Question

The equilibrium constant for the reaction is :
$In ^{2+}+ Cu ^{2+} \longrightarrow In ^{3+}+ Cu ^{+}$at $298\, K$
Given, $E_{ Cu ^{2+} / Cu ^{+}}^{\circ}=0.15 \,V$
$E_{\text{In}^{3+} / In ^{+}}^{\circ}=-0.42 \,V , E_{\text{In}^{2+} / In ^{+}}^{\circ}=-0.40 \,V$

• A. $10^{5}$
• B. $10^{3}$
• C. $10^{10}$
• D. $10^{6}$

Answer 1

Answer: (C)

Given, $\text{In}^{3+}+2 e ^{-} \longrightarrow \operatorname{In}^{+} ; E_{1}^{\circ}=-0.42 V \ldots$ (i)
$In ^{2+}+ e ^{-} \longrightarrow \text { In }^{+} ; E_{2}^{\circ}=-0.40 V \ldots \text { (ii) }$
By subtracting. Eq. (ii) from Eq. (i) a third half-cell reaction can be obtained as:

$In ^{3+}+ e ^{-} \longrightarrow \operatorname{In}^{2+} ; E_{3}^{\circ}=-0.44\, V$
For the reactions: $Cu ^{2+}+ e ^{-} \longrightarrow Cu ^{+} ; E_{4}^{\circ}=0.15$
$In ^{2+} \longrightarrow \operatorname{In}^{3+}+ e ^{-} ; E_{3}^{\circ}=+0.44$
The net redox change: $Cu ^{2+}+ In ^{2+} \longrightarrow Cu ^{+}+ In ^{3+}$;
$E_{\text {cell }}^{\circ}=E_{4}^{\circ}+E_{3}^{\circ}=0.15+0.44=0.59 V$
Also. $E_{\text {cell }}^{\circ}=\frac{0.059}{1} \log K_{c}$
$0.59=\frac{0.059}{1} \log K_{c} $
$ \therefore K_{c}=10^{10}$