Question

The following reaction takes place at $298K$ in an electrochemical cell involving two metals $A$ and $B,A_2+(aq)+B(s)\to B^{2+}(aq)+A(s)$ with $[A^{2+}]=4x10^{-3}M$ and $[B^{2+}]=2\times 10^{-3}M$ in the respective half-cells, the cell $EMF$ is $1.091V.$ The equilibrium constant of the reaction is closest to

• A. $4\times 10^{36}$
• B. $2\times 10^{37}$
• C. $2\times 10^{34}$
• D. $4\times 10^{37}$

Answer 1

Answer: (B)

For the reaction,
$A^{2+}(aq)+B(s)⟶B^{2+}(aq)+A(s)$
According to Nernst equation,
$E_{\text{cell }}=E_{\text{cell }}^{\circ }-\frac{0.0591}{n}\log \frac{[P]}{[R]}$
$1.091=E_{\text{cell }}^{\circ }-\frac{0.0591}{2}\log \frac{\left[B^{2+}\right]}{\left[A^{2+}\right]}$
$1.091=E_{\text{cell }}^{\circ }-\frac{0.0591}{2}\log \frac{2\times 10^{-3}}{4\times 10^{-3}}$
$E_{\text{cell }}^{\circ }=1.099$
$\Delta G=-2.303RT\log K_{\text{eq }}$ ...(i)
Also, $\Delta G=-nF{E}^{\circ }$ ...(ii)
Equating Eqs. (i) and (ii)
$-nF{E}^{\circ }=-2.303RT\log K_{eq}$
$=-2\times 96500\times 1.099$
$=-2.303\times 8.314\times 298\log K_{eq}$
$\log K_{eq}=\frac{2\times 96500\times 1.099}{2.303\times 8.314\times 298}$
$\log K_{eq}=37.17$
$\therefore K_{eq}=2\times 10^{37}$