Electrochemistry - ANSWERS



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Electrochemistry - ANSWERS 1. Using a table of standard electrode potentials, predict if the following reactions will occur spontaneously as written. a) Al 3+ + Ni Ni 2+ + Al Al 3+ + 3e - Al E = -1.68 V Ni Ni 2+ + 2e - E = +0.26 V Total Cell Potential: E = -1.42 V non-spontaneous b) Ag + + Co Co 2+ + Ag Ag + + 1e - Ag E = +0.80 V Cp Co 2+ + 2e - E = +0.28 V Total Cell Potential: E = +1.08 V spontaneous c) Sn 2+ + Pb 2+ Sn 4+ + Pb Sn 2+ Sn 4+ + 2e - E = -0.15 V Pb 2+ + 2e - Pb E = -0.13 V Total Cell Potential: E = -0.28 V non-spontaneous d) Au 3+ + Fe Au + Fe 2+ Au 3+ + 3e - Au E = +1.50 V Fe Fe 2+ + 2e - E = +0.44 V Total Cell Potential: E = +1.94 V spontaneous e) Cr + 3Ag + Cr 3+ + 3Ag Cr Cr 3+ + 3e - E = +0.74 V Ag + + 1e - Ag E = +0.80 V Total Cell Potential: E = +1.54 V spontaneous f) Cu + + Fe 3+ Cu 2+ + Fe 2+ Cu + Cu 2+ + 1e - E = -0.16 V Fe 3+ + 1e - Fe 2+ E = +0.77 V Total Cell Potential: E = +0.61 V spontaneous g) Mn 2+ + 2H 2 O + I 2 MnO 2 + 4H + + 2I - Mn 2+ + 2H 2 O MnO 2 + 4H + + 2e - I 2 + 2e - 2I - E = -1.23 V E = +0.54 V Total Cell Potential: E = -1.23 V non-spontaneous h) 2Cr 3+ + 7H 2 O + 3Pb 2+ Cr 2 O 7 2- + 14H + + 3Pb 2Cr 3+ + 7H 2 O Cr 2 O 7 2- + 14H + + 6e - E = -1.33 V Pb 2+ + 2e - Pb E = -0.13 V Total Cell Potential: E = -1.46 V non-spontaneous

2. Predict the voltages produced by the following cells. a) Zn Zn 2+ Fe 2+ Fe E = (+0.76) + (-0.44 V) = 0.32 V b) Mn Mn 2+ Br 2 Br - E = (+1.18) + (+1.06 V) = 2.24 V c) H 2 SO 3 SO 2-4 MnO4-2+ Mn E = (-0.17) + (+1.51 V) = 1.34 V d) Ni Ni 2+ Ag + Ag E = (+0.26) + (+0.80 V) = 1.06 V e) Cu Cu 2+ Au 3+ Au E = (-0.34) + (+1.50 V) = 1.16 V f) Pb Pb 2+ Cl 2 Cl - E = (+0.13) + (+1.36 V) = 1.49 V 3. A cell is made up as follows. A piece of nickel foil is immersed in a beaker of nickel(ii) chloride solution, and a strip of copper foil is immersed in a beaker of copper(ii) sulfate solution. A wire then connects the metal electrodes and the beakers connected by a salt bridge. a) Write the half reactions that occur. Identify the oxidation half and the reduction half. Oxidation ½ = Ni Ni 2+ + 2e - E = +0.26 V Reduction ½ = Cu 2+ + 2e - Cu E = +0.34 V b) Which electrode is the anode? Nickel is the anode c) Toward which electrode do the sulfate ions migrate? The sulfate ions are anions (negative ions). They migrate towards the nickel anode. d) Which way do the electrons migrate in the wire? The electrons always migrate from the anode to the cathode. In this cell, they would migrate from the nickel (anode) to the copper (cathode). e) If 0.025 mol of copper is produced in the reaction, how many moles of electrons flow through the wire? Cu 2+ + 2e - Cu 2 mol e- 0.025 mol Cu 0.050 050 mol e - 1 mol Cu f) Toward which electrode do the nickel ions migrate? The nickel ions are cations (positive ions). They migrate towards the copper cathode.

