Question

Find the equilibrium constant for the reaction: $I{n}^{2+}+C{u}^{2+}\to I{n}^{3+}+C{u}^{+}$ at $298$K.
Given: $E_{C{u}^{2+}/C{u}^{+}}^0=0.15$; $E_{I{n}^{2+}/I{n}^{+}}^0=-0.40V$, $E_{I{n}^{3+}/I{n}^{+}}^0=-0.42$V.

Answer 1

(i) $C{u}^2++e^{-}⟶C{u}^{+};E_{C{u}^2+/C{u}^{+}}^o=0.15V$

(ii) $I{n}^{2+}+e^{-}I{n}^{+};E_{I{n}^{2+}/In}^o=-0.40V$

$I{n}^{3+}+2e^{-}⟶I{n}^{+};E_{I{n}^{3+}/I{n}^{+}}^o=-0.42V$

(iii) $I{n}^{+}⟶I{n}^{3+}+2e^{-};E_{I{n}^{+}/I{n}^{3+}}^o=+0.42V$

Adding )i), (ii) & (iii), we get,

$I{n}^{2+}+C{u}^{2+}⟶I{n}^{3+}+C{u}^{+};E^o=0.15-0.40+0.42$

$E^o=0.17V$

Now, $\Delta G^o=-RTln{K}_{eq}$

$\therefore -nF{E}^o=-RTln{K}_{eq}$

$\therefore ln\left(K_{eq}\right)=\frac{nF{E}^o}{RT}$

$=\frac{2\times 96500\times 0.17}{8.314\times 298}$

$\Rightarrow ln{K}_{eq}=13.24$

$\therefore K_{eq}=e^{13.24}$

$\therefore K^{eq}=562417.6$

Therefore, equilibrium constant is $562417.6$