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Reduction half-reactions with corresponding standard half-cell potentials are shown below. Zn2+(aq) + 2 e- ? Zn(s)...

Reduction half-reactions with corresponding standard half-cell potentials are shown below.
Zn2+(aq) + 2 e- ? Zn(s) E

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Answer #1

Since Aluminium is above Zinc in the potentials:

Al is oxidized (anode) and Zn+2 is reduced (cathode)
The e-s leave the Al electrode and flow to the Zn electrode.

Oxidation: 2(Al0 ? Al+3 + 3e-)

Recution: 3(Zn+2 + 2e- ? Zn0)

Overall reaction: 2Al0 + 3Zn+2 + 6e- ? 6e- + 2Al+3 + 3Zn0

E cell = E ox + E red

E0cell = (1.66v) + (-0.76v*) = +0.90v

NOTE that the sign must be inverted *

Always change the sign of the reduction half-cell. All voltaic cells have positive voltage values.

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