Electrochemistry Course Materials

Electrochemistry 1970 Account for the observation that silver dissolves in 1 molar hydroiodic acid despite the fact that the standard electrode potential for the change, Ag → Ag+ + e- is –0.80 volt. Answer: The standard potential is based upon [Ag+] = 1.0 M. But because AgI is insoluble (very small KSP) the concentration of Ag+ never reaches 1.0 M. Therefore, the equilibrium is shifted in favor of the production of the ion and the potential under these conditions is > 0 volts. 1970 Why are solutions of thiosulfate for analysis not acidic? Refer to the following standard electrode potentials and write the balanced equation that would account for this fact. E°_ 22 S + 6 OH → S2O3 + 3 H2O + 4e +0.74v S2O32- + 6 OH- → 2 SO32- + 3 H2O + 4e+0.58v S2O32- + H2O → 2 SO2 + 2 H+ + 4e–0.40v 2 S + 3 H2O → S2O32- + 6 H+ + 4e–0.50v Answer: S2O32- + 6 H+ + 4e- → 2 S + 3 H2O E° = +0.50v S2O32- + H2O → 2 SO2 + 2 H+ + 4eE° = –0.40v --------------------------------------------------------------- page 1 Carbon-12 has been given the standard atomic mass of 12.000… (b) the fluorine/fluoride potential is 2.65 v better than the half reaction, 2 H+ + 2 e- → H2, which has been given a value of 0.00 v. 1972 Br2 + 2 Fe2+(aq) → 2 Br-(aq) + 2 Fe3+(aq) For the reaction above, the following data are available: 2 Br-(aq) → Br2(l) + 2eE° = -1.07 volts Fe2+(aq) → Fe3+(aq) + eE° = -0.77 volts ------------------------------------------------------------------------------------- Br2(l) -27.1 Br-(aq) -70.1 S°, cal/mole.K 58.6 19.6 Fe2+(aq) Fe3+(aq) ------------------------------------------------------------------------------------- (a) Determine ΔS° (b) Determine ΔG° (c) Determine ΔH° Answer: (a) = [(19.6)(2)+(-70.1)(2)]-[58.6+(-27.1)(2)] cal = -105.4 cal (b) ΔG° = -nℑE° = -(2)(23060cal/v)(0.30v) = -13800 cal. 2 S2O3 + 4 H → 2 S + 2 H2O + 2 SO2 E (c) ΔG° = ΔH° - TΔS° ; ΔH° = ΔG° + TΔS° = +0.10v = -13800 + (298)(-105.4) = -45.4 kcal The thiosulfate decomposes into sulfur, water, and sulfur dioxide. 1973 B 1971 Sn + 2 Ag+ → Sn2+ + 2 Ag Quantitative chemical data are often based on arbitrary (a) Calculate the standard voltage of a cell involving standards. Discuss this statement with the following the system above. data for fluorine (b) What is the equilibrium constant for the system (a) The atomic weight of fluorine is 19.00 above? (b) E°, the standard electrode potential, is +2.65 volts (c) Calculate the voltage at 25°C of a cell involving for the half reaction: F2 + 2e- → 2 Fthe system above when the concentration of Ag+ is Answer: 0.0010 molar and that of Sn2+ is 0.20 molar. (a) The atomic mass of fluorine is 19.00 times the mass of 1/12 of the mass of a carbon-12 atom. Answer: 2- + Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry page 2 (a) E° = [0.80v - (-0.14v)] = 0.94v (b) 1976 B (a) Calculate the value of ΔG° for the standard cell (c) reaction Zn + Cu2+(1M) → Zn2+(1M) + Cu (b) One half cell of an electrochemical cell is made by placing a strip of pure zinc in 500 milliliters of 0.10 molar ZnCl2 solution. The other half cell is made by placing a strip of pure copper in 500 milliliters of 0.010 molar Cu(NO3)2 solution. Calculate the initial voltage of this cell when the two half cells are joined by a salt bridge and the two metal strips are joined by a wire. (c) Calculate the final concentration of copper ion, Cu2+, in the cell described in part (b) if the cell were allowed to produce an average current of 1.0 ampere for 3 minutes 13 seconds. Answer: (a) E° = 0.76 + 0.34 = 1.10 volts 1974 B A steady current of 1.00 ampere is passed through an electrolytic cell containing a 1 molar solution of AgNO3 and having a silver