Chemistry - Electrochemistry Question with Solution | TestHub

Given that,

For a reaction, 

$\mathrm{A}+{\mathrm{B}}^{2+}⟶\mathrm{B}+{\mathrm{A}}^{2+},\mathrm{E}°=0.2955\mathrm{V}$

$$\begin{gathered} \mathrm{T}={25}^{°}\mathrm{C}, \Rightarrow \mathrm{T}=298 \mathrm{K} \\ \mathrm{n}=2 \end{gathered}$$

Using Nernst equation,

$\mathrm{E}°=\frac{2.303\mathrm{RT}}{\mathrm{nF}}{\mathrm{logk}}_{\mathrm{eq}}$

$\mathrm{E}°=\frac{0.0591}{\mathrm{n}}{\mathrm{logk}}_{eq}\left(\mathrm{at} {25}^{\mathrm{o}}\mathrm{C}\right)$

$\Rightarrow 0.2955=\frac{0.0591}{2}\log k_{\mathrm{eq}}$

$k_{\mathrm{eq}}={10}^{10}$

Therefore, the value of ${\mathrm{K}}_{\mathrm{eq}}$is ${10}^{10}$.