Galvanic and Electrolytic Cells Question Bank Set 4

CHEM 121 - GENERAL CHEMISTRY I - Galvanic and Electrolytic Cells Question Bank - Set 4 Liberty University Question 1 Question Define a galvanic cell and an electrolytic cell. Explain the main differences between the two types of cells. Solution Galvanic Cell: Step 1: A galvanic cell, also known as a voltaic cell, is an electrochemical cell that generates electrical energy from spontaneous redox reactions. Step 2: In a galvanic cell, chemical energy is converted into electrical energy; the redox reaction releases energy that can be harnessed as electricity. Step 3: Electrons flow from the anode (where oxidation occurs) to the cathode (where reduction occurs) through an external circuit. Step 4: A classic example of a galvanic cell is the Daniell cell, which consists of a zinc electrode (anode), a copper electrode (cathode), and an electrolyte solution. Step 5: The voltage produced by a galvanic cell is positive because the reaction is spontaneous. Electrolytic Cell: Step 6: An electrolytic cell is an electrochemical cell that uses electrical energy to drive a non-spontaneous redox reaction. Step 7: In an electrolytic cell, electrical energy is converted into chemical energy; the non-spontaneous redox reaction is made to occur by external electrical energy. Step 8: Electrons flow from the external power source to the anode (where oxidation occurs) and from the cathode (where reduction occurs) back to the power source. Step 9: A classic example of an electrolytic cell is the electrolysis of water, which breaks down water into hydrogen and oxygen gas. Step 10: The voltage required by an electrolytic cell is negative because the reaction is non-spontaneous and requires an external energy source. Main differences between Galvanic and Electrolytic Cells: Step 11: Galvanic cells produce electrical energy from spontaneous reactions, while electrolytic cells require electrical energy to drive non-spontaneous reactions. Step 12: In galvanic cells, electrons flow from the anode to the cathode; in electrolytic cells, electrons flow from the external source to the anode and from the cathode back to the external source. Step 13: Galvanic cells have a positive cell potential, while electrolytic cells have a negative cell potential. Step 14: Galvanic cells are commonly used in batteries to generate electricity, while electrolytic cells are used in processes such as electroplating, electrorefining, and electrolysis. Step 15: Overall, galvanic cells convert chemical energy into electrical energy, while electrolytic cells convert electrical energy into chemical energy. Question 2 Question Consider a galvanic cell with the following half-reactions: ◦ Zn2+ (aq) + 2e− → Zn(s) Ered = −0.76 V ◦ Cu2+ (aq) + 2e− → Cu(s) Ered = +0.34 V If the initial concentrations are [Zn2+ ] = 1.0 M and [Cu2+ ] = 1.0 M, calculate the cell potential when the cell is operating. Solution Step 1: Write the overall cell reaction by adding the two half-reactions together: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Step 2: Determine the cell potential using the standard reduction potentials provided: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = (+0.34 V) − (−0.76 V) = 1.10 V Step 3: Apply the Nernst equation to account for non-standard conditions: The Nernst equation is given by:   [Zn2+ ] 0.0592 ◦ log Ecell = Ecell − n [Cu2+ ] Since this is a redox reaction involving the transfer of 2 electrons, n = 2. Substitute the given concentrations:   1.0 0.0592 Ecell = 1.10 V − log 2 1.0 Ecell = 1.10 V Therefore, the cell potential when the cell is in operation is 1.10 V. 2 Question 3 Question ◦ Consider a galvanic cell with a standard reduction potential of Ecell = 0.30 V. If the cell potential is measured to be 0.45 V under non-standard conditions where the concentration of Cu2+ ions in the cathode compartment is 0.10 M, determine the concentration of Cu2+ ions in the anode compartment. Solution Step 1: Write the half-reactions for the cell. The overall cell reaction can be split into two half-reactions: Cathode (Reduction): Cu2+ (aq) + 2e− → Cu(s) Anode (Oxidation): Zn(s) → Zn2+ (aq) + 2e− Step 2: Calculate the cell potential under non-standard conditions. By using the Nernst equation:   [Cu2+ ]anode 0.0592 ◦ log Ecell = Ecell − 2 [Cu2+ ]cathode Substitute the given values: 0.45 = 0.30 − 0.0592 log 2  [Cu2+ ]anode 0.10  Step 3: Solve for the concentration of Cu2+ ions in the anode compartment. Simplify the equation:   [Cu2+ ]anode 0.15 = −0.0296 log 0.10   2+ [Cu ]anode 5.07 = log 0.10 105.07 = [Cu2+ ]anode [Cu2+ ]anode 0.10 = 105.07 × 0.10 [Cu2+ ]anode ≈ 1.23 × 105 M Therefore, the concentration of Cu2+ ions in the anode compartment is approximately 1.23 × 105 M. Question 4 Question Consider a galvanic cell that consists of a copper electrode in a Cu2+ solution connected by a salt bridge to a silver electrode in a Ag + solution. The standard ◦ ◦ reduction potentials are ECu 2+ /Cu = 0.34 V and EAg + /Ag = 0.80 V. Calculate the cell potential for this galvanic cell. 3 Solution Step 1: Write the half-reactions for the redox reactions occurring at each electrode. Reduction at the copper electrode: Cu2+ + 2e− → Cu Reduction at the silver electrode: Ag + + e− → Ag E ◦ = 0.34 V E ◦ = 0.80 V Step 2: Write the overall cell reaction by summing the two half-reactions. The cell potential is the difference in standard electrode potentials. Cu2+ + 2Ag → Cu + 2Ag + ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.80 V − 0.34 V ◦ Ecell = 0.46 V Therefore, the cell potential for this galvanic cell is 0.46 V. Question 5 Question Consider a galvanic cell with the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V Calculate the cell potential at 25°C when the concentrations of Zn2+ and Cu2+ are both 0.10 M. Assume both half-reactions are at standard conditions. Solution Step 1: Write the overall cell reaction by subtracting the reduction potential of the Zn half-reaction from the reduction potential of the Cu half-reaction: Zn2+ (aq) + Cu(s) → Cu2+ (aq) + Zn(s) ◦ Step 2: Determine the standard cell potential (Ecell ) using the Nernst equation: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = (0.34 V) − (−0.76 V) ◦ Ecell = 1.10 V Step 3: Calculate the cell potential (Ecell ) at 25°C using the Nernst equation:   0.0592 [Cu2+ ] ◦ Ecell = Ecell − log n [Zn2+ ] 4 Step 4: Calculate the cell potential by plugging in the given values:   0.0592 0.10 Ecell = 1.10 V − log 2 0.10 Ecell = 1.10 V Therefore, the cell potential at 25°C when the concentrations of Zn2+ and Cu2+ are both 0.10 M is 1.10 V . Question 6 Question Consider a galvanic cell based on the following half-reactions: Anode: Zn(s) → Zn2+ (aq) + 2e− Cathode: 2H+ (aq) + 2e− → H2 (g) Calculate the standard cell potential at 25°C for this galvanic cell. Solution Step 1: Write the cell reaction by reversing the anode half-reaction and adding the cathode half-reaction. Zn2+ (aq) + 2H+ (aq) → Zn(s) + H2 (g) Step 2: Write the standard cell potential in terms of standard reduction potentials (E ◦ ) for the half-reactions involved. The standard cell potential is given by ◦ ◦ ◦ Ecell = Ecathode − Eanode Step 3: Find the standard reduction potentials for the half-reactions from a standard reduction potential table. The values are: ◦ EZn 2+ /Zn = −0.76 V ◦ E2H + /H = 0.00 V 2 Step 4: Substitute the values into the formula for the standard cell potential. ◦ Ecell = 0.00 V − (−0.76 V) ◦ Ecell = 0.76 V Therefore, the standard cell potential at 25°C for the given galvanic cell is 0.76 V. 5 Question 7 Question Consider a galvanic cell with the following half-reactions: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) ◦ ◦ If the standard reduction potentials are ECu 2+ /Cu = 0.34 V and EZn2+ /Zn = ◦ −0.76 V, calculate the standard cell potential (Ecell ) of the galvanic cell. Solution Step 1: Write the overall cell reaction by combining the two half-reactions, ensuring that the number of electrons lost equals the number of electrons gained: Zn(s) + Cu2+ (aq) −−→ Zn2+ (aq) + Cu(s) ◦ ) using the standard reStep 2: Calculate the standard cell potential (Ecell duction potentials of the half-reactions. The standard cell potential is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ Substitute the values of ECu 2+ /Cu = 0.34 V and EZn2+ /Zn = −0.76 V into the equation: ◦ Ecell = 0.34 V − (−0.76 V) = 1.10 V ◦ = 1.10 V. Therefore, the standard cell potential of the galvanic cell is Ecell Question 8 Question Consider an electrochemical cell consisting of a copper electrode immersed in 1.0 M Cu2+ solution and a silver electrode immersed in 1.0 M Ag+ solution. Write the balanced half-reactions and the overall cell reaction. Is this cell a galvanic cell or an electrolytic cell? Justify your answer. Solution Step 1: Write the balanced half-reactions: At the copper electrode: Cu2+ + 2e− → Cu At the silver electrode: Ag+ + e− → Ag Step 2: Write the overall cell reaction: Cu2+ + 2Ag → Cu + 2Ag+ 6 Step 3: Determine if the cell is a galvanic or electrolytic cell: In a galvanic cell, the redox reaction is spontaneous and produces electrical energy. In this case, the cell reaction is proceeding from left to right (oxidation at copper electrode, reduction at silver electrode), which means it is a spontaneous process. Therefore, this cell is a galvanic cell. Question 9 Question Consider a galvanic cell constructed with a silver electrode in a 1.0 M AgNO3 solution and a copper electrode in a 1.0 M Cu(NO3 )2 solution. The standard reduction potentials are: Ag+ (aq) + e− → Ag(s) (E ◦ = 0.80 V) and Cu2+ (aq) + 2e− → Cu(s) (E ◦ = 0.34 V). Determine the cell potential at 25°C. Solution Step 1: Write out the overall reaction for the galvanic cell: Ag(s) + Cu2+ (aq) → Cu(s) + Ag+ (aq) ◦ Step 2: Determine the standard cell potential (Ecell ) using the standard reduction potentials provided: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EAg+ /Ag ◦ Ecell = 0.34 V − 0.80 V ◦ Ecell = −0.46 V Step 3: Calculate the non-standard cell potential using the Nernst equation: E = E◦ − 0.0592 log(Q) n where n is the number of moles of electrons transferred and Q is the reaction quotient. Step 4: Calculate the reaction quotient Q: Q= [Cu][Ag+ ] [products] = [reactants] [Ag][Cu2+ ] Since both concentrations are 1.0 M, Q = 1. Step 5: Substitute the values into the Nernst equation: E = −0.46 V − 0.0592 log(1) 2 E = −0.46 V Therefore, the cell potential at 25°C for the given galvanic cell is −0.46 V. 7 Question 10 Question Consider a Galvanic cell with the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V Pb2+ (aq) + 2e− → Pb(s) E ◦ = −0.13 V If the initial concentrations are [Zn2+ ] = 1.0 M, [Pb2+ ] = 0.1 M, and the cell operates until equilibrium is reached, determine the cell potential at equilibrium. Solution Step 1: Write the cell reaction using the given half-reactions: Zn2+ (aq) + Pb(s) → Zn(s) + Pb2+ (aq) ◦ Step 2: Calculate the standard cell potential (Ecell ) using the standard reduction potentials of the half-reactions: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = −0.13 V − (−0.76 V) = 0.63 V Step 3: Calculate the reaction quotient (Q) at equilibrium using the initial concentrations given: [Zn](Pb2+ ) Q= [Zn2+ ][Pb] Q= (1.0)(0.1) =1 (1.0)(0.1) Step 4: Use the Nernst equation to find the cell potential at equilibrium (Ecell ): 0.0592 ◦ Ecell = Ecell − log(Q) n Since the reaction involves the transfer of 2 electrons, n = 2. Ecell = 0.63 V − 0.0592 log(1) = 0.63 V 2 Therefore, the cell potential at equilibrium is 0.63 V. 8 Question 11 Question Consider the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V Determine if a spontaneous reaction will occur when a strip of copper metal is placed in a solution of zinc sulfate. If so, write the overall balanced cell reaction and calculate the cell potential. Solution Step 1: Write the individual half-reactions and find the standard cell potential. Given the half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V The overall cell reaction will involve the reduction of Cu2+ and the oxidation of Zn. To determine if a spontaneous reaction will occur, we calculate the overall ◦ standard cell potential, Ecell , using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode where ◦ Ecathode = 0.34 V ◦ Eanode = −0.76 V Plugging these values into the formula, we get: ◦ Ecell = 0.34 V − (−0.76 V) = 1.10 V Since the standard cell potential is positive, a spontaneous reaction will occur. Step 2: Write the overall balanced cell reaction. The overall cell reaction is the sum of the two half-reactions. Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Step 3: Calculate the cell potential. The standard cell potential was already calculated as 1.10 V in Step 1. 9 Question 12 Question A voltaic cell is constructed using a silver/silver chloride electrode and a zinc electrode. The standard reduction potentials are as follows: ◦ AgCl(s) + e− → Ag(s) + Cl− (aq) Ecell = 0.22 V ◦ Zn2+ (aq) + 2e− → Zn(s) Ecell = −0.76 V What is the cell potential when the concentration of Zn2+ is 1.0 M and the solid AgCl is in contact with a 0.10 M AgNO3 solution? Use R = 8.314 J/K mol, F = 96, 485 C/mol, and T = 298 K. Solution Step 1: Write the cell reaction and cell potential equation using the given halfreactions. The cell reaction is the sum of the two half-reactions: AgCl(s) + Zn2+ (aq) + 2e− → Ag(s) + Zn(s) + Cl− (aq) The cell potential equation is: 0.0592 log Q n ◦ Ecell = Ecell − where n is the number of moles of electrons transferred in the balanced reaction, and Q is the reaction quotient. Step 2: Calculate the cell potential First, calculate Q using the concentrations provided. Since there are no concentrations for the solid electrodes (Ag and Zn), their activities are assumed to be 1. Therefore, Q= [Ag+ ][Cl− ] [Zn2+ ] 0.102 = 0.01 1.0 Step 3: Plug the values into the formula Q= Ecell = 0.22V − 0.0592 log 0.01 2 Ecell = 0.22V − 0.0296 × 2 × 2 Ecell = 0.22V − 0.1184 = 0.1016 V Therefore, the cell potential is 0.1016 V. 10 Question 13 Question Consider a galvanic cell where the reaction occurring is: Cd2+ (aq) + 2Ag(s) → Cd(s) + 2Ag+ (aq) ◦ Determine the cell potential at standard conditions given that ECd 2+ /Cd = ◦ −0.40 V and EAg+ /Ag = 0.80 V. Solution Step 1: Write the half-reactions and standard electrode potentials. The halfreactions involved in the cell are: Reduction half-reaction: ◦ Cd2+ (aq) + 2e− → Cd(s) ECd 2+ /Cd = −0.40 V Oxidation half-reaction: 2Ag(s) → 2Ag+ (aq) + 2e− ◦ EAg + /Ag = 0.80 V ◦ Step 2: Calculate the cell potential. 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 that ECd 2+ /Cd = −0.40 V and EAg+ /Ag = 0.80 V, we have: ◦ ◦ ◦ Ecell = EAg + /Ag − ECd2+ /Cd = 0.80 − (−0.40) = 1.20 V Therefore, the cell potential at standard conditions is 1.20 V. Question 14 Question Consider a galvanic cell with the following half-reactions: Cu2+ + 2e− → Cu Fe3+ + e− → Fe2+ ◦ Ecell = 0.34 V ◦ Ecell = −0.77 V Calculate the standard Gibbs free energy change (∆G◦ ) for the overall cell reaction and state whether the cell is spontaneous at standard conditions. 11 Solution ◦ Step 1: Calculate the standard cell potential (Ecell ) for the overall reaction. The overall cell reaction is the sum of the two half-reactions: Cu2+ + 2Fe3+ → 2Fe2+ + Cu The standard cell potential is the sum of the standard reduction potentials: ◦ ◦ ◦ Ecell = Ecathode − Eanode Substitute the given values: ◦ ◦ ◦ Ecell = ECu − EFe ◦ Ecell = 0.34 V − (−0.77 V) = 1.11 V Step 2: Calculate the standard Gibbs free energy change (∆G◦ ) for the overall reaction. The relationship between standard cell potential and standard Gibbs free energy change is given by: ◦ ∆G◦ = −nF Ecell where n is the number of moles of electrons transferred in the balanced cell reaction, and F is the Faraday constant (F = 96485 C/mol). Since 2 moles of electrons are transferred in the balanced cell reaction, n = 2. Substitute the values: ∆G◦ = −2 × 96485 C/mol × 1.11 V ∆G◦ = −213.59 kJ/mol Step 3: Determine if the cell is spontaneous at standard conditions. For a reaction to be spontaneous at standard conditions, ∆G◦ must be negative. Since ∆G◦ = −213.59 kJ/mol (which is negative), the cell reaction is spontaneous at standard conditions. Question 15 Question ◦ Consider a galvanic cell with a standard reduction potential of Ecell = 1.23 V. Assuming the reaction in the cell is spontaneous, calculate the standard Gibbs free energy change, ∆G◦ , for the cell reaction. Also, if the concentration of Mn2+ ions in the cell’s cathode compartment is 0.1 M, determine the cell potential when the concentration of Mn2+ ions in the cathode compartment is increased to 1.0 M. 12 Solution Step 1: Recall the relationship between standard Gibbs free energy change, standard cell potential, and the Faraday constant: ◦ ∆G◦ = −nF Ecell where n is the number of moles of electrons transferred in the balanced redox ◦ reaction, F is the Faraday constant (96485 C/mol), and Ecell is the standard cell potential. Step 2: Calculate the number of moles of electrons transferred. The stan◦ dard reduction potential (Ecell ) is related to the number of moles of electrons transferred (n) by the equation: ◦ Ecell = ∆G◦ −nF ◦ Given Ecell = 1.23 V, and F = 96485 C/mol, we can solve for n. ◦ ∆G◦ = −nF Ecell n= −∆G◦ ◦ F Ecell Step 3: Substitute the values into the equation to solve for n. n= −∆G◦ −∆G◦ = ◦ F Ecell (96485 C/mol)(1.23 V) Step 4: Calculate the standard Gibbs free energy change (∆G◦ ). We know ◦ that ∆G◦ = −nF Ecell and we have found the value for n in the previous step. ◦ ∆G = ◦ −nF Ecell  =− −∆G◦ (96485 C/mol)(1.23 V)  (96485 C/mol)(1.23 V) 1 = ∆G◦ Therefore, ∆G◦ = 1 J. Step 5: Calculate the cell potential when the concentration of Mn2+ ions in the cathode compartment is 1.0 M. Assuming the cell reaction is occurring at equilibrium, we can use the Nernst equation:   0.0592 [Mn2+ ]cathode ◦ E=E − log n [Mn2+ ]anode where [Mn2+ ]cathode is the concentration of Mn2+ ions in the cathode compartment and [Mn2+ ]anode is the concentration of Mn2+ ions in the anode compartment. Step 6: Calculate the cell potential with [Mn2+ ]cathode = 1.0 M and [Mn2+ ]anode = ◦ 0.1 M. Given that Ecell = 1.23 V, n = 2 (as indicated by the balanced redox reaction), and substituting [Mn2+ ]cathode = 1.0 M and [Mn2+ ]anode 13 Question 16 Question Consider a galvanic cell where the following reaction takes place: Zn(s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag(s) Determine the cell potential at 25°C if the standard reduction potentials are E ◦ (Zn2+ /Zn) = −0.76 V and E ◦ (Ag+ /Ag) = 0.80 V. Solution Step 1: Write the half-reactions and find the cell potential. The standard cell ◦ potential (Ecell ) can be calculated using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode We have the half-reactions: Zn2+ (aq) + 2e− → Zn(s) ◦ with EZn 2+ /Zn = −0.76 V 2Ag+ (aq) + 2e− → 2Ag(s) with ◦ EAg + /Ag = 0.80 V The cell potential is: ◦ ◦ ◦ ◦ ◦ Ecell = Ecathode − Eanode = EAg + /Ag − EZn2+ /Zn = 0.80V − (−0.76V ) = 1.56 V Therefore, the cell potential at 25°C is 1.56 V. Question 17 Question Consider a galvanic cell that consists of a standard hydrogen electrode (SHE) and a standard iron (III) ion electrode with an Fe3+ /Fe2+ half-cell. The standard reduction potentials are as follows: E◦ (H+ /H2 ) = 0.00 V and E◦ (Fe3+ /Fe2+ ) = -0.77 V. Calculate the cell potential (E◦cell ) for this galvanic cell. Is the reaction spontaneous or non-spontaneous? Solution Step 1: Write the overall cell reaction as the sum of the two half-cell reactions. Overall reaction: 2H+ + Fe3+ → H2 + Fe2+ Step 2: Write the two half-cell reactions, making sure the number of electrons match to cancel out in the overall reaction. Anode (oxidation): Fe 3+ + e− → Fe2+ 14 E ◦ = -0.77 V + Cathode (reduction): 2H + 2e− → H2 E ◦ = 0.00 V Step 3: Calculate the overall cell potential using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.00 V − (−0.77 V) ◦ Ecell = 0.77 V Step 