anode and a platinum cathode until 1.54 grams of silver is deposited. (a) How long does the current flow to obtain this deposit? (b) What weight of chromium would be deposited in a second cell containing 1–molar chromium(III) nitrate and having a chromium anode and a platinum cathode by the same current in the same ΔG° = -nℑE = - (2)(23.06kcal/v)(1.10v) = -50.7 time as was used in the silver cell? (c) If both electrodes were platinum in this second kcal cell, what volume of O2 gas measured at standard temperature and pressure would be released at the anode while the chromium is being deposited at (b) = 1.07 v the cathode? The current and the time are the same (c) (1.0 amp)(193 sec.)(1 farad./96500 coul) = as in (b) = 0.0020 faraday Answer: (a) 0.0020 faraday/2 = 0.0010 mol Cu2+ reduced (0.0050 - 0.0010) mol = 0.0040 mol Cu2+ remaining 0.0040 mol / 0.500 L = 0.0080 M final [Cu2+] = 1380 sec. (b) (c) 2 H2O → O2 + 4 H+ + 4e- 1978 B (a) When 300.0 milliliters of a solution of 0.200 molar AgNO3 is mixed with 100.0 milliliters of a 0.0500 molar CaCl2 solution, what is the concentration of silver ion after the reaction has gone to completion? (b) Write the net cell reaction for a cell formed by placing a silver electrode in the solution remaining from the reaction above and connecting it to a standard hydrogen electrode. Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry (c) Calculate the voltage of a cell of this type in which the concentration of silver ion is 4×10-2 M. (d) Calculate the value of the standard free energy change ΔG° for the following half reaction: Ag+ (1M) + e- → Ag° Answer: (a) Ag+ + Cl- → AgCl mol Ag+ added = (0.300 L)(0.200 mol/L) = = 0.0600 mol mol CaCl2 added = (0.100 L)(0.0500 mol/L) = = 0.00500 mol mol Cl- added = 2(0.00500 mol) = 0.0100 mol mol Ag+ remaining = 0.0600 - 0.0100 = 0.0500 mol (b) H2 + 2 Ag+ → 2 H+ + 2 Ag Cu (aq) + 2e- → Cu(s) 2+ page 3 0.340 v --------------------------------------------------------- Cu2+(aq) + M(s) → M2+(aq) + Cu(s) 0.740 v X (oxid. potential) = 0.740 v - 0.340 v = 0.400 v M2+(aq) + 2e- → M(s) E° = -0.400 v 2+ (b) M + Cu → M2+ + Cu initial: 1.00 M 1.00 M change: -0.80 M +0.80 M final: 0.20 M +1.80 M (c) At equilibrium, Ecell = 0 (c) E = (0.80 - 0.08) v = 0.72 volt (d) ΔG° = -nℑE° = -(1)(96.5kJ/mol)(0.80v) = -77 kJ or -18 kcal 1980 B M(s) + Cu2+(aq) → M2+(aq) + Cu(s) For the reaction above E° = 0.740 volt at 25°C (a) Determine the standard electrode potential for the reduction half reaction: M2+(aq) + 2e- → M(s) (b) A cell is constructed in which the reaction above occurs. All substances are initially in their standard states, and equal volumes of the solutions are used. The cell is then discharged. Calculate the value of the cell potential E, when [Cu2+] has dropped to 0.20 molar. (c) Find the ratio [M2+]aq/[Cu2+]aq when the cell reaction above reaches equilibrium. Answer: (a) M(s) → M2+(aq) + 2eXv 1981 D A solution of CuSO4 was electrolyzed using platinum electrodes by passing a current through the solution. As a result, there was a decrease in both [Cu2+] and the solution pH; one electrode gained in weight a gas was evolved at the other electrode. (a) Write the cathode half reaction that is consistent with the observations above. (b) Write the anode half reaction that is consistent with the observations above. (c) Sketch an apparatus that can be used for such an experiment and label its necessary components. Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry page 4 (d) List the experimental measurements that would be needed in order to determine from such an (c) experiment the value of the faraday. =0.155 mol O2 Answer: (d) from ΔHf° for H2O: (a) Cu2+ + 2e- → Cu (b) 2 H2O → O2 + 4 H+ + 4eH2(g) + 1/2 O2(g) → H2O(g) (c) ΔH = -242 kJ when the H2 is burned. (d) current; time; mass of cathode before and its mass after passage of current - or - volume of O2 released with its temperature and pressure. 