4. An electrochemical cell was made as follows. A weighed strip of tin was immersed in a beaker of one molar tin(ii) sulfate and a weighed strip of silver was immersed in a second beaker containing one molar silver nitrate. A wire then connected the metal strips and a salt bridge connected the beakers. a) What is the balanced redox equation for the reaction? Oxidation ½ = Sn Sn 2+ + 2e - E = +0.14 V Reduction ½ = Ag + + 1e - Ag E = +0.80 V Redox Reaction = Sn + 2 Ag + Sn 2+ + 2Ag b) Which electrode is the cathode? Silver is the cathode. c) Towards which electrode do the silver ions migrate? The silver ions are cations and migrate towards the silver cathode. d) Which way do the electrons flow in the wire? The electrons flow from the tin anode to the silver cathode. e) Did the silver electrode gain or lose mass? Silver is the cathode, reduction takes place at the cathode. Ag + + 1e - Ag Silver ions change to silver atoms and deposit on the cathode. The silver electrode (cathode) gains mass. f) If 0.010 mol of tin goes into solution, how many moles of electrons flow through the wire? Sn 2+ + 2e - Sn 2 mol e- 0.010 mol Sn 0.020 020 mol e - 1 mol Sn g) If 0.020 mol of tin goes into solution, how many moles of silver are involved in the reaction? Sn + 2 Ag + Sn 2+ + Ag 2 mol Ag 0.020 mol Sn 0.040 040 mol Ag 1 mol Sn i) How many moles of electrons flow through the salt bridge in part (g)? Zero. Electrons do not flow through the salt bridge, they flow through the wire (external circuit).

5. A galvanic cell involves the overall reaction of iodide ions with acidified permanganate ions to form manganese(ii) ions and iodine. The salt bridge contains potassium nitrate. a) Write the half-reactions, and the overall cell reaction. Oxidation ½ 2I - I 2 + 2e - E = -0.54 V Reduction ½ MnO - 4 + 8H + + 5e - Mn 2+ + 4H 2 O E = +1.57 V Overall Cell Reaction 2MnO 4 - + 16H + + 10I - 2Mn 2+ + 8H 2 O + 5I 2 E = 1.03 V b) Identify the oxidizing agent and the reducing agent. Oxidizing Agent = MnO 4 - Reducing Agent = I - c) The inert anode and cathode are both made of graphite. Solid iodine forms on one of them. Which one? Solid iodine is a product of the oxidation half reaction. Oxidation occurs at the anode, therefore solid iodine forms on the anode. 6. a) Which is the stronger oxidizing agent? i) Zn 2+ or Ca 2+ Zn 2+ ii) Br 2 or I 2 Br 2 b) Which is the stronger reducing agent? i) Mn or Pb Mn ii) Ba or Sn Ba 7. Predict the reaction that will be favoured in the following cells. a) A mixture of Br 2 and Cl 2 is added to a beaker containing copper(ii) sulfate solution and a copper rod. Cu Cu 2+ + 2e - E = -0.34 V Cl 2 + 2e - 2Cl - E = 1.36 V Cu Cu + + 1e - E = -0.52 V Br 2 + 2e - 2Br - E = 1.06 V Cu 2+ + 1e - Cu + E = +0.16 V Cu 2+ + 2e - Cu E = +0.34 V Reaction = Cu + Cl 2 Cu 2+ + 2Cl -