4: Determine the spontaneity of the reaction based on the cell potential. Since the cell potential is positive (0.77 V), the reaction is spontaneous as the flow of electrons is favored from the anode to the cathode. Question 18 Question A galvanic cell is constructed with a standard hydrogen electrode (Pt|H2 (g)|H+ (1M)) on one side and an iron electrode (Fe3+ (1M)|Fe2+ (1M)) on the other side. The standard reduction potential of the iron electrode is −0.77 V and the standard reduction potential of the hydrogen electrode is 0 V. Determine the cell potential, and identify the cathode and anode in this galvanic cell. Also, determine if the reaction is spontaneous or not. Solution Step 1: Write the half-reactions for the two electrodes. The half-reaction of the hydrogen electrode is the reduction of hydrogen ions to hydrogen gas: 2H+ + 2e− → H2 (g) The half-reaction of the iron electrode is the reduction of iron(III) ions to iron(II) ions: Fe3+ + e− → Fe2+ Step 2: Determine the standard cell potential by finding the difference between the standard reduction potentials of the two half-reactions. The standard ◦ cell potential (Ecell ) can be calculated as: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ where Ecathode is the standard reduction potential of the cathode and Eanode is the standard reduction potential of the anode. ◦ ◦ Given that Ecathode = 0 V and Eanode = −0.77 V, we have: ◦ Ecell = 0 V − (−0.77 V) = 0.77 V Step 3: Identify the cathode and anode in the galvanic cell. The electrode with the higher standard reduction potential is the cathode, while the electrode 15 with the lower standard reduction potential is the anode. So, in this case, the hydrogen electrode is the cathode and the iron electrode is the anode. Step 4: Determine if the reaction is spontaneous or not. The reaction is ◦ ◦ spontaneous if Ecell is positive. Since Ecell = 0.77 V, the reaction is spontaneous. Question 19 Question A cell is constructed with a standard hydrogen electrode (SHE) and a zinc electrode. The standard reduction potential for the hydrogen electrode is 0 V, and for the zinc electrode is -0.76 V. Calculate the standard cell potential for this galvanic cell. Solution Step 1: Write the half-reactions for the hydrogen electrode and zinc electrode. The half-reactions are: 1 H+ (aq) + e− → H2 (g) (Reduction at SHE) 2 Zn2+ (aq) + 2e− → Zn(s) (Reduction at Zn electrode) Step 2: Determine the standard cell potential. The standard cell potential ◦ (Ecell ) is given by 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. In this case, the cathode is the SHE (standard hydrogen electrode) with ◦ ◦ Ecathode = 0 V, and the anode is the zinc electrode with Eanode = −0.76 V. Therefore, the standard cell potential is: ◦ Ecell = 0 V − (−0.76 V) = 0.76 V So, the standard cell potential for this galvanic cell is 0.76 V. Question 20 Question Consider the following cell reaction: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + 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 ) and state whether this cell is galvanic or electrolytic. 16 Solution ◦ To calculate the standard cell potential (Ecell ), we use the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ where Ecathode is the standard reduction potential of the cathode (Cu2+ /Cu) ◦ and Eanode is the standard reduction potential of the anode (Zn2+ /Zn). We are given: ◦ EZn 2+ /Zn = −0.76 V, ◦ ECu 2+ /Cu = 0.34 V Therefore, ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EZn2+ /Zn = 0.34 V − (−0.76 V) = 1.10 V ◦ Since the standard cell potential (Ecell ) is positive, the cell is galvanic. Question 21 Question Consider a galvanic cell that consists of a zinc electrode in a 1.0 M solution of Zn2+ ions and a copper electrode in a 1.0 M solution of Cu2+ ions. The ◦ ◦ standard reduction potentials are EZn 2+ /Zn = −0.76 V and ECu2+ /Cu = 0.34 V, respectively. Determine the cell potential at standard conditions and identify the anode and cathode. Solution Step 1: Write the reduction half-reactions and standard cell potential. The reduction half-reactions are: Zn2+ + 2e− → Zn (oxidation) Cu2+ + 2e− → Cu (reduction) The standard cell potential can be calculated using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EZn2+ /Zn ◦ Ecell = 0.34 V − (−0.76 V) Step 2: Calculate the standard cell potential. ◦ Ecell = 1.10 V Therefore, the standard cell potential is 1.10 V. 17 Step 3: Identify the anode and cathode. Since the cell potential is positive, the reaction is spontaneous, with the copper electrode acting as the cathode (where reduction occurs) and the zinc electrode acting as the anode (where oxidation occurs). Thus, in the given galvanic cell, zinc is the anode and copper is the cathode. Question 22 Question Consider a galvanic cell with a zinc electrode in a 1.0 M Zn2+ solution and a silver electrode in a 1.0 M Ag + solution. The standard reduction potentials ◦ ◦ are EZn 2+ /Zn = −0.76 V and EAg+ /Ag = 0.80 V. Calculate the cell potential at standard conditions. Solution Step 1: Write the two half-reactions representing the oxidation and reduction processes. The oxidation half-reaction occurs at the zinc electrode: Zn(s) → Zn2+ (aq) + 2e− The reduction half-reaction occurs at the silver electrode: Ag+ (aq) + e− → Ag(s) ◦ Step 2: Determine the standard cell potential (Ecell ) using the standard reduction potentials. The cell potential is calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.80 V − (−0.76 V) = 1.56 V Therefore, the cell potential under standard conditions is 1.56 V. Question 23 Question Consider a galvanic cell that consists of a copper electrode in a 1.0 M copper(II) sulfate solution and a platinum electrode in a 1.0 M hydrogen ion solution. Suppose the standard reduction potential for the copper half-cell is +0.34 V, and the standard reduction potential for the hydrogen half-cell is 0.00 V. Determine the cell potential at 25°C for this galvanic cell. 18 Solution Step 1: Write the half-reactions for each electrode based on the given information. Copper electrode: Cu2+ (aq) + 2e− → Cu(s) E ◦ = +0.34 V Platinum electrode: 2H+ (aq) + 2e− → H2 (g) E ◦ = 0.00 V Step 2: Write the overall cell reaction by summing the two half-reactions. Cu2+ (aq) + 2H + (aq) → Cu(s) + H2 (g) Step 3: Determine the cell potential by adding the standard reduction potentials of the two half-cells. ◦ ◦ ◦ Cell potential: Ecell = Ecathode − Eanode = 0.00 V − (+0.34 V) = −0.34 V Therefore, the cell potential at 25°C for this galvanic cell is -0.34 V. Question 24 Question Consider a galvanic cell constructed with a standard hydrogen electrode (SHE) as the anode and a copper electrode as the cathode. If the standard reduction potential of the copper electrode is +0.34 V, determine the cell potential at 25◦ C. Is this cell galvanic or electrolytic? Solution Step 1: Write the two half-reactions for the cell: The half-reaction at the anode involving the standard hydrogen electrode (SHE) is: 2H+ + 2e− → H2 (g) The half-reaction at the cathode involving the copper electrode is: Cu2+ + 2e− → Cu(s) Step 2: Calculate the standard cell potential (E ◦ ) using the standard reduction potentials of the half-reactions: Given that the standard reduction potential of the copper electrode is +0.34 V, and the standard reduction potential of the standard hydrogen electrode (SHE) is 0 V, the cell potential can be calculated as: ◦ ◦ E ◦ = Ecathode − Eanode E ◦ = +0.34 V − 0 V E ◦ = +0.34 V 19 Step 3: Calculate the cell potential at 25◦ C using the Nernst equation: The Nernst equation relates the cell potential (E) to the standard cell potential (E ◦ ), the reaction quotient (Q), the gas constant (R), the temperature in Kelvin (T ), and the number of electrons transferred in the reaction (n): E = E◦ − RT ln(Q) nF At 25◦ C, T = 298 K. Since the reaction quotient Q = 1 for standard conditions, the equation simplifies to: E = E◦ Therefore, the cell potential at 25◦ C is +0.34 V. Step 4: Determine if the cell is galvanic or electrolytic: A cell is galvanic if the cell potential is positive, indicating a spontaneous redox reaction. Since the cell potential at 25◦ C is +0.34 V (positive), the cell is galvanic. Question 25 Question Consider a galvanic cell with the following half-reactions: Cathode: Ag+ (aq) + e− → Ag(s) E ◦ = 0.80 V Anode: Zn(s) → Zn2+ (aq) + 2e− E ◦ = −0.76 V ◦ Calculate the cell potential (Ecell ) and the equilibrium constant (Keq ) for the overall reaction of the galvanic cell. Solution ◦ Step 1: Calculate Ecell by using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode Substitute the given values: ◦ Ecell = 0.80 V − (−0.76 V) ◦ Ecell = 1.56 V Step 2: Calculate Keq using the formula: ◦ Ecell = RT ln Keq nF Solving for Keq : Keq = e ◦ nF Ecell RT 20 Given that R = 8.314 J/mol K, T = 298 K, F = 96485 C/mol, and n = 2 (from balancing the half-reactions), we have: Keq = e 2×96485×1.56 8.314×298 Keq = e7.77 Keq ≈ 2417.89 Question 26 Question Consider a galvanic cell with the following half-reactions: Anode: Cathode: Cd(s) → Cd2+ (aq) + 2e− Ag+ (aq) + e− → Ag(s) Calculate the cell potential at standard conditions (T = 298 K, P = 1 atm), given that [Cd2+ ] = 0.20 M and [Ag+ ] = 0.15 M. Solution Step 1: Write the overall cell reaction by adding the two half-reactions. + Cd(s) + 2Ag (aq) → Cd2+ (aq) + 2Ag(s) Step 2: Write the Nernst equation for cell potential: E = E◦ − RT ln(Q) nF Where: - E is the cell potential, - E ◦ is the standard cell potential, - R is the ideal gas constant (8.314 J/(mol·K)), - T is the temperature in Kelvin (298 K), - n is the number of electrons transferred in the balanced cell reaction, - F is the Faraday constant (96485 C/mol), - Q is the reaction quotient. Step 3: Calculate the standard cell potential E ◦ using the Standard Reduction Potentials table. ◦ ◦ E ◦ = Ecathode − Eanode ◦ ◦ = EAg + /Ag − ECd2+ /Cd = 0.80 V − (−0.40 V) = 1.20 V Step 4: Calculate the reaction quotient Q using the given concentrations. Q= [Cd2+ ] 0.20 = 0.152 [Ag+ ]2 21 Step 5: Substitute all values into the Nernst equation and solve for the cell potential E.   (8.314 J/(mol · K)) · (298 K) 0.20 E = 1.20 V − ln 2 · (96485 C/mol) 0.152   0.20 E = 1.20 V − 0.0268 V · ln 0.152 E ≈ 1.16 V Therefore, the cell potential at standard conditions is approximately 1.16 V. Question 27 Question Consider a galvanic cell with the following half-reactions: Anode: Pb(s) → Pb2+ (aq) + 2e− Cathode: 2Ag+ (aq) + 2e− → 2Ag(s) If the standard reduction potentials are E ◦ (Pb2+ /Pb) = −0.13 V and E ◦ (Ag+ /Ag) = ◦ 0.80 V, calculate the standard cell potential (Ecell ) for this galvanic cell. Solution Step 1: Determine the overall cell reaction by adding the half-reactions. Pb(s) + 2Ag+ (aq) → Pb2+ (aq) + 2Ag(s) ◦ Step 2: Calculate the standard cell potential (Ecell ) using the standard reduction potentials. The standard cell potential is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode From the data given, we have: ◦ Eanode = −0.13 V and ◦ Ecathode = 0.80 V Therefore, ◦ Ecell = 0.80 V − (−0.13 V) = 0.93 V Hence, the standard cell potential for this galvanic cell is 0.93 V. 22 Question 28 Question Consider the reaction between silver ions and copper metal: Ag+ (aq) + Cu(s) → Ag(s) + Cu2+ (aq) Given that the standard reduction potentials are E ◦ (Ag+ /Ag) = 0.80 V and E ◦ (Cu2+ /Cu) = 0.34 V, determine whether the reaction is spontaneous under standard conditions. If not, calculate the voltage that would be required to drive the reaction forward. Solution ◦ Step 1: Calculate the standard cell potential (Ecell ) using the standard reduction potentials. ◦ Ecell = E ◦ (cathode) − E ◦ (anode) ◦ Ecell = E ◦ (Ag+ /Ag) − E ◦ (Cu2+ /Cu) ◦ Ecell = 0.80 V − 0.34 V ◦ Ecell = 0.46 V Step 2: Determine if the reaction is spontaneous under standard conditions. ◦ Since Ecell = 0.46 V, the reaction is spontaneous under standard conditions ◦ because Ecell > 0. Therefore, the given reaction proceeds spontaneously to the right. Step 3: If the reaction is not spontaneous, calculate the voltage required to drive the reaction forward. Since the reaction is spontaneous under standard conditions, we do not need to calculate the additional voltage required in this case. Question 29 Question Consider a galvanic cell with the following half-reactions: Ag+ (aq) + e− → Ag(s) E ◦ = 0.80 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V If the concentration of Ag+ is 1.0 M and Cu2+ is 0.1 M, determine the cell potential for this galvanic cell at 25◦ C. 23 Solution Step 1: Identify the anode and cathode reactions. The half-reaction with the higher standard reduction potential (E ◦ ) will be reduced (cathode) and the other will be oxidized (anode). Here, the reduction potential for the reduction half-reaction of Ag+ is 0.80 V, which is higher than 0.34 V for Cu2+ . Thus, the reduction half-reaction for Ag+ will occur at the cathode and the oxidation half-reaction for Cu2+ will occur at the anode. Step 2: Write the overall cell reaction by adding the anode and cathode half-reactions. Ensure that the same number of electrons are involved in both half-reactions so that they can cancel out. The overall reaction for the galvanic cell is: 2Ag+ (aq) + Cu(s) → 2Ag(s) + Cu2+ (aq) Step 3: Calculate the cell potential, Ecell , using the Nernst equation: Ecell = E ◦ − 0.0592 log(Q) n where: - E ◦ is the standard cell potential, - n is the number of moles of electrons transferred in the balanced cell reaction, - Q is the reaction quotient (ratio of product concentrations to reactant concentrations). In this case, n = 2 since 2 moles of electrons are transferred in the overall reaction. Step 4: Calculate Q using the given concentrations. Remember, the concentrations of solid species do not appear in the expression for Q. Q= [Ag+ ]2 (1.0)2 2+ = 0.1 = 10 [Cu ] Step 5: Substitute the values of E ◦ , n, and Q into the Nernst equation and solve for Ecell . Ecell = 0.80 − 0.0592 log(10) = 0.80 − 0.0296 × 1 = 0.7704 V 2 Therefore, the cell potential for this galvanic cell at 25◦ C is 0.7704 V. Question 30 Question Consider the following Galvanic and Electrolytic cells: Galvanic Cell Zn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu(s) Electrolytic Cell Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s) 24 For each cell, determine the following: 1. The half-reaction that occurs at the cathode and the anode. 2. The overall cell reaction. 3. The direction of electron flow. 4. The direction of cation flow. 5. Whether the cell functions as a galvanic cell or an electrolytic cell. Solution 1. For the Galvanic Cell: ˆ Anode (Zn(s)): Zn(s) → Zn2+ (aq) + 2e− ˆ Cathode (Cu2+ (aq)): Cu2+ (aq) + 2e− → Cu(s) For the Electrolytic Cell: ˆ Anode (Cu(s)): Cu(s) → Cu2+ (aq) + 2e− ˆ Cathode (Ag+ (aq)): Ag+ (aq) + e− → Ag(s) 2. The overall cell reaction for the Galvanic Cell is the sum of the halfreactions: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) The overall cell reaction for the Electrolytic Cell is the sum of the halfreactions: Cu(s) + Ag+ (aq) → Cu2+ (aq) + Ag(s) 3. In both cells, electrons flow from the anode to the cathode. 4. In the Galvanic Cell, cations flow from the anode to the cathode. In the Electrolytic Cell, cations flow from the cathode to the anode. 5. The Galvanic Cell functions as a spontaneous chemical reaction generating electricity, while the Electrolytic Cell requires an external electrical current to drive a non-spontaneous chemical reaction. Question 31 Question A voltaic cell is constructed with a standard hydrogen electrode and a standard silver electrode. The standard cell potential is measured to be 0.80 V at 25°C. Calculate the standard free energy change for the cell reaction. Solution Step 1: Write the balanced cell reaction for the voltaic cell. The cell reaction for a voltaic cell with a standard hydrogen electrode (H+ (aq), H2 (g)) and a standard silver electrode (Ag+ (aq), Ag(s)) is: 2H+ (aq) + 2e− → H2 (g) E ◦ = 0 V 25 2Ag+ (aq) + 2e− → 2Ag(s) E ◦ = 0.80 V Overall cell reaction: 2Ag+ (aq) + 2H+ (aq) → 2Ag(s) + H2 (g) ◦ Step 2: Calculate the standard cell potential (Ecell ) using the standard re◦ duction potentials. The standard cell potential (Ecell ) is the difference between the standard reduction potentials of the two half-reactions: ◦ ◦ ◦ Ecell = Ecathode − Eanode = 0.80 V − 0 V = 0.80 V Step 3: Calculate the standard free energy change (∆G◦ ) using the equation: ◦ ∆G◦ = −nF Ecell where n is the number of moles of electrons transferred in the balanced cell reaction, F is the Faraday constant (96, 485 C/mol). For the given cell reaction, n = 2 moles of electrons transferred. Thus, ∆G◦ = −2 × 96485 C/mol × 0.80 V = −155136 J/mol Therefore, the standard free energy change for the cell reaction is −155136 J/mol. Question 32 Question Consider a galvanic cell consisting of a standard hydrogen electrode (SHE) and a copper electrode with the following half-reactions: Cathode: Cu2+ + 2 e− −−→ Cu Anode: 2 H+ + 2 e− −−→ H2 E◦ = 0.34 V E◦ = 0.00 V Calculate the cell potential at 298 K when the concentration of Cu2+ is 0.10 M, and the pressure of H2 is 1.0 atm. Solution Step 1: Write the overall cell reaction. The overall cell reaction is the sum of the two half-reactions: Cu2+ + 2 H+ −−→ Cu + H2 Step 2: Calculate the cell potential at standard conditions. ◦ The standard cell potential Ecell can be calculated using the standard reduction potentials: ◦ ◦ ◦ Ecell = Ecathode − Eanode 26 ◦ Ecell = 0.34V − 0.00V = 0.34 V Step 3: Consider the non-standard conditions. Since the concentrations of Cu2+ and the pressure of H2 are given, we can calculate the cell potential using the Nernst equation: 0.0592 ◦ Ecell = Ecell − log(Q) n where n is the number of electrons transferred in the cell reaction and Q is the reaction quotient. Step 4: Calculate the reaction quotient Q. [Cu] Q= 2+ [Cu ][H+ ]2 1.0 Q= = 10 0.10 × 1.02 Step 5: Substitute values into the Nernst equation and solve for the cell potential. 0.0592 Ecell = 0.34 − log(10) 2 Ecell = 0.34 − 0.0296 × 1 Ecell = 0.3104 V Therefore, the cell potential at 298 K when the concentration of Cu2+ is 0.10 M and the pressure of H2 is 1.0 atm is 0.3104 V. Question 33 Question Consider a galvanic cell with a standard cell potential of E ◦ = 0.78 V that operates under standard conditions. The cell reaction is given by: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Calculate the standard Gibbs free energy change for the cell reaction and determine if the reaction is spontaneous. Solution Step 1: The standard Gibbs free energy change (∆G◦ ) for the cell reaction can be calculated using the equation: ∆G◦ = −nF E ◦ where n is the number of moles of electrons transferred in the balanced redox equation, and F is the Faraday constant (96500 C/mol). From the given cell reaction, we see that 2 moles of electrons are transferred, so n = 2. Substituting into the equation: ∆G◦ = −2 × 96500 C/mol × 0.78 V 27 ∆G◦ = −188220 J/mol Step 2: To determine if the reaction is spontaneous, we can use the relationship between ∆G◦ and the equilibrium constant (K): ∆G◦ = −RT ln K where R is the gas constant (8.314 J/mol · K) and T is the temperature in Kelvin. At standard conditions, T = 298 K and R = 8.314 J/mol · K. Thus, we have: −188220 J/mol = −8.314 J/mol · K × 298 K ln K Solving for ln K: ln K = −188220 J/mol −8.314 J/mol · K × 298 K ln K = 27.37 Step 3: Exponentiating both sides of the equation to eliminate the natural logarithm: K = e27.37 K = 1.17 × 1011 Since K > 1, the reaction is spontaneous. Question 34 Question ◦ A student sets up a galvanic cell with a standard hydrogen electrode (EH + /H = 2 ◦ 0.00 V) and a copper electrode (ECu2+ /Cu = 0.34 V), connected by a salt bridge. The initial concentrations are [H+ ] = 1.0 M and [Cu2+ ] = 0.1 M. Calculate the cell potential at 25 Step 1: Write the balanced redox reaction and the cell notation. The balanced redox reaction is: Cu2+ (aq) + 2H+ (aq) → Cu(s) + H2 (g) The cell notation is: Pt|H2 (1 M)|H+ (1 M)||Cu2+ (0.1 M)|Cu ◦ Step 2: Calculate the cell potential (Ecell ) at standard conditions. 