Simple question: How can this REDOX rxn. occur?

[zut212]

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Calculate the voltage of this cell:
Zn|1 M Zn2+||1 M Ni2+|Ni

Given:
Zn2+ + 2e- ----> Zn V=-0.763
Ni2+ + 2e- ----> Ni V=-0.250

The way I look at it:
1. One thing gets further oxidized, and the other thing gets reduced.
2. Since both potentials are < 0, this reaction should NOT occur. Furthermore,
3. one of the reactions CAN'T be Zn --> Zn2+ + 2e-, simply because you don't have any Zn to start off with! You only have Zn2+.

However, the solution says that this occurs with a volate of 0.513, which is +0.763 - 0.250.

Please explain.

 

[vsl5]

By IUPAC Conventions, the anode rxn occurs on the left, and cathode reaction occurs on the right. On the right, you have Ni2+ --> Ni as the cathode rxn (reduction at cathode obviously). Therefore Zn is being oxidized since it is the anode. In this case, the potential is the standard cell potential (1M concentrations). In ALL cases, potential of a cell is calculated as E (cathode) - E (anode) = -.250 - (-0.763) = 0.513 V.

OR Reduction potential of the reduction rxn + Oxidation potential of oxidation rxn (-0.250 + 0.763). Both potentials are given as reduction potentials, so to get the oxidation potential, simply flip the negative sign. Do not multiply by any coefficients since E is an intensive property. And yes you do have zinc at the electrode.