Question

Find the equilibrium constant for the reaction ${In}^{+2}+{Cu}^{+2}\to {In}^{+3}+{Cu}^{+}$ at 298k.
Given $E_{{cu}^{+2}}^0/Cu=0.15V,E_{{cu}^{+2}}^0/In=-0.42V,E_{{cu}^{+2}}^0/{In}^{+}=-0.40V$

• A. ${10}^{10}$
• B. ${10}^{11}$
• C. ${10}^{12}$
• D. ${10}^{14}$

Answer 1

The correct option is A ${10}^{10}$

$\begin{array}{c}C{u}^{+2}⟶C{u}^{+};E_1^0=0.15V-\left(1\right)\\ I{n}^{3+}+2e⟶I{n}^{+};E_2^0=-0.42V-\left(2\right)\\ I{n}^{2+}+2e⟶I{n}^{+};E_3^0=-0.40V-\left(3\right)\end{array}$

$\begin{array}{c}\text{By eqs (2)}\&\text{(3) , a third half cell reaction}\\ \text{can be given as}In{}^{3+}+e⟶In{}^{2+}\end{array}$

$\begin{array}{c}1\times F\times E_4^0=E_2^0\times 2F-e_3^0\times 1\times F\\ \text{or}{E}_4^0=2\times (-0.42)-(-0.40)\times 1=-0.44V\end{array}$

$\text{For half reactions, the net redox change as asked is}$

${Cu}^{2+}+{In}^{+}⟶{Cu}^{+}+{In}^{3+}$

$E_{\text{cell}}^0=E_{RP\left(C{u}^{2+}/C{u}^{+}\right)}^0-E_{OP\left(I{n}^{2+}/I{n}^{3+}\right)}^0$

$[E_{\text{OP}I{n}^{2+}/J{n}^{3+}}=-E_{RP}^0=+0.44]$

$\begin{array}{cc}& =0.15+0.44=0.59V\\ \text{Also},E^{\circ }cell& =\frac{0.059}{1}\log K_C\\ 0.59& =\frac{0.059}{1}\log K_C\\ \therefore & K_C={10}^{10}\end{array}$