The equilibrium constant for the reaction
Zn(s)+Sn²⁺(aq) ⇌ Zn²⁺(aq)+Sn(s) is 1×10²⁰ at 298 K. The magnitude of standard electrode potential of Sn/Sn²⁺ if E_(Zn²⁺/Zn)^Θ = - 0.76 V is __________ × 10⁻² V. (Nearest integer)
Given : 2.303RT/F=0.059 V

Solution

Chemistry ma’am · 3 steps

Turn K into E°_(cell) first, then peel off the zinc electrode to leave tin's potential.

1. Link K and E°. At equilibrium E_(cell) = 0 and Q = K, so the Nernst equation reduces to a direct link between E°_(cell) and K. Two electrons move from Zn to Sn²⁺, so n = 2.

   E°_(cell) = 0.059/n log K = 0.059/2 × log(10²⁰) = 0.0295 × 20 = 0.59 V

2. Find tin's potential. Zn is oxidised, so it is the anode, and Sn²⁺ is reduced at the cathode. From E°_(cell) = E°_(cathode) - E°_(anode): 0.59 = E°_(Sn²⁺/Sn) - (-0.76), so E°_(Sn²⁺/Sn) = -0.17 V.
3. So. The magnitude is 0.17 V = 17 × 10⁻² V, so the answer is 17.

The common slip: 42. Taking n = 1 instead of 2 gives E°_(cell) = 1.18 V and a tin potential of magnitude 0.42 V, so 42 instead of 17.

The idea to keep. A large K means a large positive E°_(cell), through E°_(cell) = 0.059/n log K. Always count n from the electrons actually transferred.

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