For indium, E deg(In^3+/In+) = -0.44 V and E deg(In+/In) = -0.14 V at 298 K. Taking (2.303RT)/(F) = 0.0591 V, which statement about In+ in aqueous solution is correct?

In+ disproportionates (3In+ -> In^3+ + 2In) with K approx 10^10

In+ is stable to disproportionation, K approx 10^-10

In+ disproportionates with K approx 10^5

In+ disproportionates with K approx 10^-10

JEE Advanced levelSingle correct optionDifficulty 3 of 3ec076

For indium, $E^{\circ} (In X^{3 +} / In X^{+}) = - 0.44 V$ and $E^{\circ} (In X^{+} / In) = - 0.14 V$ at 298 K. Taking $\frac{2.303 RT}{F} = 0.0591 V$ , which statement about $In X^{+}$ in aqueous solution is correct?

A $In X^{+}$ disproportionates ( $3 In X^{+} In X^{3 +} + 2 In$ ) with $K \approx 1 0^{10}$
B $In X^{+}$ is stable to disproportionation, $K \approx 1 0^{- 10}$
C $In X^{+}$ disproportionates with $K \approx 1 0^{5}$
D $In X^{+}$ disproportionates with $K \approx 1 0^{- 10}$

A species disproportionates when its reduction potential (to the lower state) exceeds the potential for its oxidation to the higher state. Here

E^{\circ} (In X^{+} / In) = - 0.14 > E^{\circ} (In X^{3 +} / In X^{+}) = - 0.44 V

, so

E_{cell}^{\circ} = - 0.14 - (- 0.44) = + 0.30 V

and the reaction is spontaneous. For

3 In X^{+} In X^{3 +} + 2 In

the electrons transferred are

n = 2

, so

lo g K = 2 \times 0.30/0.0591 = 10.2

K \approx 1 0^{10}

. Option C uses

n = 1