1983 C Ti3+ + HOBr → TiO2+ + Br- (in acid solution) (a) Write the correctly balanced half-reactions and net ionic equation for the skeletal equation shown above. (b) Identify the oxidizing agent and the reducing agent in this reaction. (c) A galvanic cell is constructed that utilizes the reaction above. The concentration of each species is 0.10 molar. Compare the cell voltage that will be observed with the standard cell potential. Explain your reasoning. (d) Give one example of a property of this reaction, other than the cell voltage, that can be calculated from the standard cell potential, E°. State the relationship between E° and the property you have specified. Answer: (a) Ti3+ + H2O → TiO2+ + 2 H+ + eH+ + HOBr + 2e- → Br- + H2O 2 Ti3+ + HOBr + H2O → 2 TiO2+ + 3 H+ + Br(b) HOBr is the oxidizing agent and Ti3+ is the reducing agent. (c) The observed voltage will be greater than the E° value since: 1982 B When a dilute solution of H2SO4 is electrolyzed, O2(g) is produced at the anode and H2(g) is produced at the cathode. (a) Write the balanced equations for the anode, cathode, and overall reactions that occur in this cell. (b) Compute the coulombs of charge passed through the cell in 100. minutes at 10.0 amperes. (c) What number of moles of O2 is produced by the cell when it is operated for 100. minutes at 10.0 amperes? (d) The standard enthalpy of formation of H2O(g) is -242 kilojoules per mole. How much heat is liberated by the complete combustion, at 298K and 1.00 atmospheres, of the hydrogen produced by the cell operated as in (c)? Answer: (a) cathode: 2 H2O + 2e- → H2 + 2 OH(d) Identification or 2 H+ + 2e- → H2 K, pH anode: 2 H2O → O2 + 4 H+ + 4e(e) ΔG° = -nℑE° or 2 O2– → O2 + 4e1985 B overall: 2 H2O → 2 H2 + O2 (a) Titanium can TiO2+ to Ti3+ (b) (10.0amp)(10.0 min)(60 ) = 6.00×104 coul. of a property from the group: ΔG, or E° = log K be reduced in an acid solution from with zinc metal. Write a balanced Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry page 5 equation for the reaction of TiO with zinc in acid solution. (b) What mass of zinc metal is required for the reduction of a 50.00 millilitre sample of a 0.115 molar solution of TiO2+? (c) Alternatively, the reduction of TiO2+ to Ti3+ can be carried out electrochemically. What is the minimum time, in seconds, required to reduce another 50.000 millilitre sample of the 0.115 molar TiO2+ solution with a direct current of 1.06 amperes? (d) The standard reduction potential, E°, for TiO2+ to Ti3+ is +0.060 volt. The standard reduction potential, E°, for Zn2+ to Zn(s) is -0.763 volt. Calculate the standard cell potential, E°, and the between platinum electrodes for a period of 1.100 hours. Oxygen gas was produced at the anode. The only change at the cathode was a slight change in the color of the solution. At the end of the electrolysis, the electrolyte was acidified with sulfuric acid and was titrated with an aqueous solution of potassium permanganate. The volume of the KMnO4 solution required to reach the end point was 24.65 millilitres. (a) How many faradays were passed through the solution? (b) Write a balanced half-reaction for the process that occurred at the cathode during the electrolysis. (c) Write a balanced net ionic equation for the reaction that occurred during the titration with potassium permanganate. standard free energy change, ΔG°, for the reaction (d) Calculate the molarity of the KMnO4 solution. Answer: described