ecell . ◦ Ecell, reverse = −1.20 V Question 25 Question Consider the following reaction in an electrochemical cell: Pb2+ (aq) + 2Ag(s) → Pb(s) + 2Ag+ (aq) Determine which electrode acts as the anode and which as the cathode. Explain your reasoning. 22 Solution Step 1: Write the standard reduction potentials for the half-reactions involved. The standard reduction potentials for the half-reactions involved are: Ag+ (aq) + e− → Ag(s) E ◦ = +0.80 V Pb2+ (aq) + 2e− → Pb(s) E ◦ = −0.13 V Step 2: Identify the half-reaction with the more positive standard reduction potential as the reduction half-reaction (cathode). Since the reduction potential for the reduction of Ag+ to Ag is higher (+0.80 V), this is the reduction halfreaction. Therefore, the Ag electrode acts as the cathode. Step 3: Deduce the anode based on the cathode. The other electrode, the Pb electrode, must therefore act as the anode. Therefore, in the given cell setup: - The Ag electrode acts as the cathode. The Pb electrode acts as the anode. Question 26 Question Consider a galvanic cell and an electrolytic cell, both involving the same redox reaction. The galvanic cell produces an electric current when the reaction occurs spontaneously, while the electrolytic cell requires an external electric current to drive the non-spontaneous reaction. Explain the key differences between these two types of cells, focusing on their operating principles and applications. Solution Step 1: Galvanic Cell A galvanic cell, also known as a voltaic cell, converts chemical energy into electrical energy. It consists of two half-cells connected by a salt bridge or porous barrier. In the spontaneous redox reaction, electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs), generating an electric current. This current can be used to power external devices. Step 2: Electrolytic Cell An electrolytic cell uses an external electric current to drive a non-spontaneous redox reaction. It consists of two electrodes (anode and cathode) immersed in an electrolyte solution. When an electric current is applied, electrons are forced to move against their natural direction. This process allows for the reduction of cations at the cathode and the oxidation of anions at the anode. Step 3: Key Differences 1. Energy Conversion: Galvanic cells convert chemical energy into electrical energy, while electrolytic cells require electrical energy to drive a non-spontaneous reaction. 2. Spontaneity: Galvanic cells run spontaneously, while electrolytic cells need an external power source to operate. 3. 23 Electrode Reactions: In galvanic cells, oxidation occurs at the anode and reduction at the cathode. In electrolytic cells, the reactions are reversed due to the external electric current. 4. Applications: Galvanic cells are used in batteries to power devices, while electrolytic cells are used in processes like electroplating, metal refining, and electrolysis of water. In summary, galvanic cells produce electrical energy from chemical reactions, while electrolytic cells use electrical energy to drive non-spontaneous reactions. They have different applications and operate based on opposite principles of redox reactions. Question 27 Question ◦ Consider a galvanic cell that consists of a standard hydrogen electrode (Ecell = 0.00 V) and a silver electrode in a 1.0 M solution of silver nitrate. Part a: Write the overall cell reaction for this galvanic cell. ◦ Part b: If the standard reduction potential of silver is EAg + /Ag = 0.80 V, calculate the cell potential at standard conditions. Part c: Will the cell potential increase or decrease if the concentration of silver nitrate is increased? Justify your answer. Solution Part a: The overall cell reaction for the galvanic cell involving a standard hydrogen electrode and a silver electrode is given by: 2H+ (aq) + 2e− → H2 (g) E ◦ = 0.00 V Ag+ (aq) + e− → Ag(s) E ◦ = 0.80 V Adding these two half-reactions