#Chem 162-2008 homework 13th, 14th, 15th week

#Chem 162-2008 homework 13th, 14th, 15th week - CHAPTER 18...

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Electrolysis 75. Write a net ionic equation for the expected reaction when the electrolysis of Cu(NO 3 ) 2 (aq) is conducted using (a) a copper anode and a copper cathode, (b) an inert platinum anode and an iron cathode, (c) inert platinum for both electrodes. Cu 2+ + 2e - → Cu +0.340 NO 3 - + 4H + + 3e - → NO(g) + 2H 2 O +0.956 Fe 2+ + 2e - → Fe -0.440 H 2 O 2 + 2H + + 2e - → 2H 2 O +1.763 O 2 + 4H + + 4e - → 2H 2 O +1.229 2H 2 O + 2e - → H 2 + 2OH - -0.828 (a) If it was a galvanic cell, not an electrolysis cell, then the reaction would be: NO 3 - + 4H + + 3e - → NO(g) + 2H 2 O +0.956 Cu → Cu 2+ + 2e - -0.340 A plausible electrolytic reaction of low negative voltage would be: NO 3 - + 4H + + 3e - → NO(g) + 2H 2 O +0.956 2H 2 O → O 2 + 4H + + 4e - -1.229 However, a reaction of even lower negative voltage would be, where a small amount of battery help is necessary to drive the electrons through the wire: Anode: Cu → Cu 2+ + 2e - -0.340V Cathode: Cu 2+ + 2e - → Cu +0.340V Net reaction. The Cu anode gets thinner, while the Cu cathode gets thicker. (b) an inert platinum anode and an iron cathode. Cu 2+ + 2e - → Cu +0.340 NO 3 - + 4H + + 3e - → NO(g) + 2H 2 O +0.956 Fe 2+ + 2e - → Fe -0.440 H 2 O 2 + 2H + + 2e - → 2H 2 O +1.763 O 2 + 4H + + 4e - → 2H 2 O +1.229 2H 2 O + 2e - → H 2 + 2OH - -0.828 Platinum electrodes are inert. Iron cathodes are inert, because iron does not capture electrons. So the only thing that is left to react is what is in the solution. A plausible electrolytic reaction of the least negative voltage would be: 4(NO 3 - + 4H + + 3e - → NO(g) + 2H 2 O) +0.956 3(2H 2 O → O 2 + 4H + + 4e - ) -1.229 4NO 3 - + 4H + → 2H 2 O + 4NO + 3O 2 -0.273V The Solutions Manual shows the answer as: 2(Cu 2+ + 2e - → Cu) +0.340 2H 2 O → O 2 + 4H + + 4e - -1.229 2Cu 2+ + 2H 2 O → 2Cu + O 2 + 4H + -0.889V That’s because they consider information such as overvoltage of the NO, which is not provided in this problem. The General Chemistry student is not responsible for this. Ignore the Solution Manual’s answer. 1
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(c) inert platinum for both electrodes Cu 2+ + 2e - → Cu +0.340 NO 3 - + 4H + + 3e - → NO(g) + 2H 2 O +0.956 Fe 2+ + 2e - → Fe -0.440 H 2 O 2 + 2H + + 2e - → 2H 2 O +1.763 O 2 + 4H + + 4e - → 2H 2 O +1.229 2H 2 O + 2e - → H 2 + 2OH - -0.828 Platinum electrodes are inert. A plausible electrolytic reaction of low negative voltage would be: 4(NO 3 - + 4H + + 3e - → NO(g) + 2H 2 O) +0.956 3(2H 2 O → O 2 + 4H + + 4e - ) -1.229 4NO 3 - + 4H + → 2H 2 O + 4NO + 3O 2 -0.273V The Solutions Manual shows the answer as: 2(Cu 2+ + 2e - → Cu) +0.340 2H 2 O → O 2 + 4H + + 4e - -1.229 2Cu 2+ + 2H 2 O → 2Cu + O 2 + 4H + -0.889V That’s because they consider information such as overvoltage of the NO, which is not provided in this problem. The General Chemistry student is not responsible for this. Ignore the Solution Manual’s answer. 77.
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This note was uploaded on 09/23/2009 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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#Chem 162-2008 homework 13th, 14th, 15th week - CHAPTER 18...

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