CHEM 1062 Lecture Notes - Lecture 30: Lead
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Zn2+ + 2e- zn (s) (oxidized) reduction = -0. 76 v. Half cell goes in the opposite direction, which would make it + 0. 76 v. Cu2+ + 2e- cu (s) (reduced) eo. H2 2h+ + 2e- red = 0. 34 v (electrode) Zn (s) zn2+ + 2e- eo. Zn (s) + 2e- + cu2+ cu (s) + 2e- + zn2+ Eo red = 0. 76 v (from zn) + 0. 34 v (cu) red = +0. 76 v. Some standard reductions are positive, some are negative. Eo red = 2. 8 v (very favorable) red = -3. 05 v. Li wants to get rid of the electrons, not accept them which is why the reduction potential is very negative. Or take individual equations (seen above) and put them in correct directions and add anode reduction potentials. Overall equation: pb2+ + mn mn2+ + pb. E = 1. 18 v - 0. 13 v = 1. 05 v. Ag+ + e- ag eo = 0. 80 v.
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