in part (a). (a) (1.100 hr)(3600 sec/hr) = 3960 sec. Answer: 2+ (3960 sec)(0.125 amp) = 495 coul (a) Zn → Zn + 2e2+ + 3+ 2 TiO + 4 H + 2e- → 2 Ti + 2 H2O 2+ ------------------------------------------------------------------------------ 495 coul × 2 TIO2+ + 4 H+ + Zn → Zn2+ + 2Ti3+ + 2 H2O (b) 0.0500 L × × × = 5.13×10-3 faraday (b) Fe3+ + 1e- → Fe2+ (c) MnO4- + 8 H+ + 5e- → Mn2+ + 4 H2O 5 Fe2+ → 5 Fe3+ + 5e--------------------------------------------------------------------------------- MnO4- + 8 H+ + 5 Fe2+ → 5 Fe3+ + Mn2+ + 4 H2O = 0.188 g Zn (c) (d) 1987 D A dilute solution of sodium sulfate, Na2SO4, was electrolyzed using inert platinum electrodes. In a separate experiment, a concentrated solution of sodium chloride, NaCl, was electrolyzed also using inert ΔG° = -nℑE° platinum electrodes. In each experiment, gas formation was observed at both electrodes. (a) Explain why metallic sodium is not formed in 5 either experiment. = -1.59×10 J (b) Write balanced equations for the half–reactions that occur at the electrodes during electrolysis of 1986 B the dilute sodium sulfate solution. Clearly indicate A direct current of 0.125 ampere was passed through which half–reaction occurs at each electrode. 200 millilitres of a 0.25 molar solution of Fe2(SO4)3 (d) Zn → Zn2+ + 2eE° = 0.763 v 2+ + 3+ 2TiO + 4H + 2e- → 2Ti + 2H2O E° = 0.060 v E° = 0.823 v Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry (c) Write balanced equations for the half–reactions that occur at the electrodes during electrolysis of the concentrated sodium chloride solution. Clearly indicate which half–reaction occurs at each electrode. (d) Select two of the gases obtained in these experiments, and for each gas, indicate one experimental procedure that can be used to identify it. Answer: (a) Na+ is not reduced as easily as H2O (or H+ or OH–). OR If Na(s) were formed, it would rapidly react with water to reform Na+. (b) Anode: 2 H2O → O2 + 4 H+ + 4ecathode: 2 H2O + 2e- → H2 + 2 OHor: 2 H+ + 2e- → H2 (c) anode: 2 Cl- → Cl2 + 2ecathode: 2 H2O + 2e- → H2 + 2 OHor: 2 H+ + 2e- → H2 (d) H2 - “pop” with a lit splint; O2 - ignites a glowing splint; Cl2 - yellowish-green color (other suitable tests accepted) page 6 (b) 2 Ag + Sn → 2 Ag + Sn E° = [0.80v - (-0.14v)] = 0.94v + 2+ (c) (d) 1989 B The electrolysis of an aqueous solution of potassium iodide, KI, results in the formation of hydrogen gas at the cathode and iodine at the anode. A sample of 80.0 millilitres of a 0.150 molar solution of KI was electrolyzed for 3.00 minutes, using a constant current. At the end of this time, the I2 produced was titrated against a 0.225 molar solution of sodium thiosulfate, which reacts with iodine according to the equation below. The end point of the titration was reached when 37.3 millilitres of the Na2S2O3 solution had been added. I2 + 2 S2O32- → 2 I- + S4O621988 B (a) How many moles of I2 was produced during the An electrochemical cell consists of a tin electrode in an electrolysis? acidic solution of 1.00 molar Sn2+ connected by a salt (b) The hydrogen gas produced at the cathode during bridge to a second compartment with a silver electrode the electrolysis was collected over water at 25°C in an acidic solution of 1.00 molar Ag+. (a) Write the equation for the half–cell reaction at a total pressure of 752 millimetres of mercury. occurring at each electrode. Indicate which Determine the volume of hydrogen collected. (The half–reaction occurs at the anode. vapor pressure of water at 25°C is 24 millimetres of (b) Write the balanced chemical equation for the mercury.) overall spontaneous cell reaction that occurs when (c) Write the equation for the half–reaction that the circuit is