together, we get the overall cell reaction: 2H+ (aq) + 2Ag+ (aq) → H2 (g) + 2Ag(s) Part b: ◦ The cell potential Ecell at standard conditions can be calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.80 V − 0.00 V = 0.80 V Therefore, the cell potential at standard conditions is 0.80 V. 24 Part c: If the concentration of silver nitrate is increased, the concentration of Ag+ in the cathode compartment increases. According to the Nernst equation, an increase in the concentration of a reactant shifts the equilibrium towards products resulting in a higher cell potential. Therefore, increasing the concentration of silver nitrate will increase the cell potential. Question 28 Question Consider a galvanic cell with the half-reactions: Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V Fe2+ (aq) + 2e− → Fe(s) E ◦ = −0.44 V Suppose the standard cell potential is measured to be 0.78 V. Determine the concentrations of Cu2+ and Fe2+ if the cell operates under standard conditions. Solution Step 1: Write the cell reaction. The cell reaction is obtained by adding the half-reactions together: Cu2+ (aq) + Fe(s) → Cu(s) + Fe2+ (aq) Step 2: Calculate the standard cell potential. The standard cell potential, E◦ , is the difference between the standard reduction potentials of the two halfreactions: E◦ = E ◦ (cathode) − E ◦ (anode) E ◦ = 0.34 V − (−0.44 V) E ◦ = 0.78 V Step 3: Determine the reaction quotient. Since the reaction is at equilibrium, the reaction quotient, Q, is equal to 1. Step 4: Use the Nernst equation to find the ratio of concentrations. The Nernst equation relates the standard cell potential to the reaction quotient and the concentrations of species involved in the cell reaction: E = E◦ − 0.0592 log Q n where n is the number of moles of electrons transferred in the cell reaction. Step 5: Solve for the concentrations of Cu2+ and Fe2+ . Given that Q = 1 in this case and n = 2, we can write: 0.78 V = 0.34 V − 0.0592 log 1 2 0.44 = 0.0592 log 1 25 0 = log 1 Thus, at standard conditions, the concentrations of Cu2+ and Fe2+ do not change. Question 29 Question A student is studying galvanic and electrolytic cells in their electrochemistry course. They are given a cell diagram but are unsure whether the cell is galvanic or electrolytic. The diagram shows a zinc metal electrode dipping into a solution of zinc sulfate and a silver metal electrode dipping into a solution of silver nitrate. The two electrodes are connected by a wire. The student must determine whether the cell is galvanic or electrolytic, and explain their reasoning. Solution To determine whether the given cell is galvanic or electrolytic, we need to consider the standard electrode potentials of the half-reactions involved. Step 1: Write the half-reactions The half-reaction at the zinc electrode is the oxidation of zinc metal: Zn(s) → Zn2+ (aq) + 2e− The half-reaction at the silver electrode is the reduction of silver ions: Ag+ (aq) + e− → Ag(s) Step 2: Determine the standard electrode potentials Given standard reduction potentials: ◦ EZn 2+ /Zn = −0.76 V ◦ EAg + /Ag = 0.80 V Step 3: Determine if the cell is galvanic or electrolytic To determine if the cell is galvanic or electrolytic, we can calculate the cell potential using the standard reduction potentials. The overall cell reaction is the sum of the two half-reactions: Zn(s) + Ag+ (aq) → Zn2+ (aq) + Ag(s) ◦ The overall standard cell potential (Ecell ) can be calculated as follows: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = EAg + /Ag − EZn2+ /Zn 26 ◦ Ecell = 0.80 V − (−0.76 V) = 1.56 V Since the cell potential is positive (1.56 V), the cell is galvanic. In a galvanic cell, the cell potential is positive, indicating that the spontaneous electrochemical reaction occurs. Question 30 Question Consider a galvanic cell involving the reaction: Cr3+ (aq) + Zn(s) → Zn2+ (aq) + Cr2+ (aq) The standard reduction potentials are: ◦ ECr 3+ /Cr2+ = −0.74 V ◦ EZn 2+ /Zn = −0.76 V (a) Calculate the standard cell potential for this galvanic cell. (b) Will the reaction in this galvanic cell proceed spontaneously at standard conditions? (c) If a voltage of 1.5 V is applied to the cell in the reverse direction, what reactions will take place at the electrodes? Solution (a) To calculate the standard cell potential, we use the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode Given that reduction occurs at the cathode and oxidation occurs at the anode, we have: ◦ ◦ ◦ Ecell = ECr 2+ /Cr3+ − EZn2+ /Zn ◦ Ecell = (−0.74 V) − (−0.76 V) ◦ Ecell = 0.02 V Therefore, the standard cell potential for this galvanic cell is 0.02 V. (b) For a reaction to proceed spontaneously, the standard cell potential must be positive. Since the standard cell potential for this galvanic cell is 0.02 V (which is positive), the reaction will proceed spontaneously at standard conditions. (c) When a voltage of 1.5 V is applied to the cell in the reverse direction, the reactions at the electrodes will also be reversed. This means that Zn will be plated at the cathode and Cr will be oxidized at the anode. The reactions will be: 27 At the cathode (Zn): Zn2+ (aq) + 2e− → Zn(s) At the anode (Cr): Cr2+ (aq) → Cr3+ (aq) + e− Question 31 Question ◦ Consider a galvanic cell with a standard cell potential of Ecell = 1.10 V. If the 3+ concentration of Fe in one of the half-cells is 0.10 M and the concentration of Fe2+ in the other half-cell is 1.0 M at 25◦ C, determine the cell potential when the concentration of Fe3+ in the first half-cell is reduced to 0.050 M and the concentration of Fe2+ in the second half-cell is increased to 4.0 M. Solution Step 1: Write the overall cell reaction for the galvanic cell: The half-reactions for the cell are: Fe3+ + 3e− → Fe2+ E ◦ = −0.77 V Fe2+ → Fe + 2e− E ◦ = −0.44 V The overall cell reaction is the sum of the two half-reactions: Fe3+ + Fe2+ → Fe + Fe2+ Step 2: Calculate the initial cell potential using the Nernst equation: The standard cell potential is E ◦ = 1.10 V. The initial concentrations of Fe3+ and Fe2+ are 0.10 M and 1.0 M, respectively. The Nernst equation is:  2+  0.0592 [Fe ] ◦ Ecell = Ecell − log n [Fe3+ ] Substitute the values and calculate the initial cell potential:   1.0 0.0592 log Ecell = 1.10 − 2 0.10 Ecell = 1.10 − 0.0296 × 1 ≈ 1.0704 V Step 3: Calculate the final cell potential after changing the concentrations: The new concentrations are 0.050 M for Fe3+ and 4.0 M for Fe2+ . Use the Nernst equation again to calculate the new cell potential:   0.0592 4.0 Ecell = 1.10 − log 2 0.050 Ecell = 1.10 − 0.0296 × 3.322 ≈ 1.0068 V Therefore, the cell potential after changing the concentrations is approximately 1.0068 V. 28 Question 32 Question ◦ Consider a galvanic cell with a standard cell potential of Ecell = 0.80 V. If the reaction in the cell is spontaneous, what can you conclude about the standard Gibbs free energy change for the cell reaction? Explain your answer. Solution Step 1: Recall the relationship between standard Gibbs free energy change ◦ (∆G◦ ) and standard cell potential (Ecell ): ◦ ∆G◦ = −nF Ecell where n is the number of electrons transferred in the cell reaction and F is the Faraday constant (96,485 C/mol). Step 2: Since the cell reaction is spontaneous, it means that the standard Gibbs free energy change for the reaction is negative: ∆G◦ < 0 Step 3: Substituting the given standard cell potential into the equation, we have: ∆G◦ = −nF (0.80 V) Step 4: To determine the sign of ∆G◦ , we need to consider the signs of n ◦ and F . Since Ecell is positive, n must also be positive to ensure a negative value ◦ for ∆G . Step 5: Therefore, we can conclude that the standard Gibbs free energy change for the cell reaction is negative, indicating that the reaction is spontaneous and capable of performing work. Question 33 Question Consider a galvanic cell constructed with a copper