Answers:
(a) +3.160 V; (b) +1.613 V; (c) +2.783 V
Step-by-step explanation:
We must look up the standard reduction potentials for the half-reactions.
The half-reaction on the left-hand side of the cell diagram is an oxidation, so we reverse the sign of E°, and then add the two half-reactions and their E° values.
(a)
Al ⟶ Al³⁺ + 3e⁻; +1.662 V
Au³⁺ + 3e⁻ ⟶ Au; +1.498 V
Al + Au³⁺ ⟶ Al³⁺ + Au; +3.160 V
(b)
Zn ⟶ Zn²⁺ + 2e⁻; +0.762 V
Hg²⁺ + 2e⁻ ⟶ Hg; +0.851 V
Zn + Hg²⁺ ⟶ Zn²⁺ + Hg; +1.613 V
(c)
2×[Li ⟶ Li⁺ + e⁻]; +3.040 V
1×[Ni²⁺ + 2e⁻ ⟶ Ni]; -0.257 V
2Li + Ni²⁺ ⟶ 2Li⁺ + Ni; +2.783 V