28 The standard cell potential can be calculated using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ Given that EH + /H = 0.00 V, and ECu2+ /Cu = 0.34 V, we have: 2 ◦ Ecell = 0.00 V − 0.34 V = −0.34 V Step 3: Calculate the reaction quotient (Q). The reaction quotient Q is calculated using the concentrations of the species involved in the reaction: [Cu](solid) Q= [Cu2+ ][H+ ]2 Plugging in the values, we get: Q= 1 = 10 0.1 × 12 Step 4: Calculate the cell potential (Ecell ) under non-standard conditions. The Nernst equation relates the cell potential under non-standard conditions ◦ (Ecell ) to the standard cell potential (Ecell ) and the reaction quotient (Q): ◦ Ecell = Ecell − 0.0592 log Q n Since the reaction involves the transfer of 2 moles of electrons, n = 2. Plugging in the values, we get: Ecell = −0.34 V − 0.0592 log 10 2 Ecell = −0.34 V − 0.0296 log 10 Ecell = −0.34 V − 0.0296 × 1 Ecell = −0.34 V − 0.0296 Ecell = −0.3696 V Therefore, the cell potential at 25 Question Consider a galvanic cell that uses the following half-reactions: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) If the standard reduction potential of Cu2+ /Cu is 0.34 V, calculate the stan◦ dard cell potential (Ecell ) of this galvanic cell. 29 Question 3 Question ◦ Consider a galvanic cell with a standard reduction potential of Ecell = 0.30 V. If the cell potential is measured to be 0.45 V under non-standard conditions where the concentration of Cu2+ ions in the cathode compartment is 0.10 M, determine the concentration of Cu2+ ions in the anode compartment. Solution Step 1: Write the half-reactions for the cell. The overall cell reaction can be split into two half-reactions: Cathode (Reduction): Cu2+ (aq) + 2e− → Cu(s) Anode (Oxidation): Zn(s) → Zn2+ (aq) + 2e− Step 2: Calculate the cell potential under non-standard conditions. By using the Nernst equation:   [Cu2+ ]anode 0.0592 ◦ log Ecell = Ecell − 2 [Cu2+ ]cathode Substitute the given values: 0.45 = 0.30 − 0.0592 log 2  [Cu2+ ]anode 0.10  Step 3: Solve for the concentration of Cu2+ ions in the anode compartment. Simplify the equation:   [Cu2+ ]anode 0.15 = −0.0296 log 0.10   2+ [Cu ]anode 5.07 = log 0.10 105.07 = [Cu2+ ]anode [Cu2+ ]anode 0.10 = 105.07 × 0.10 [Cu2+ ]anode ≈ 1.23 × 105 M Therefore, the concentration of Cu2+ ions in the anode compartment is approximately 1.23 × 105 M. Question 4 Question Consider a galvanic cell that consists of a copper electrode in a Cu2+ solution connected by a salt bridge to a silver electrode in a Ag + solution. The standard ◦ ◦ reduction potentials are ECu 2+ /Cu = 0.34 V and EAg + /Ag = 0.80 V. Calculate the cell potential for this galvanic cell. 3 Solution Step 1: Write the half-reactions for the redox reactions occurring at each electrode. Reduction at the copper electrode: Cu2+ + 2e− → Cu Reduction at the silver electrode: Ag + + e− → Ag E ◦ = 0.34 V E ◦ = 0.80 V Step 2: Write the overall cell reaction by summing the two half-reactions. The cell potential is the difference in standard electrode potentials. Cu2+ + 2Ag → Cu + 2Ag + ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.80 V − 0.34 V ◦ Ecell = 0.46 V Therefore, the cell potential for this galvanic cell is 0.46 V. Question 5 Question Consider a galvanic cell with the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V Calculate the cell potential at 25°C when the concentrations of Zn2+ and Cu2+ are both 0.10 M. Assume both half-reactions are at standard conditions. Solution Step 1: Write the overall cell reaction by subtracting the reduction potential of the Zn half-reaction from the reduction potential of the Cu half-reaction: Zn2+ (aq) + Cu(s) → Cu2+ (aq) + Zn(s) ◦ Step 2: Determine the standard cell potential (Ecell ) using the Nernst equation: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = (0.34 V) − (−0.76 V) ◦ Ecell = 1.10 V Step 3: Calculate the cell potential (Ecell ) at 25°C using the Nernst equation:   0.0592 [Cu2+ ] ◦ Ecell = Ecell − log n [Zn2+ ] 4 Step 4: Calculate the cell potential by plugging in the given values:   0.0592 0.10 Ecell = 1.10 V − log 2 0.10 Ecell = 1.10 V Therefore, the cell potential at 25°C when the concentrations of Zn2+ and Cu2+ are both 0.10 M is 1.10 V . Question 6 Question Consider a galvanic cell based on the following half-reactions: Anode: Zn(s) → Zn2+ (aq) + 2e− Cathode: 2H+ (aq) + 2e− → H2 (g) Calculate the standard cell potential at 25°C for this galvanic cell. Solution Step 1: Write the cell reaction by reversing the anode half-reaction and adding the cathode half-reaction. Zn2+ (aq) + 2H+ (aq) → Zn(s) + H2 (g) Step 2: Write the standard cell potential in terms of standard reduction potentials (E ◦ ) for the half-reactions involved. The standard cell potential is given by ◦ ◦ ◦ Ecell = Ecathode − Eanode Step 3: Find the standard reduction potentials for the half-reactions from a standard reduction potential table. The values are: ◦ EZn 2+ /Zn = −0.76 V ◦ E2H + /H = 0.00 V 2 Step 4: Substitute the values into the formula for the standard cell potential. ◦ Ecell = 0.00 V − (−0.76 V) ◦ Ecell = 0.76 V Therefore, the standard cell potential at 25°C for the given galvanic cell is 0.76 V. 5 Question 7 Question Consider a galvanic cell with the following half-reactions: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) ◦ ◦ If the standard reduction potentials are ECu 2+ /Cu = 0.34 V and EZn2+ /Zn = ◦ −0.76 V, calculate the standard cell potential (Ecell ) of the galvanic cell. Solution Step 1: Write the overall cell reaction by combining the two half-reactions, ensuring that the number of electrons lost equals the number of electrons gained: Zn(s) + Cu2+ (aq) −−→ Zn2+ (aq) + Cu(s) ◦ ) using the standard reStep 2: Calculate the standard cell potential (Ecell duction potentials of the half-reactions. The standard cell potential is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ Substitute the values of ECu 2+ /Cu = 0.34 V and EZn2+ /Zn = −0.76 V into the equation: ◦ Ecell = 0.34 V − (−0.76 V) = 1.10 V ◦ = 1.10 V. Therefore, the standard cell potential of the galvanic cell is Ecell Question 8 Question Consider an electrochemical cell consisting of a copper electrode immersed in 1.0 M Cu2+ solution and a silver electrode immersed in 1.0 M Ag+ solution. Write the balanced half-reactions and the overall cell reaction. Is this cell a galvanic cell or an electrolytic cell? Justify your answer. Solution Step 1: Write the balanced half-reactions: At the copper electrode: Cu2+ + 2e− → Cu At the silver electrode: Ag+ + e− → Ag Step 2: Write the overall cell reaction: Cu2+ + 2Ag → Cu + 2Ag+ 6 Step 3: Determine if the cell is a galvanic or electrolytic cell: In a galvanic cell, the redox reaction is spontaneous and produces electrical energy. In this case, the cell reaction is proceeding from left to right (oxidation at copper electrode, reduction at silver electrode), which means it is a spontaneous process. Therefore, this cell is a galvanic cell. Question 9 Question Consider a galvanic cell constructed with a silver electrode in a 1.0 M AgNO3 solution and a copper electrode in a 1.0 M Cu(NO3 )2 solution. The standard reduction potentials are: Ag+ (aq) + e− → Ag(s) (E ◦ = 0.80 V) and Cu2+ (aq) + 2e− → Cu(s) (E ◦ = 0.34 V). Determine the cell potential at 25°C. Solution Step 1: Write out the overall reaction for the galvanic cell: Ag(s) + Cu2+ (aq) → Cu(s) + Ag+ (aq) ◦ Step 2: Determine the standard cell potential (Ecell ) using the standard reduction potentials provided: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EAg+ /Ag ◦ Ecell = 0.34 V − 0.80 V ◦ Ecell = −0.46 V Step 3: Calculate the non-standard cell potential using the Nernst equation: E = E◦ − 0.0592 log(Q) n where n is the number of moles of electrons transferred and Q is the reaction quotient. Step 4: Calculate the reaction quotient Q: Q= [Cu][Ag+ ] [products] = [reactants] [Ag][Cu2+ ] Since both concentrations are 1.0 M, Q = 1. Step 5: Substitute the values into the Nernst equation: E = −0.46 V − 0.0592 log(1) 2 E = −0.46 V Therefore, the cell potential at 25°C for the given galvanic cell is −0.46 V. 7 Question 10 Question Consider a Galvanic cell with the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V Pb2+ (aq) + 2e− → Pb(s) E ◦ = −0.13 V If the initial concentrations are [Zn2+ ] = 1.0 M, [Pb2+ ] = 0.1 M, and the cell operates until equilibrium is reached, determine the cell potential at equilibrium. Solution Step 1: Write the cell reaction using the given half-reactions: Zn2+ (aq) + Pb(s) → Zn(s) + Pb2+ (aq) ◦ Step 2: Calculate the standard cell potential (Ecell ) using the standard reduction potentials of the half-reactions: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = −0.13 V − (−0.76 V) = 0.63 V Step 3: Calculate the reaction quotient (Q) at equilibrium using the initial concentrations given: [Zn](Pb2+ ) Q= [Zn2+ ][Pb] Q= (1.0)(0.1) =1 (1.0)(0.1) Step 4: Use the Nernst equation to find the cell potential at equilibrium (Ecell ): 0.0592 ◦ Ecell = Ecell − log(Q) n Since the reaction involves the transfer of 2 electrons, n = 2. Ecell = 0.63 V − 0.0592 log(1) = 0.63 V 2 Therefore, the cell potential at equilibrium is 0.63 V. 8 Question 11 Question Consider the following half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V Determine if a spontaneous reaction will occur when a strip of copper metal is placed in a solution of zinc sulfate. If so, write the overall balanced cell reaction and calculate the cell potential. Solution Step 1: Write the individual half-reactions and find the standard cell potential. Given the half-reactions: Zn2+ (aq) + 2e− → Zn(s) E ◦ = −0.76 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V The overall cell reaction will involve the reduction of Cu2+ and the oxidation of Zn. To determine if a spontaneous reaction will occur, we calculate the overall ◦ standard cell potential, Ecell , using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode where ◦ Ecathode = 0.34 V ◦ Eanode = −0.76 V Plugging these values into the formula, we get: ◦ Ecell = 0.34 V − (−0.76 V) = 1.10 V Since the standard cell potential is positive, a spontaneous reaction will occur. Step 2: Write the overall balanced cell reaction. The overall cell reaction is the sum of the two half-reactions. Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Step 3: Calculate the cell potential. The standard cell potential was already calculated as 1.10 V in Step 1. 9 Question 12 Question A voltaic cell is constructed using a silver/silver chloride electrode and a zinc electrode. The standard reduction potentials are as follows: ◦ AgCl(s) + e− → Ag(s) + Cl− (aq) Ecell = 0.22 V ◦ Zn2+ (aq) + 2e− → Zn(s) Ecell = −0.76 V What is the cell potential when the concentration of Zn2+ is 1.0 M and the solid AgCl is in contact with a 0.10 M AgNO3 solution? Use R = 8.314 J/K mol, F = 96, 485 C/mol, and T = 298 K. Solution Step 1: Write the cell reaction and cell potential equation using the given halfreactions. The cell reaction is the sum of the two half-reactions: AgCl(s) + Zn2+ (aq) + 2e− → Ag(s) + Zn(s) + Cl− (aq) The cell potential equation is: 0.0592 log Q n ◦ Ecell = Ecell − where n is the number of moles of electrons transferred in the balanced reaction, and Q is the reaction quotient. Step 2: Calculate the cell potential First, calculate Q using the concentrations provided. Since there are no concentrations for the solid electrodes (Ag and Zn), their activities are assumed to be 1. Therefore, Q= [Ag+ ][Cl− ] [Zn2+ ] 0.102 = 0.01 1.0 Step 3: Plug the values into the formula Q= Ecell = 0.22V − 0.0592 log 0.01 2 Ecell = 0.22V − 0.0296 × 2 × 2 Ecell = 0.22V − 0.1184 = 0.1016 V Therefore, the cell potential is 0.1016 V. 10 Question 13 Question Consider a galvanic cell where the reaction occurring is: Cd2+ (aq) + 2Ag(s) → Cd(s) + 2Ag+ (aq) ◦ Determine the cell potential at standard conditions given that ECd 2+ /Cd = ◦ −0.40 V and EAg+ /Ag = 0.80 V. Solution Step 1: Write the half-reactions and standard electrode potentials. The halfreactions involved in the cell are: Reduction half-reaction: ◦ Cd2+ (aq) + 2e− → Cd(s) ECd 2+ /Cd = −0.40 V Oxidation half-reaction: 2Ag(s) → 2Ag+ (aq) + 2e− ◦ EAg + /Ag = 0.80 V ◦ Step 2: Calculate the cell potential. 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 that ECd 2+ /Cd = −0.40 V and EAg+ /Ag = 0.80 V, we have: ◦ ◦ ◦ Ecell = EAg + /Ag − ECd2+ /Cd = 0.80 − (−0.40) = 1.20 V Therefore, the cell potential at standard conditions is 1.20 V. Question 14 Question Consider a galvanic cell with the following half-reactions: Cu2+ + 2e− → Cu Fe3+ + e− → Fe2+ ◦ Ecell = 0.34 V ◦ Ecell = −0.77 V Calculate the standard Gibbs free energy change (∆G◦ ) for the overall cell reaction and state whether the cell is spontaneous at standard conditions. 11 Solution ◦ Step 1: Calculate the standard cell potential (Ecell ) for the overall reaction. The overall cell reaction is the sum of the two half-reactions: Cu2+ + 2Fe3+ → 2Fe2+ + Cu The standard cell potential is the sum of the standard reduction potentials: ◦ ◦ ◦ Ecell = Ecathode − Eanode Substitute the given values: ◦ ◦ ◦ Ecell = ECu − EFe ◦ Ecell = 0.34 V − (−0.77 V) = 1.11 V Step 2: Calculate the standard Gibbs free energy change (∆G◦ ) for the overall reaction. The relationship between standard cell potential and standard Gibbs free energy change is given by: ◦ ∆G◦ = −nF Ecell where n is the number of moles of electrons transferred in the balanced cell reaction, and F is the Faraday constant (F = 96485 C/mol). Since 2 moles of electrons are transferred in the balanced cell reaction, n = 2. Substitute the values: ∆G◦ = −2 × 96485 C/mol × 1.11 V ∆G◦ = −213.59 kJ/mol Step 3: Determine if the cell is spontaneous at standard conditions. For a reaction to be spontaneous at standard conditions, ∆G◦ must be negative. Since ∆G◦ = −213.59 kJ/mol (which is negative), the cell reaction is spontaneous at standard conditions. Question 15 Question ◦ Consider a galvanic cell with a standard reduction potential of Ecell = 1.23 V. Assuming the reaction in the cell is spontaneous, calculate the standard Gibbs free energy change, ∆G◦ , for the cell reaction. Also, if the concentration of Mn2+ ions in the cell’s cathode compartment is 0.1 M, determine the cell potential when the concentration of Mn2+ ions in the cathode compartment is increased to 1.0 M. 12 Solution Step 1: Recall the relationship between standard Gibbs free energy change, standard cell potential, and the Faraday constant: ◦ ∆G◦ = −nF Ecell where n is the number of moles of electrons transferred in the balanced redox ◦ reaction, F is the Faraday constant (96485 C/mol), and Ecell is the standard cell potential. Step 2: Calculate the number of moles of electrons transferred. The stan◦ dard reduction potential (Ecell ) is related to the number of moles of electrons transferred (n) by the equation: ◦ Ecell = ∆G◦ −nF ◦ Given Ecell = 1.23 V, and F = 96485 C/mol, we can solve for n. ◦ ∆G◦ = −nF Ecell n= −∆G◦ ◦ F Ecell Step 3: Substitute the values into the equation to solve for n. n= −∆G◦ −∆G◦ = ◦ F Ecell (96485 C/mol)(1.23 V) Step 4: Calculate the standard Gibbs free energy change (∆G◦ ). We know ◦ that ∆G◦ = −nF Ecell and we have found the value for n in the previous step. ◦ ∆G = ◦ −nF Ecell  =− −∆G◦ (96485 C/mol)(1.23 V)  (96485 C/mol)(1.23 V) 1 = ∆G◦ Therefore, ∆G◦ = 1 J. Step 5: Calculate the cell potential when the concentration of Mn2+ ions in the cathode compartment is 1.0 M. Assuming the cell reaction is occurring at equilibrium, we can use the Nernst equation:   0.0592 [Mn2+ ]cathode ◦ E=E − log n [Mn2+ ]anode where [Mn2+ ]cathode is the concentration of Mn2+ ions in the cathode compartment and [Mn2+ ]anode is the concentration of Mn2+ ions in the anode compartment. Step 6: Calculate the cell potential with [Mn2+ ]cathode = 1.0 M and [Mn2+ ]anode = ◦ 0.1 M. Given that Ecell = 1.23 V, n = 2 (as indicated by the balanced redox reaction), and substituting [Mn2+ ]cathode = 1.0 M and [Mn2+ ]anode 13 Question 16 Question Consider a galvanic cell where the following reaction takes place: Zn(s) + 2Ag+ (aq) → Zn2+ (aq) + 2Ag(s) Determine the cell potential at 25°C if the standard reduction potentials are E ◦ (Zn2+ /Zn) = −0.76 V and E ◦ (Ag+ /Ag) = 0.80 V. Solution Step 1: Write the half-reactions and find the cell potential. The standard cell ◦ potential (Ecell ) can be calculated using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode We have the half-reactions: Zn2+ (aq) + 2e− → Zn(s) ◦ with EZn 2+ /Zn = −0.76 V 2Ag+ (aq) + 2e− → 2Ag(s) with ◦ EAg + /Ag = 0.80 V The cell potential is: ◦ ◦ ◦ ◦ ◦ Ecell = Ecathode − Eanode = EAg + /Ag − EZn2+ /Zn = 0.80V − (−0.76V ) = 1.56 V Therefore, the cell potential at 25°C is 1.56 V. Question 17 Question Consider a galvanic cell that consists of a standard hydrogen electrode (SHE) and a standard iron (III) ion electrode with an Fe3+ /Fe2+ half-cell. The standard reduction potentials are as follows: E◦ (H+ /H2 ) = 0.00 V and E◦ (Fe3+ /Fe2+ ) = -0.77 V. Calculate the cell potential (E◦cell ) for this galvanic cell. Is the reaction spontaneous or non-spontaneous? Solution Step 1: Write the overall cell reaction as the sum of the two half-cell reactions. Overall reaction: 2H+ + Fe3+ → H2 + Fe2+ Step 2: Write the two half-cell reactions, making sure the number of electrons match to cancel out in the overall reaction. Anode (oxidation): Fe 3+ + e− → Fe2+ 14 E ◦ = -0.77 V + Cathode (reduction): 2H + 2e− → H2 E ◦ = 0.00 V Step 3: Calculate the overall cell potential using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.00 V − (−0.77 V) ◦ Ecell = 0.77 V Step 4: Determine the spontaneity of the reaction based on the cell potential. Since the cell potential is positive (0.77 V), the reaction is spontaneous as the flow of electrons is favored from the anode to the cathode. Question 18 Question A galvanic cell is constructed with a standard hydrogen electrode (Pt|H2 (g)|H+ (1M)) on one side and an iron electrode (Fe3+ (1M)|Fe2+ (1M)) on the other side. The standard reduction potential of the iron electrode is −0.77 V and the standard reduction potential of the hydrogen electrode is 0 V. Determine the cell potential, and identify the cathode and anode in this galvanic cell. Also, determine if the reaction is spontaneous or not. Solution Step 1: Write the half-reactions for the two electrodes. The half-reaction of the hydrogen electrode is the reduction of hydrogen ions to hydrogen gas: 2H+ + 2e− → H2 (g) The half-reaction of the iron electrode is the reduction of iron(III) ions to iron(II) ions: Fe3+ + e− → Fe2+ Step 2: Determine the standard cell potential by finding the difference between the standard reduction potentials of the two half-reactions. The standard ◦ cell potential (Ecell ) can be calculated as: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ where Ecathode is the standard reduction potential of the cathode and Eanode is the standard reduction potential of the anode. ◦ ◦ Given that Ecathode = 0 V and Eanode = −0.77 V, we have: ◦ Ecell = 0 V − (−0.77 V) = 0.77 V Step 3: Identify the cathode and anode in the galvanic cell. The electrode with the higher standard reduction potential is the cathode, while the electrode 15 with the lower standard reduction potential is the anode. So, in this case, the hydrogen electrode is the cathode and the iron electrode is the anode. Step 4: Determine if the reaction is spontaneous or not. The reaction is ◦ ◦ spontaneous if Ecell is positive. Since Ecell = 0.77 V, the reaction is spontaneous. Question 19 Question A cell is constructed with a standard hydrogen electrode (SHE) and a zinc electrode. The standard reduction potential for the hydrogen electrode is 0 V, and for the zinc electrode is -0.76 V. Calculate the standard cell potential for this galvanic cell. Solution Step 1: Write the half-reactions for the hydrogen electrode and zinc electrode. The half-reactions are: 1 H+ (aq) + e− → H2 (g) (Reduction at SHE) 2 Zn2+ (aq) + 2e− → Zn(s) (Reduction at Zn electrode) Step 2: Determine the standard cell potential. The standard cell potential ◦ (Ecell ) is given by 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. In this case, the cathode is the SHE (standard hydrogen electrode) with ◦ ◦ Ecathode = 0 V, and the anode is the zinc electrode with Eanode = −0.76 V. Therefore, the standard cell potential is: ◦ Ecell = 0 V − (−0.76 V) = 0.76 V So, the standard cell potential for this galvanic cell is 0.76 V. Question 20 Question Consider the following cell reaction: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + 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 ) and state whether this cell is galvanic or electrolytic. 