For the given reactions
Sn²⁺ + 2e⁻ → Sn
Sn⁴⁺ + 4e⁻ → Sn
the electrode potentials are ; E_(Sn²⁺/Sn)ᵒ = - 0.140 V and E_(Sn⁴⁺/Sn)ᵒ = + 0.010 V. The magnitude of standard electrode potential for Sn⁴⁺/Sn²⁺ i.e. E_(Sn⁴⁺/Sn²⁺)ᵒ is _____________ × 10⁻² V. (Nearest integer)

Solution

Chemistry ma’am · 4 steps

Potentials do not add, but nE° values do, so combine the two half-reactions through Δ G°.

1. Use free energy. For any half-reaction, Δ G° = -nFE°. Free energies add when reactions add, but potentials do not, because n changes from one reaction to the next.
2. Combine the reactions. Take (Sn⁴⁺ + 4e⁻ → Sn) minus (Sn²⁺ + 2e⁻ → Sn). This gives Sn⁴⁺ + 2e⁻ → Sn²⁺, so n = 2 for the target reaction.

   -2FE° = -4F(0.010) - [-2F(-0.140)]

3. Solve. Cancel -F from every term: 2E° = 4(0.010) - 2(-0.140) = 0.040 + 0.280 = 0.320. So E° = +0.160 V.
4. So. E°_(Sn⁴⁺/Sn²⁺) = 0.16 V = 16 × 10⁻² V, so the answer is 16.

The common slip: 15. Subtracting the two potentials directly, 0.010 - (-0.140) = 0.150 V, gives 15. This ignores the different electron counts in the two half-reactions.

The idea to keep. When you get a new half-reaction by adding or subtracting others, add or subtract nE°, never E° alone. Then divide by the new n.

Answer from the NTA final answer key. Solution by PrepWiser.