Question

For an electrochemical cell
\(\mathrm{Sn(s)\,|\,Sn^{2+}(aq,1M)\,\|\,Pb^{2+}(aq,1M)\,|\,Pb(s)}\)
the ratio \(\frac{[\mathrm{Sn^{2+}}]}{[\mathrm{Pb^{2+}}]}\) when this cell attains equilibrium is _________.
(Given \(E^0_{\mathrm{Sn^{2+}|Sn}} = -0.14\,\mathrm{V}\), \(E^0_{\mathrm{Pb^{2+}|Pb}} = -0.13\,\mathrm{V}\), \(\frac{2.303RT}{F} = 0.06\))

Hint:

At equilibrium E_cell = 0, so E°_cell = (0.06/n) log K with n = 2 and K = [Sn²⁺]/[Pb²⁺].

Answer 1

Correct Answer: 2.15

Cell reaction: Sn(s) + Pb²⁺ → Sn²⁺ + Pb(s), n = 2.
E°_cell = E°_cathode - E°_anode = (-0.13) - (-0.14) = 0.01 V.
At equilibrium E_cell = 0, so 0 = E° - (0.06/2) log Q with Q = K = [Sn²⁺]/[Pb²⁺].
log K = 2(0.01)/0.06 = 1/3 = 0.3333.
K = 10^1/3 = 2.154 ≈ 2.15.
Answer: 2.15