16 Solution ◦ To calculate the standard cell potential (Ecell ), we use the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ where Ecathode is the standard reduction potential of the cathode (Cu2+ /Cu) ◦ and Eanode is the standard reduction potential of the anode (Zn2+ /Zn). We are given: ◦ EZn 2+ /Zn = −0.76 V, ◦ ECu 2+ /Cu = 0.34 V Therefore, ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EZn2+ /Zn = 0.34 V − (−0.76 V) = 1.10 V ◦ Since the standard cell potential (Ecell ) is positive, the cell is galvanic. Question 21 Question Consider a galvanic cell that consists of a zinc electrode in a 1.0 M solution of Zn2+ ions and a copper electrode in a 1.0 M solution of Cu2+ ions. The ◦ ◦ standard reduction potentials are EZn 2+ /Zn = −0.76 V and ECu2+ /Cu = 0.34 V, respectively. Determine the cell potential at standard conditions and identify the anode and cathode. Solution Step 1: Write the reduction half-reactions and standard cell potential. The reduction half-reactions are: Zn2+ + 2e− → Zn (oxidation) Cu2+ + 2e− → Cu (reduction) The standard cell potential can be calculated using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ ◦ Ecell = ECu 2+ /Cu − EZn2+ /Zn ◦ Ecell = 0.34 V − (−0.76 V) Step 2: Calculate the standard cell potential. ◦ Ecell = 1.10 V Therefore, the standard cell potential is 1.10 V. 17 Step 3: Identify the anode and cathode. Since the cell potential is positive, the reaction is spontaneous, with the copper electrode acting as the cathode (where reduction occurs) and the zinc electrode acting as the anode (where oxidation occurs). Thus, in the given galvanic cell, zinc is the anode and copper is the cathode. Question 22 Question Consider a galvanic cell with a zinc electrode in a 1.0 M Zn2+ solution and a silver electrode in a 1.0 M Ag + solution. The standard reduction potentials ◦ ◦ are EZn 2+ /Zn = −0.76 V and EAg+ /Ag = 0.80 V. Calculate the cell potential at standard conditions. Solution Step 1: Write the two half-reactions representing the oxidation and reduction processes. The oxidation half-reaction occurs at the zinc electrode: Zn(s) → Zn2+ (aq) + 2e− The reduction half-reaction occurs at the silver electrode: Ag+ (aq) + e− → Ag(s) ◦ Step 2: Determine the standard cell potential (Ecell ) using the standard reduction potentials. The cell potential is calculated by subtracting the standard reduction potential of the anode from the standard reduction potential of the cathode: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ Ecell = 0.80 V − (−0.76 V) = 1.56 V Therefore, the cell potential under standard conditions is 1.56 V. Question 23 Question Consider a galvanic cell that consists of a copper electrode in a 1.0 M copper(II) sulfate solution and a platinum electrode in a 1.0 M hydrogen ion solution. Suppose the standard reduction potential for the copper half-cell is +0.34 V, and the standard reduction potential for the hydrogen half-cell is 0.00 V. Determine the cell potential at 25°C for this galvanic cell. 18 Solution Step 1: Write the half-reactions for each electrode based on the given information. Copper electrode: Cu2+ (aq) + 2e− → Cu(s) E ◦ = +0.34 V Platinum electrode: 2H+ (aq) + 2e− → H2 (g) E ◦ = 0.00 V Step 2: Write the overall cell reaction by summing the two half-reactions. Cu2+ (aq) + 2H + (aq) → Cu(s) + H2 (g) Step 3: Determine the cell potential by adding the standard reduction potentials of the two half-cells. ◦ ◦ ◦ Cell potential: Ecell = Ecathode − Eanode = 0.00 V − (+0.34 V) = −0.34 V Therefore, the cell potential at 25°C for this galvanic cell is -0.34 V. Question 24 Question Consider a galvanic cell constructed with a standard hydrogen electrode (SHE) as the anode and a copper electrode as the cathode. If the standard reduction potential of the copper electrode is +0.34 V, determine the cell potential at 25◦ C. Is this cell galvanic or electrolytic? Solution Step 1: Write the two half-reactions for the cell: The half-reaction at the anode involving the standard hydrogen electrode (SHE) is: 2H+ + 2e− → H2 (g) The half-reaction at the cathode involving the copper electrode is: Cu2+ + 2e− → Cu(s) Step 2: Calculate the standard cell potential (E ◦ ) using the standard reduction potentials of the half-reactions: Given that the standard reduction potential of the copper electrode is +0.34 V, and the standard reduction potential of the standard hydrogen electrode (SHE) is 0 V, the cell potential can be calculated as: ◦ ◦ E ◦ = Ecathode − Eanode E ◦ = +0.34 V − 0 V E ◦ = +0.34 V 19 Step 3: Calculate the cell potential at 25◦ C using the Nernst equation: The Nernst equation relates the cell potential (E) to the standard cell potential (E ◦ ), the reaction quotient (Q), the gas constant (R), the temperature in Kelvin (T ), and the number of electrons transferred in the reaction (n): E = E◦ − RT ln(Q) nF At 25◦ C, T = 298 K. Since the reaction quotient Q = 1 for standard conditions, the equation simplifies to: E = E◦ Therefore, the cell potential at 25◦ C is +0.34 V. Step 4: Determine if the cell is galvanic or electrolytic: A cell is galvanic if the cell potential is positive, indicating a spontaneous redox reaction. Since the cell potential at 25◦ C is +0.34 V (positive), the cell is galvanic. Question 25 Question Consider a galvanic cell with the following half-reactions: Cathode: Ag+ (aq) + e− → Ag(s) E ◦ = 0.80 V Anode: Zn(s) → Zn2+ (aq) + 2e− E ◦ = −0.76 V ◦ Calculate the cell potential (Ecell ) and the equilibrium constant (Keq ) for the overall reaction of the galvanic cell. Solution ◦ Step 1: Calculate Ecell by using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode Substitute the given values: ◦ Ecell = 0.80 V − (−0.76 V) ◦ Ecell = 1.56 V Step 2: Calculate Keq using the formula: ◦ Ecell = RT ln Keq nF Solving for Keq : Keq = e ◦ nF Ecell RT 20 Given that R = 8.314 J/mol K, T = 298 K, F = 96485 C/mol, and n = 2 (from balancing the half-reactions), we have: Keq = e 2×96485×1.56 8.314×298 Keq = e7.77 Keq ≈ 2417.89 Question 26 Question Consider a galvanic cell with the following half-reactions: Anode: Cathode: Cd(s) → Cd2+ (aq) + 2e− Ag+ (aq) + e− → Ag(s) Calculate the cell potential at standard conditions (T = 298 K, P = 1 atm), given that [Cd2+ ] = 0.20 M and [Ag+ ] = 0.15 M. Solution Step 1: Write the overall cell reaction by adding the two half-reactions. + Cd(s) + 2Ag (aq) → Cd2+ (aq) + 2Ag(s) Step 2: Write the Nernst equation for cell potential: E = E◦ − RT ln(Q) nF Where: - E is the cell potential, - E ◦ is the standard cell potential, - R is the ideal gas constant (8.314 J/(mol·K)), - T is the temperature in Kelvin (298 K), - n is the number of electrons transferred in the balanced cell reaction, - F is the Faraday constant (96485 C/mol), - Q is the reaction quotient. Step 3: Calculate the standard cell potential E ◦ using the Standard Reduction Potentials table. ◦ ◦ E ◦ = Ecathode − Eanode ◦ ◦ = EAg + /Ag − ECd2+ /Cd = 0.80 V − (−0.40 V) = 1.20 V Step 4: Calculate the reaction quotient Q using the given concentrations. Q= [Cd2+ ] 0.20 = 0.152 [Ag+ ]2 21 Step 5: Substitute all values into the Nernst equation and solve for the cell potential E.   (8.314 J/(mol · K)) · (298 K) 0.20 E = 1.20 V − ln 2 · (96485 C/mol) 0.152   0.20 E = 1.20 V − 0.0268 V · ln 0.152 E ≈ 1.16 V Therefore, the cell potential at standard conditions is approximately 1.16 V. Question 27 Question Consider a galvanic cell with the following half-reactions: Anode: Pb(s) → Pb2+ (aq) + 2e− Cathode: 2Ag+ (aq) + 2e− → 2Ag(s) If the standard reduction potentials are E ◦ (Pb2+ /Pb) = −0.13 V and E ◦ (Ag+ /Ag) = ◦ 0.80 V, calculate the standard cell potential (Ecell ) for this galvanic cell. Solution Step 1: Determine the overall cell reaction by adding the half-reactions. Pb(s) + 2Ag+ (aq) → Pb2+ (aq) + 2Ag(s) ◦ Step 2: Calculate the standard cell potential (Ecell ) using the standard reduction potentials. The standard cell potential is given by: ◦ ◦ ◦ Ecell = Ecathode − Eanode From the data given, we have: ◦ Eanode = −0.13 V and ◦ Ecathode = 0.80 V Therefore, ◦ Ecell = 0.80 V − (−0.13 V) = 0.93 V Hence, the standard cell potential for this galvanic cell is 0.93 V. 22 Question 28 Question Consider the reaction between silver ions and copper metal: Ag+ (aq) + Cu(s) → Ag(s) + Cu2+ (aq) Given that the standard reduction potentials are E ◦ (Ag+ /Ag) = 0.80 V and E ◦ (Cu2+ /Cu) = 0.34 V, determine whether the reaction is spontaneous under standard conditions. If not, calculate the voltage that would be required to drive the reaction forward. Solution ◦ Step 1: Calculate the standard cell potential (Ecell ) using the standard reduction potentials. ◦ Ecell = E ◦ (cathode) − E ◦ (anode) ◦ Ecell = E ◦ (Ag+ /Ag) − E ◦ (Cu2+ /Cu) ◦ Ecell = 0.80 V − 0.34 V ◦ Ecell = 0.46 V Step 2: Determine if the reaction is spontaneous under standard conditions. ◦ Since Ecell = 0.46 V, the reaction is spontaneous under standard conditions ◦ because Ecell > 0. Therefore, the given reaction proceeds spontaneously to the right. Step 3: If the reaction is not spontaneous, calculate the voltage required to drive the reaction forward. Since the reaction is spontaneous under standard conditions, we do not need to calculate the additional