b) A mixture of powdered Al and Fe is added to a beaker of Cr 3+ solution. Al Al 3+ + 3e - E = +1.68 V Cr 3+ + 1e - Cr 2+ E = -0.42 V Fe Fe 3+ + 3e - E = +0.04 V Cr 3+ + 3e - Cr E = -0.74 V Fe Fe 2+ + 2e - E = +0.44 V 2H 2 O + 2e - H 2 + 2OH - E = -0.83 V H 2 O O 2 + 4H + + 4e - E = -1.23 V 2Cr 3+ + 7H 2 O Cr 2 O 72- + 14H + + 6e - E = -1.33 V Reaction = Al + 3Cr 3+ Al 3+ + 3Cr 2+ c) A tin strip is immersed in nitric acid. Sn Sn 2+ + 2e - E = +0.14 V 2H + + 2e - H 2 E = 0.00 V H 2 O O 2 + 4H + + 4e - E = -1.23 V NO - 3 + 4H + + 3e - NO + 2H 2 O E = +0.96 V 2H 2 O + 2e - H 2 + 2OH - E = -0.83 V Reaction = 3Sn + 2NO 3 - + 8H + 3Sn 2+ + 2NO + 4H 2 O d) A copper rod is immersed in a 1 M solution of HCl, through which is bubbled O 2 at 101.3 kpa. Cu Cu 2+ + 2e - E = -0.34 V O 2 + 4H + + 2e - H 2 O 2 E = +0.70 V Cu Cu + + 1e - E = -0.52 V O 2 + 2H + + 4e - H 2 O E = +1.23 V 2Cl - Cl 2 + 2e - E = -1.36 V 2H 2 O + 2e - H 2 + 2OH - E = -0.83 V H 2 O O 2 + 4H + + 4e - E = -1.23 V 2H + + 2e - H 2 E = 0.00 V Reaction = 2Cu + O 2 + 4H + 2Cu 2+ + H 2 O 8. Predict the overall reactions that you would expect when the following are electrolyzed. Calculate the minimum voltage that must be applied to the cell. a) 1 M NaI 2I - I 2 + 2e - E = -0.54 V Na + + 1e - Na E = -2.71 V Reaction = 2I - + 2H 2 O I 2 + H 2 + 2OH - E = (-0.54 V) + (-0.83 V) = -1.37 V Minimum applied voltage = 1.37 V

b) 1M KCl 2Cl - Cl 2 + 2e - E = -1.36 V K + + 1e - K E = -2.91 V Reaction = H 2 O + 4H 2 O O 2 + 4H + + H 2 + 4OH - = Note 4H + + 4OH - equals 4H 2 O = H 2 O O 2 + H 2 E = (-1.23 V) + (-0.83 V) = -2.06 V Minimum applied voltage = 2.06 V c) 1 M CuF 2 2F - F 2 + 2e - E = -2.87 V Cu 2+ + 1e - Cu + E = +0.16 V H 2 O O 2 + 4H + + 4e - E = -1.23 V Cu 2+ + 2e - Cu E = +0.34 V 2H 2 O + 2e - H 2 + 2OH - E = -0.83 V Reaction = H 2 O + 2Cu 2+ O 2 + 4H + + 2Cu E = (-1.23 V) + (+0.34 V) = -0.89 V Minimum applied voltage = 0.89 V d) 1 M CoI 2 and 1 M HCl 2Cl - Cl 2 + 2e - E = -1.36 V Co 2+ + 2e - Co E = -0.28 V 2I - I 2 + 2e - E = -0.54 V 2H + + 2e - H 2 E = 0.00 V Reaction = 2I - + 2H + I 2 + H 2 E = (-0.54 V) + (+0.00 V) = -0.54 V Minimum applied voltage = 0.54 V 9. For the electrolysis of molten lithium bromide, write: a) the half-reaction that occurs at the negative electrode The negative electrode is the cathode in an electrolytic cell. Reduction occurs at the cathode. Li + + e - Li b) the half-reaction that occurs at the positive electrode The positive electrode is the anode in an electrolytic cell. Oxidation occurs at the anode. 2Br - Br 2 + 2e - c) the net ionic equation for the overall cell reaction 2Li + + 2Br - 2Li + Br 2

10. How many moles of Cu can be produced by passing a 2.50 A current through 1 M CuSO 4 for 4000 s? 11. How many coulombs are required to produce 0.325 g of I 2 by electrolysis of an aqueous solution of KI? 12. A 12.5 A current was passed through a 1.00 M solution of lead(ii) nitrate for 1.00 h using inert electrodes. How much will the mass of the cathode have increased after 1.00 h? 13. What quantity of electricity (in coulombs) is required to produce 1.00 kg of Br 2 by electrolysis of KBr solution? 14. What mass of tin can be plated out of a Sn 2+ solution if we pass 3.00 A for 6.00 h? 15. A certain amount of electricity deposits 45.0 g of silver from a solution containing silver ions. What mass of copper will this amount of electricity deposit from a solution of copper(ii) sulfate?

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