complete. Calculate the standard occurs at the anode during the electrolysis. voltage, E°, for this cell reaction. (d) Calculate the current used during the electrolysis. (c) Calculate the equilibrium constant for this cell Answer: reaction at 298K. (d) A cell similar to the one described above is constructed with solutions that have initial (a) concentrations of 1.00 molar Sn2+ and 0.0200 = 4.20×10-3 mol I2 molar Ag+. Calculate the initial voltage, E°, of this cell. (b) PH2 = Ptotal - PH2O Answer: = (752 - 24) mm Hg = 728 mm Hg 2+ (a) Sn → Sn + 2e- anode reaction 1 mol H2 = 1 mol I2 + Ag + e- → Ag PV = nRT ; V = nRT/P Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry = 0.107 L (c) At anode: 2 I– → I2 + 2e(d) 1991 D Explain each of the following. (a) When an aqueous solution of NaCl is electrolyzed, Cl2(g) is produced at the anode, but no Na(s) is produced at the cathode. (b) The mass of Fe(s) produced when 1 faraday is used to reduce a solution of FeSO4 is 1.5 times the mass of Fe(s) produced when 1 faraday is used to reduce a solution of FeCl3. (c) Zn + Pb2+ (1–molar) → Zn2+ (1–molar) + Pb The cell that utilizes the reaction above has a higher potential when [Zn2+] is decreased and [Pb2+] is held constant, but a lower potential when [Pb2+] is decreased and [Zn2+] is held constant. (d) The cell that utilizes the reaction given in (c) has the same cell potential as another cell in which [Zn2+] and [Pb2+] are each 0.1–molar. Answer: (a) Cl– is more easily oxidized than water water is more easily reduced than Na+ (b) Fe2+ req. 2 farad/mol Fe(s) or 1 farad → 1/2 mol Fe(s) Fe3+ req. 3 farad/mol Fe(s) or 1 farad → 1/3 mol Fe(s) for equal numbers of farad 1/2:1/3 :: 1.5:1 (c) using LeChatelier’s principle if [Zn2+] ↓, reaction shifts →, ∴ cell potential ↑ if [Pb2+] ↓, reaction shifts ←, ∴ cell potential ↓ or using the Nernst Equation Ecell = E° - RT lnQ, where Q = [Zn2+]/[Pb2+], if [Zn2+] ↓, Q<1, Ecell > E° if [Pb2+] ↓, Q>1, Ecell < E° (d) [Zn2+]/[Pb2+] does not change regardless of the values, ∴ Ecell = E° or [Zn2+]/[Pb2+] = 1; ln Q = 0; Ecell = E° page 7 1992 B An unknown metal M forms a soluble compound, M(NO3)2. (a) A solution of M(NO3)2 is electrolyzed. When a constant current of 2.50 amperes is applied for 35.0 minutes, 3.06 grams of the metal M is deposited. Calculate the molar mass of M and identify the metal. (b) The metal identified in (a) is used with zinc to construct a galvanic cell, as shown below. Write the net ionic equation for the cell reaction and calculate the cell potential, E°. (c) Calculate the value of the standard free energy change, ΔG°, at 25°C for the reaction in (b). (d) Calculate the potential, E, for the cell shown in (b) if the initial concentration of ZnSO4 is 0.10-molar, but the concentration of the M(NO3)2 solution remains unchanged. Answer: (a) 3.06 g/2.72×10-2 mol = 112 g/mol; ∴ metal is Cd (b) Zn → Zn2+ + 2eCd2+ + 2e- → Cd E° = +0.76 v E° = –0.40 v ------------------------------------------ Cd2+ + Zn → Cd + Zn2+ E° = +0.36 v (c) ΔG° = -nℑE° = -(2)(96.5 kJ/v)(0.36v) = -69 kJ (d) Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry page 8 Consider the reaction represented above that occurs at 25°C. All reactants and products are in their standard states. The value of the equilibrium constant, Keq, for 1993 D 17 A galvanic cell is constructed using a chromium the reaction is 4.2×10 at 25°C. electrode in a 1.00-molar solution of Cr(NO3)3 and a (a) Predict the sign of the standard cell potential, E°, copper electrode in a 1.00-molar solution of Cu(NO3)2. for a cell based on the reaction. Explain your prediction. Both solutions are at 25°C. (b) Identify the oxidizing agent for the spontaneous (a) Write a balanced net ionic equation for the reaction. spontaneous reaction that occurs as the cell operates. Identify the oxidizing agent and the (c) If the reaction were carried out at 60°C instead of reducing agent. 