Answer:
E° = 0.29 V
Explanation:
Let's consider the following redox reaction.
Sn(s) + Sn⁴⁺(aq) → 2 Sn²⁺(aq)
We can identify both half-reactions:
Reduction (cathode): Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V
Oxidation (anode): Sn(s) → Sn²⁺(aq) + 2 e⁻ Ered = -0.14 V
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = 0.15 V - (-0.14 V) = 0.29 V