Question

The following reaction takes place at $298\, K$ in an electrochemical cell involving two metals $A$ and $B,\,A_{2}+ ( aq) +B (s) \rightarrow B^{2+} (aq) + A(s)$ with $[A^{2+} ] = 4 x 10^{-3 }M$ and $[B^{2+}] = 2\times 10^{-3 }M$ in the respective half-cells, the cell $EMF$ is $1.091\, V.$ 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,
$A^{2+}(a q)+B(s) \longrightarrow B^{2+}(a q)+A(s)$
According to Nernst equation,
$E_{\text {cell }}=E_{\text {cell }}^{\circ}-\frac{0.0591}{n} \log \frac{[P]}{[R]}$
$1.091=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{\left[B^{2+}\right]}{\left[A^{2+}\right]}$
$1.091=E_{\text {cell }}^{\circ}-\frac{0.0591}{2} \log \frac{2 \times 10^{-3}}{4 \times 10^{-3}}$
$E_{\text {cell }}^{\circ} =1.099$
$\Delta G =-2.303\, R T\, \log K_{\text {eq }}$ ...(i)
Also, $\Delta G =-n F E^{\circ}$ ...(ii)
Equating Eqs. (i) and (ii)
$-n F E^{\circ} =-2.303\, R T\, \log K_{ eq }$
$=-2 \times 96500 \times 1.099$
$=-2.303 \times 8.314 \times 298 \log K_{ eq }$
$\log K_{ eq } =\frac{2 \times 96500 \times 1.099}{2.303 \times 8.314 \times 298}$
$\log K_{ eq } =37.17$
$\therefore K_{ eq }=2 \times 10^{37}$