25°C, how would the cell potential change? (b) A partial diagram of the cell is shown below. Justify your answer. (d) How would the cell potential change if the reaction were carried out at 25°C with a 1.0-molar solution of Mg(NO3)2 and a 0.10-molar solution of Sr(NO3)2 ? Explain. (e) When the cell reaction in (d) reaches equilibrium, what is the cell potential? Answer: (a) (+); K > 1 OR reaction is spontaneous OR E° for Sr2+ is more positive OR E° for Sr is more negative OR E° = +0.52 v (b) Mg2+ (i) Which metal is the cathode? (ii) What additional component is necessary to make the cell operate? (c) increase; E° = ln Keq, as T increases so does (iii) What function does the component in (ii) E°. serve? (c) How does the potential of this cell change if the concentration of Cr(NO3)3 is changed to (d) increase; Ecell = E° – ln Q, when Q = 3.00-molar at 25°C? Explain. Answer: (a) 2 Cr + 3 Cu2+ → 2 Cr3+ + 3 Cu Cr = reducing agent; Cu2+ = oxidizing agent (b) (i) Cu is cathode (ii) salt bridge (iii) transfer of ions or charge but not electrons (c) Ecell decreases. use the Nernst equation to explain answer 1996 D Sr(s) + Mg2+ → Sr2+ + Mg(s) then Ecell becomes larger. OR an explanation using LeChâtelier’s Principle. (e) Ecell = 0 1997 B In an electrolytic cell, a current of 0.250 ampere is passed through a solution of a chloride of iron, producing Fe(s) and Cl2(g). (a) Write the equation for the half-reaction that occurs at the anode. (b) When the cell operates for 2.00 hours, 0.521 gram of iron is deposited at one electrode. Determine Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry page 9 the formula of the chloride of iron in the original solution. (c) Write the balanced equation for the overall reaction that occurs in the cell. (d) How many liters of Cl2(g), measured at 25°C and 750 mm Hg, are produced when the cell operates as described in part (b) ? (e) Calculate the current that would produce chlorine gas from the solution at a rate of 3.00 grams per hour. Answer: (a) 2 Cl– – 2 e- → Cl2 Answer the following questions regarding the electrochemical cell shown. (a) Write the balanced net-ionic equation for the spontaneous reaction that occurs as the cell 1800 coul × = 0.0187 mol eoperates, and determine the cell voltage. (b) In which direction do anions flow in the salt bridge as the cell operates? Justify your answer. 0.521 g Fe × = 0.00933 mol Fe (c) If 10.0 mL of 3.0-molar AgNO3 solution is added to the half-cell on the right, what will happen to the cell voltage? Explain. 2+ Fe + 2e → Fe ; ∴ FeCl2 (d) If 1.0 gram of solid NaCl is added to each 2+ – half-cell, what will happen to the cell voltage? (c) Fe + 2 Cl → Fe + Cl2 Explain. (e) If 20.0 mL of distilled water is added to both (d) 0.0187 mol e- × = 0.00933 mol Cl2 half-cells, the cell voltage decreases. Explain. Answer (a) 2 Ag+ + 2 e- → 2 Ag E° = +0.80 v Cd – 2 e-- → Cd2+ E° = +0.40 v V= = + 2+ 2 Ag + Cd → 2 Ag + Cd E = +1.20v = 0.231 L (b) Anions flow into the cadmium half-cell. As the cell operates, Cd2+ cations increase in number and (e) need to be balanced by an equal number of anion charges from the salt bridge. (c) Cell voltage will increase. An increase in silver × = 2.27 amp ion concentration will result in faster forward OR reaction and a higher cell potential. (d) Cell voltage will decrease. As the salt dissolves, the Cl– ion will cause the Ag+ ion to precipitate as = AgCl and decrease the [Ag+]. This will result in a slower forward