Given E deg(Cu^2+/Cu) = +0.34 V and E deg(Cu+/Cu) = +0.52 V at 298 K. The disproportionation 2Cu+ -> Cu^2+ + Cu has equilibrium constant K. Find log_10 K (nearest integer). Take (2.303RT)/(F) = 0.0591 V.

JEE Advanced levelNumerical answerDifficulty 3 of 3ec01

Given $E^{\circ} (Cu X^{2 +} / Cu) = + 0.34 V$ and $E^{\circ} (Cu X^{+} / Cu) = + 0.52 V$ at 298 K. The disproportionation $2 Cu X^{+} Cu X^{2 +} + Cu$ has equilibrium constant $K$ . Find $lo g_{10} K$ (nearest integer). Take $\frac{2.303 RT}{F} = 0.0591 V$ .

Numerical answer type.

First get

E^{\circ} (Cu X^{2 +} / Cu X^{+})

from

Δ G^{\circ}

additivity:

2 (0.34) = 1 (x) + 1 (0.52)

, so

x = 0.16 V

. For the disproportionation, cathode is

Cu X^{+} Cu

(0.52 V) and anode is

Cu X^{+} Cu X^{2 +}

(0.16 V):

E^{\circ} = 0.52 - 0.16 = 0.36 V

with

n = 1

lo g K = \frac{n E ^{\circ}}{0.0591} = \frac{0.36}{0.0591} = 6.09 \approx 6

Cu X^{+}

is unstable in water for this reason.