Question

Find the equilibrium constant for the reaction $I{n}^{2+}+C{u}^{2+}\to I{n}^{3+}+C{u}^{+}$ at 298 K. Given, $E{{}^o}_{C{u}^{2+}/C{u}^{+}}=0.15V,E{{}^o}_{I{n}^{3+}/I{n}^{+}}=-0.42V$ $E{{}^o}_{I{n}^{2+}/I{n}^{+}}=-0.40V$ is approximately:

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

Answer 1

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

The relationship between standard gibbs energy change for a reaction and its cell potential is ${\Delta G}^o=-nF{E}_{cell}^o$.

For indium electrodes,

${\Delta G}_{I{n}^{3+}|I{n}^{+}}^o={\Delta G}_{I{n}^{3+}|I{n}^{2+}}^o+{\Delta G}_{I{n}^{2+}|I{n}^{+}}^o$

$-nF{E}_{I{n}^{3+}|I{n}^{+}}^o=-nF{E}_{I{n}^{3+}|I{n}^{2+}}^o-nF{E}_{I{n}^{3+}|I{n}^{2+}}^o$

$2E_{I{n}^{3+}|I{n}^{+}}^o=E_{I{n}^{3+}|I{n}^{2+}}^o+E_{I{n}^{3+}|I{n}^{2+}}^o$

Substitute values in the above equation.

$2\times (-0.42V)=E_{I{n}^{3+}|I{n}^{2+}}^0-0.40V$

$E_{I{n}^{3+}|I{n}^{2+}}^o=-0.44V$

Also,

$E_{cell}^o=E_{C{u}^{2+}|C{u}^{+}}^o-E_{I{n}^{3+}|I{n}^{+}}^o$

$E_{cell}^o=0.15V-(-0.44V)=0.59V$

The relationship between equilibrium constant for a reaction and its equilibrium constant is $E_{cell}^o=\frac{0.0592}{n}log{K}_c.$

Substitute values in the above equation.

$0.59V=\frac{0.059}{1}log{K}_c$

$log{K}_c=10$

$K_c\cong {10}^{10}.$