reaction and a decrease in cell ; X = 2.27 amp potential. Since cadmium chloride is a soluble salt, it will not affect the cadmium half-cell. 1998 D (b) 0.250 amp × 7200 sec = 1800 coulomb Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry page 10 current of 100. amperes passed through 5.00 L of 2.00 M CuSO4(aq) for a period of 1.00 hour. (e) E = 1.20v – ; while both Answer: concentrations are 1.0M, the cell potential is (a) (i) Zn(s) – 2 e- → Zn2+(aq) 1.20v. But if each solution’s concentration is cut in (ii) Zn(s) + Co2+(aq) → Zn2+(aq) + Co(s) cell half, then, E = 1.20v – cell = 1.19v (iii) oxid: Zn(s) – 2 e- → Zn2+(aq) E° = +0.76V E° = –0.28V red: Co2+(aq)+ 2 e- → Co(s) 2000 B 2. Answer the following questions that relate to +0.48V electrochemical reactions. (b) (i) ΔG° = –nℑE° = –(2)(96500)(0.55) = –106 kJ (a) Under standard conditions at 25°C, Zn(s) reacts with Co2+(aq) to produce Co(s). (ii) Keq = e–ΔG/RT = e–(–106150/(8.31)(298)) = (i) Write the balanced equation for the oxidation 4.13×1018 half reaction. (ii) Write the balanced net-ionic equation for the (iii) 2 H2O(l) + O2(g) → 2 H2O2(aq) E° = –0.55 V overall reaction. (iii) Calculate the standard potential, E°, for the O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) E° = 1.23 V ______________________________________________________________________________________________________________________ overall reaction at 25°C. (b) At 25°C, H2O2 decomposes according to the following equation. 2 H2O2(aq) → 2 H2O(l) + O2(g) E° = 0.55 V 2 O2(g) + 4 H+(aq) + 4 e- → 2 H2O2(aq) E° = 0.68 V divide the equation by 2 but keep the E° the same. (d) 1.00 hr × × 100. amp × × × × = 119 g Cu (i) Determine the value of the standard free 2001 D energy change, ΔG°, for the reaction at 25°C. (ii) Determine the value of the equilibrium constant, Keq, for the reaction at 25°C. (iii) The standard reduction potential, E°, for the half reaction O2(g) + 4 H+(aq) + 4 e- → 2 H2O(l) has a value of 1.23 V. Using this information in addition to the information given above, determine the value of the standard reduction potential, E° for the half reaction below. O2(g) + 2 H+(aq) + 2 e- → H2O2(aq) Answer the following questions that refer to the galvanic cell shown in the diagram above. (A table of (c) In an electrolytic cell, Cu(s) is produced by the standard reduction potentials is printed on the green electrolysis of CuSO4(aq). Calculate the maximum insert and on page 4 of the booklet with the pink mass of Cu(s) that can be deposited by a direct cover.) Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry (a) Identify the anode of the cell and write the half reaction that occurs there. (b) Write the net ionic equation for the overall reaction that occurs as the cell operates and calculate the value of the standard cell potential, E°cell . (c) Indicate how the value of Ecell would be affected if the concentration of Ni(NO3)2(aq) was changed from 1.0 M to 0.10 M and the concentration of Zn(NO3)2(aq) remained at 1.0 M. Justify your answer. (d) Specify whether the value of Keq for the cell reaction is less than 1, greater than 1, or equal to 1. Justify your answer. Answer: (a) zinc; Zn(s) Zn2+(aq) + 2 e– (b) Zn(s) + Ni2+(aq) → Zn2+(aq) + Ni(s) E°cell = +0.76 + (-0.25) V = +0.51 V (c) decrease Ecell ; Ecell = E°cell – log Q, Q = , when the value of Q becomes larger than 1 then the log Q > 1 and is subtracted from the standard potential of the cell. (d) greater than 1. All spontaneous reactions (this reaction is spontaneous because the cell potential is larger than 0) have a Keq that are larger than 1, which favors the formation of products. page 11 Another galvanic cell is based on the reaction