Question

The equilibrium constant of a $2$ electron redox reaction at $298K$ is $3.8\times {10}^{-3}$. The cell potential $E^{\circ }$ (in $V$) and the free energy change $△G^{\circ }$ (in $kJmo{l}^{-1})$ for this equilibrium, respectively, are:

• A. $-0.071,-13.8$
• B. $-0.071,13.8$
• C. $0.71,-13.8$
• D. $0.071,-13.8$

Answer 1

Answer: (B)

$-0.071,13.8$

$K_{eq}=3.8\times {10}^{-3}$

$n=$ no. of electrons involved in the reaction $=2$

Now,

$\Delta G^{\circ }=-RTlnK=-8.314\times 298\times ln(3.8\times {10}^{-3})=13806J{mol}^{-1}=13.8kJ{mol}^{-1}$

Now,

$\Delta G^{\circ }=-nF{E}^{\circ }$

$\therefore E^{\circ }=-\frac{\Delta G^{\circ }}{nF}=-\frac{3.8\times {10}^3}{2\times 96500}=-0.071V$