voltage required in this case. Question 29 Question Consider a galvanic cell with the following half-reactions: Ag+ (aq) + e− → Ag(s) E ◦ = 0.80 V Cu2+ (aq) + 2e− → Cu(s) E ◦ = 0.34 V If the concentration of Ag+ is 1.0 M and Cu2+ is 0.1 M, determine the cell potential for this galvanic cell at 25◦ C. 23 Solution Step 1: Identify the anode and cathode reactions. The half-reaction with the higher standard reduction potential (E ◦ ) will be reduced (cathode) and the other will be oxidized (anode). Here, the reduction potential for the reduction half-reaction of Ag+ is 0.80 V, which is higher than 0.34 V for Cu2+ . Thus, the reduction half-reaction for Ag+ will occur at the cathode and the oxidation half-reaction for Cu2+ will occur at the anode. Step 2: Write the overall cell reaction by adding the anode and cathode half-reactions. Ensure that the same number of electrons are involved in both half-reactions so that they can cancel out. The overall reaction for the galvanic cell is: 2Ag+ (aq) + Cu(s) → 2Ag(s) + Cu2+ (aq) Step 3: Calculate the cell potential, Ecell , using the Nernst equation: Ecell = E ◦ − 0.0592 log(Q) n where: - E ◦ is the standard cell potential, - n is the number of moles of electrons transferred in the balanced cell reaction, - Q is the reaction quotient (ratio of product concentrations to reactant concentrations). In this case, n = 2 since 2 moles of electrons are transferred in the overall reaction. Step 4: Calculate Q using the given concentrations. Remember, the concentrations of solid species do not appear in the expression for Q. Q= [Ag+ ]2 (1.0)2 2+ = 0.1 = 10 [Cu ] Step 5: Substitute the values of E ◦ , n, and Q into the Nernst equation and solve for Ecell . Ecell = 0.80 − 0.0592 log(10) = 0.80 − 0.0296 × 1 = 0.7704 V 2 Therefore, the cell potential for this galvanic cell at 25◦ C is 0.7704 V. Question 30 Question Consider the following Galvanic and Electrolytic cells: Galvanic Cell Zn(s) | Zn2+ (aq) || Cu2+ (aq) | Cu(s) Electrolytic Cell Cu(s) | Cu2+ (aq) || Ag+ (aq) | Ag(s) 24 For each cell, determine the following: 1. The half-reaction that occurs at the cathode and the anode. 2. The overall cell reaction. 3. The direction of electron flow. 4. The direction of cation flow. 5. Whether the cell functions as a galvanic cell or an electrolytic cell. Solution 1. For the Galvanic Cell: ˆ Anode (Zn(s)): Zn(s) → Zn2+ (aq) + 2e− ˆ Cathode (Cu2+ (aq)): Cu2+ (aq) + 2e− → Cu(s) For the Electrolytic Cell: ˆ Anode (Cu(s)): Cu(s) → Cu2+ (aq) + 2e− ˆ Cathode (Ag+ (aq)): Ag+ (aq) + e− → Ag(s) 2. The overall cell reaction for the Galvanic Cell is the sum of the halfreactions: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) The overall cell reaction for the Electrolytic Cell is the sum of the halfreactions: Cu(s) + Ag+ (aq) → Cu2+ (aq) + Ag(s) 3. In both cells, electrons flow from the anode to the cathode. 4. In the Galvanic Cell, cations flow from the anode to the cathode. In the Electrolytic Cell, cations flow from the cathode to the anode. 5. The Galvanic Cell functions as a spontaneous chemical reaction generating electricity, while the Electrolytic Cell requires an external electrical current to drive a non-spontaneous chemical reaction. Question 31 Question A voltaic cell is constructed with a standard hydrogen electrode and a standard silver electrode. The standard cell potential is measured to be 0.80 V at 25°C. Calculate the standard free energy change for the cell reaction. Solution Step 1: Write the balanced cell reaction for the voltaic cell. The cell reaction for a voltaic cell with a standard hydrogen electrode (H+ (aq), H2 (g)) and a standard silver electrode (Ag+ (aq), Ag(s)) is: 2H+ (aq) + 2e− → H2 (g) E ◦ = 0 V 25 2Ag+ (aq) + 2e− → 2Ag(s) E ◦ = 0.80 V Overall cell reaction: 2Ag+ (aq) + 2H+ (aq) → 2Ag(s) + H2 (g) ◦ Step 2: Calculate the standard cell potential (Ecell ) using the standard re◦ duction potentials. The standard cell potential (Ecell ) is the difference between the standard reduction potentials of the two half-reactions: ◦ ◦ ◦ Ecell = Ecathode − Eanode = 0.80 V − 0 V = 0.80 V Step 3: Calculate the standard free energy change (∆G◦ ) using the equation: ◦ ∆G◦ = −nF Ecell where n is the number of moles of electrons transferred in the balanced cell reaction, F is the Faraday constant (96, 485 C/mol). For the given cell reaction, n = 2 moles of electrons transferred. Thus, ∆G◦ = −2 × 96485 C/mol × 0.80 V = −155136 J/mol Therefore, the standard free energy change for the cell reaction is −155136 J/mol. Question 32 Question Consider a galvanic cell consisting of a standard hydrogen electrode (SHE) and a copper electrode with the following half-reactions: Cathode: Cu2+ + 2 e− −−→ Cu Anode: 2 H+ + 2 e− −−→ H2 E◦ = 0.34 V E◦ = 0.00 V Calculate the cell potential at 298 K when the concentration of Cu2+ is 0.10 M, and the pressure of H2 is 1.0 atm. Solution Step 1: Write the overall cell reaction. The overall cell reaction is the sum of the two half-reactions: Cu2+ + 2 H+ −−→ Cu + H2 Step 2: Calculate the cell potential at standard conditions. ◦ The standard cell potential Ecell can be calculated using the standard reduction potentials: ◦ ◦ ◦ Ecell = Ecathode − Eanode 26 ◦ Ecell = 0.34V − 0.00V = 0.34 V Step 3: Consider the non-standard conditions. Since the concentrations of Cu2+ and the pressure of H2 are given, we can calculate the cell potential using the Nernst equation: 0.0592 ◦ Ecell = Ecell − log(Q) n where n is the number of electrons transferred in the cell reaction and Q is the reaction quotient. Step 4: Calculate the reaction quotient Q. [Cu] Q= 2+ [Cu ][H+ ]2 1.0 Q= = 10 0.10 × 1.02 Step 5: Substitute values into the Nernst equation and solve for the cell potential. 0.0592 Ecell = 0.34 − log(10) 2 Ecell = 0.34 − 0.0296 × 1 Ecell = 0.3104 V Therefore, the cell potential at 298 K when the concentration of Cu2+ is 0.10 M and the pressure of H2 is 1.0 atm is 0.3104 V. Question 33 Question Consider a galvanic cell with a standard cell potential of E ◦ = 0.78 V that operates under standard conditions. The cell reaction is given by: Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) Calculate the standard Gibbs free energy change for the cell reaction and determine if the reaction is spontaneous. Solution Step 1: The standard Gibbs free energy change (∆G◦ ) for the cell reaction can be calculated using the equation: ∆G◦ = −nF E ◦ where n is the number of moles of electrons transferred in the balanced redox equation, and F is the Faraday constant (96500 C/mol). From the given cell reaction, we see that 2 moles of electrons are transferred, so n = 2. Substituting into the equation: ∆G◦ = −2 × 96500 C/mol × 0.78 V 27 ∆G◦ = −188220 J/mol Step 2: To determine if the reaction is spontaneous, we can use the relationship between ∆G◦ and the equilibrium constant (K): ∆G◦ = −RT ln K where R is the gas constant (8.314 J/mol · K) and T is the temperature in Kelvin. At standard conditions, T = 298 K and R = 8.314 J/mol · K. Thus, we have: −188220 J/mol = −8.314 J/mol · K × 298 K ln K Solving for ln K: ln K = −188220 J/mol −8.314 J/mol · K × 298 K ln K = 27.37 Step 3: Exponentiating both sides of the equation to eliminate the natural logarithm: K = e27.37 K = 1.17 × 1011 Since K > 1, the reaction is spontaneous. Question 34 Question ◦ A student sets up a galvanic cell with a standard hydrogen electrode (EH + /H = 2 ◦ 0.00 V) and a copper electrode (ECu2+ /Cu = 0.34 V), connected by a salt bridge. The initial concentrations are [H+ ] = 1.0 M and [Cu2+ ] = 0.1 M. Calculate the cell potential at 25 Step 1: Write the balanced redox reaction and the cell notation. The balanced redox reaction is: Cu2+ (aq) + 2H+ (aq) → Cu(s) + H2 (g) The cell notation is: Pt|H2 (1 M)|H+ (1 M)||Cu2+ (0.1 M)|Cu ◦ Step 2: Calculate the cell potential (Ecell ) at standard conditions. 28 The standard cell potential can be calculated using the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ Given that EH + /H = 0.00 V, and ECu2+ /Cu = 0.34 V, we have: 2 ◦ Ecell = 0.00 V − 0.34 V = −0.34 V Step 3: Calculate the reaction quotient (Q). The reaction quotient Q is calculated using the concentrations of the species involved in the reaction: [Cu](solid) Q= [Cu2+ ][H+ ]2 Plugging in the values, we get: Q= 1 = 10 0.1 × 12 Step 4: Calculate the cell potential (Ecell ) under non-standard conditions. The Nernst equation relates the cell potential under non-standard conditions ◦ (Ecell ) to the standard cell potential (Ecell ) and the reaction quotient (Q): ◦ Ecell = Ecell − 0.0592 log Q n Since the reaction involves the transfer of 2 moles of electrons, n = 2. Plugging in the values, we get: Ecell = −0.34 V − 0.0592 log 10 2 Ecell = −0.34 V − 0.0296 log 10 Ecell = −0.34 V − 0.0296 × 1 Ecell = −0.34 V − 0.0296 Ecell = −0.3696 V Therefore, the cell potential at 25 Question Consider a galvanic cell that uses the following half-reactions: Anode: Zn(s) −−→ Zn2+ (aq) + 2 e− Cathode: Cu2+ (aq) + 2 e− −−→ Cu(s) If the standard reduction potential of Cu2+ /Cu is 0.34 V, calculate the stan◦ dard cell potential (Ecell ) of this galvanic cell. 29 Solution Step 1: Write the full cell reaction by adding the two half-reactions together. Ensure that the electrons cancel out. The full cell reaction will be: Zn(s) + Cu2+ (aq) −−→ Zn2+ (aq) + Cu(s) ◦ Step 2: Determine the standard cell potential (Ecell ) using the standard 2+ 2+ reduction potentials of Zn /Zn and Cu /Cu. The standard cell potential is given by the formula: ◦ ◦ ◦ Ecell = Ecathode − Eanode ◦ ◦ is is the standard reduction potential of the cathode and Eanode where Ecathode the standard reduction potential of the anode. ◦ ◦ Given that Ecathode (for Cu2+ /Cu) is 0.34 V, and Eanode (for Zn2+ /Zn) is −0.76 V (since it is the reverse of the reduction potential), we can plug these values into the formula. Substitute the values into the formula: ◦ Ecell = 0.34 V − (−0.76 V) = 1.10 V ◦ Therefore, the standard cell potential (Ecell ) of the galvanic cell is 1.10 V. 30