between Ag+(aq) and Cu(s), represented by the equation below. At 25˚C, the standard potential, E˚, for the cell is 0.46 V. 2 Ag+(aq) + Cu(s) → Cu2+(aq) + 2 Ag(s) (c) Determine the value of the standard free-energy change, ∆G˚, for the reaction between Ag+(aq) and Cu(s) at 25˚C. (d) The cell is constructed so that [Cu2+] is 0.045 M and [Ag+] is 0.010 M. Calculate the value of the potential, E, for the cell. (e) Under the conditions specified in part (d), is the reaction in the cell spontaneous? Justify your answer. Answer: (a) (i) AgNO3 solution 1.50 g Zn × × × = 417 mL of silver nitrate solution required to completely react the zinc, therefore, AgNO3 is the limiting reagent. (ii) 250 mL AgNO3 × × × = 0.01375 mol Zn = 0.0550 M [Zn2+] (b) 2 Ag+ + 2e- → 2 Ag E˚ = +0.80 v 2002 B E˚ = +0.76 v Zn – 2e- → Zn2+ Answer parts (a) through (e) below, which relate to +1.56 v reactions involving silver ion, Ag+. (c) ∆G˚ = -nFE˚ = -(2)(96500)(0.46 v) = -89000 J The reaction between silver ion and solid zinc is (d) Ecell = E˚ - log represented by the following equation. = 0.46 - log= 0.38 v 2 Ag+(aq) + Zn(s) → Zn2+(aq) + 2 Ag(s) (e) yes; any reaction is spontaneous with a positive voltage (a) A 1.50 g sample of Zn is combined with 250. mL of 0.110 M AgNO3 at 25˚C. (i) Identify the limiting reactant. Show 2004 D Required calculations to support your answer. (ii) On the basis of the limiting reactant that you identified in part (i), determine the value of [Zn2+] after the reaction is complete. Assume that volume change is negligible. (b) Determine the value of the standard potential, E˚, for a galvanic cell based on the reaction between AgNO3(aq) and solid Zn at 25˚C. Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited. Electrochemistry page 12 reduction (gain in electrons) and will convert metal ions into a metal. An electrochemical cell is constructed with an open switch, as shown in the diagram above. A strip of Sn and a strip of unknown metal, X are used as electrodes. (b) (see diagram) When the switch is closed, the mass of the Sn electrode increases. The half-reactions are shown (c) red: Sn2+ (aq) + 2 e– → Sn(s) below. oxid: X(s) – 3 e– → X3+(aq) E˚ = –0.14 V Sn2+ (aq) + 2 e– → Sn(s) X3+(aq) + 3 e– → X(s) E˚ = ? red: X3+(aq) + 3 e– → X(s) E˚ = –0.14 V E˚ = +0.74 V E˚cell = +0.60 V E˚ = –0.74 V (a) In the diagram above, label the electrode that is (d) Cr the cathode. Justify your answer. (b) In the diagram above, draw an arrow indicating (e) 3 Sn2+ + 2 Cr → 3 Sn + 2 Cr3+ the direction of electron flow in the external circuit (f) (i) E˚cell = 0.60 – log () when the switch is closed. (ii) greater; • (c) If the standard cell potential E˚cell is +0.60 V, what is the standard potential, in volts for the X3+/X electrode? (d) Identify metal X. (e) Write balanced net-ionic equation for the overall chemical reaction occurring in the cell. (f) In the cell, the concentration of Sn2+ is changed from 1.0 M to 0.50 M, and the concentration of X3+ is changed from 1.0 M to 0.10 M. (i) Substitute all appropriate values for determining the cell potential, Ecell, into the Nernst equation. (Do not do any calculations.) (ii) On the basis of your response in (f) (i), will the cell potential be greater than, less than, or equal to E˚cell? Justify your answer. Answer: (a) tin electrode is the cathode; cathode is the site of Copyright © 1970 to 2004 by College Entrance Examination Board, Princeton, NJ. All rights reserved. For face-to-face teaching purposes, classroom teachers are permitted to reproduce the questions. Portions copyright © 1993-2004 by Unlimited Potential Inc., Framingham, MA 01701-2619, 508/877-8723. Web or Mass distribution prohibited.