electrode in a 1.0 M Cu2+ solution and a silver electrode in a 1.0 M Ag+ solution. The standard reduction ◦ ◦ potentials are ECu 2+ /Cu = 0.34 V and EAg+ /Ag = 0.80 V. Determine the cell potential when the concentration of Cu2+ is 0.10 M and Ag+ is 0.50 M. Solution Step 1: Write the half-reactions and standard reduction potentials: The halfreactions that occur in this cell are: Cathode (Reduction at Ag+ ): Ag+ (aq) + 29 ◦ 2+ e− → Ag(s) (EAg ): Cu(s) → Cu2+ (aq)+ + /Ag = 0.80 V) Anode (Oxidation at Cu ◦ 2e− (ECu 2+ /Cu = 0.34 V) Step 2: Calculate the cell potential under standard conditions: The cell ◦ potential under standard conditions (Ecell ) is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.80 V − 0.34 V = 0.46 V Step 3: Use the Nernst equation to calculate the cell potential when the concentrations are not standard: The Nernst equation is: ◦ Ecell = Ecell − 0.0592 [Ag+ ] log n [Cu2+ ] where n is the number of electrons transferred in the balanced chemical equation. In this case, n = 1. Plugging in the values: Ecell = 0.46 V − 0.50 0.0592 log 1 0.10 Ecell = 0.46 V − 0.0592 log 5 Ecell = 0.46 V − 0.11 V = 0.35 V Therefore, the cell potential when the concentration of Cu2+ is 0.10 M and Ag is 0.50 M is 0.35 V. + Question 34 Question Consider a galvanic cell with a standard cell potential of E ◦ = 1.20 V at 25◦ C that consists of a standard hydrogen electrode (Pt—H2 (1 atm)—H+ ) and a silver electrode (Ag). Calculate the cell potential when the concentration of Ag+ is 0.10 M and the pressure of hydrogen gas is 0.5 atm at 25◦ C. Solution Step 1: Write the cell reaction for the galvanic cell: Pt(s) + 2H+ (aq) + 2e− → H2 (g) Ag(s) → Ag+ (aq) + e− Step 2: Write the balanced overall cell reaction: Pt(s) + 2Ag+ (aq) → 2Ag(s) + 2H+ (aq) 30 Step 3: Use the Nernst equation to calculate the cell potential: E = E◦ − Q 0.0592 log n K where E is the cell potential, E ◦ is the standard cell potential, n is the number of electrons transferred in the balanced cell reaction, Q is the reaction quotient, and K is the equilibrium constant. Step 4: Calculate the reaction quotient Q and the number of electrons transferred n based on the balanced cell reaction: Q= [Ag+ ]2 (0.10)2 + 2 = (1)2 [H ] n=2 Step 5: Substitute the values into the Nernst equation: E = 1.20 − 0.0592 0.01 log 2 1 E = 1.20 − 0.0296 log(0.01) E = 1.20 − 0.0296(−2) E = 1.20 + 0.0592 E = 1.2592 V Therefore, the cell potential when the concentration of Ag+ is 0.10 M and the pressure of hydrogen gas is 0.5 atm at 25◦ C is 1.2592 V. Question 35 Question Consider a galvanic cell in which the following reaction occurs: 2+ Zn(s) — Zn (aq, 1.00 M)||Cu2+ (aq, 1.00 M) — Cu(s) ◦ Calculate the cell potential at 25°C. Given: Ecell (Cu2+ /Cu) = +0.34 V and 2+ ◦ Ecell (Zn /Zn) = −0.76 V. Solution Step 1: Write the overall cell reaction. The overall cell reaction can be obtained by summing the half-reactions for each electrode. The standard cell potential, ◦ Ecell , is the difference between the standard reduction potentials of the halfreactions. The two half-reactions and their standard reduction potentials are: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V 31 Cu2+ (aq) + 2e− → Cu(s) E ◦ = +0.34 V Adding these half-reactions together gives the overall reaction: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Step 2: Calculate the cell potential, Ecell . Using the Nernst equation, the cell potential of the galvanic cell can be calculated as: ◦ Ecell = Ecell − [Zn2+ ] 0.0592 log n [Cu2+ ] The stoichiometry of the balanced reaction shows that n = 2 (2 moles of electrons transferred). ◦ ◦ Given that Ecell (Cu2+ /Cu) = +0.34 V and Ecell (Zn2+ /Zn) = −0.76 V, we substitute these values into the Nernst equation along with the concentrations of the ions: Ecell = (+0.34 V − (−0.76 V)) − 0.0592 1.00 log 2 1.00 Ecell = 1.10 V − 0 = 1.10 V Therefore, the cell potential at 25°C for the given galvanic cell is 1.10 V. 32