Department of Chemistry Chemistry with Problem Solving- II Worksheet 2 Ð Electrochemistry- Answer Key Some constants: Faraday constant F = 96,485 C/mol; R = 8.314 J/mol.K = 0.08206 L.atm/mol.K 1 bar = 0.9869 atm 1. A voltaic cell with a basic aqueous background electrolyte is based on the oxidation of Cd(s) to Cd(OH)2 (s) and the reduction of MnO4- to MnO2 (s). a. Write the half-cell reactions for the cellÕs anode and cathode. b. Write a balanced cell reaction. c. Diagram the cell. Cathode: 3e- + 2 H2O (l) + MnO4Anode: Cd (s) + 2 OH- (aq) MnO2 (s) + 4 OH- (aq) Cd(OH)2 (s) + 2e- Overall balanced equation: 3 Cd (s) + 4 H2O (l) + 2 MnO4- (aq) (aq) 2 MnO2 (s) + 3 Cd(OH)2 (s) + 2 OH- Cd(s)|Cd(OH)2(s)||MnO4-(aq)|MnO2(s)|Pt(s) 2. The positive terminal of a voltaic cell is the cathode. However, the cathode of an electrolytic cell is connected to the negative terminal of a power supply. Explain this difference in polarity. Answer: In a voltaic cell, the electrons are produced at the anode, so a negative charge builds up there; in an electrolytic cell, the electrons are being forced onto the cathode so that it builds up negative charge. The flow of electrons in the outside circuit is reversed in an electrolytic cell compared to a galvanic cell. Page 1 of 3 Department of Chemistry Chemistry with Problem Solving- II Worksheet 2 Ð Electrochemistry- Answer Key 3. A quantity of electricity equal to 9.263 x 104 C passes through a galvanic cell that has a Ni(s) anode. Compute the maximum chemical amount (in moles) of Ni2+ (aq) that can be released into solution. Answer: 0.48 moles of Ni2+ 4. A quantity of electric charge deposits 0.732 g of Ag(s) from an aqueous solution of silver nitrate. When that same quantity of charge is passed through a solution of a gold salt, 0.446 g of Au(s) is formed. What is the oxidation state of the gold ion in the salt? Answer: Oxidation state of gold = +3 5. Nuclear submarines can stay under water nearly indefinitely because they can produce their own oxygen by the electrolysis of water. a. How many liters of O2 can be produced at 298K and 1.00 bar in 1 hour in an electrolytic cell operating at a current of 0.025A? b. Could seawater be used as the source of oxygen in this electrolysis? Why/why not? Answer: (a) 5.77 mL O2 (b) Seawater contains others ions and gases (Cl2, Br2) which could interfere with production of O2 so direct oxidation will not be useful. 6. Calculate the standard reaction Gibbs Free energy for the following cell reactions: (a) 2 Ce4+ (aq) + 3 I- (aq) 2 Ce3+ (aq) + I3- (aq) (b) 6 Fe3+ (aq) + 2 Cr3+ (aq) + 7 H2O (l) + 14 H+ (aq) Eocell = - 1.29 V Eocell = 1.08 V 6 Fe2+ (aq) + Cr2O72- (aq) Answer: (a) - 2.08 x 105 Joules (b) 7.47 x 105 J 7. How much time is required to electroplate 1.50 g of silver from silver nitrate solution by using a current of 13.6 mA? Answer: 27.4 hours (or 9.86 x 104 seconds) Page 2 of 3