For two divalent metals M and N, the cell $M ∣ M X^{2 +} (0.01 M) ∣∣ N X^{2 +} (0.1 M) ∣ N$ has EMF $0.443 V$ at 298 K. The value of $lo g_{10} K$ for the reaction $M + N X^{2 +} M X^{2 +} + N$ at 298 K is (nearest integer; take $2.303 RT / F = 0.0591 V$ ).

Numerical answer type, tolerance 0.3.

Answer: 14

First recover

E^{\circ}

Q = [M X^{2 +}] / [N X^{2 +}] = 0.01/0.1 = 0.1

, so

E = E^{\circ} - \frac{0.0591}{2} lo g 0.1 = E^{\circ} + 0.02955

. Hence

E^{\circ} = 0.443 - 0.0296 = 0.4135 V

. At equilibrium

E = 0

, so

lo g K = \frac{n E ^{\circ}}{0.0591} = \frac{2 \times 0.4135}{0.0591} = 13.99 \approx 14