Question

The standard e.m.f. of a cell, involving two electrons transfer is found to be $1.182V$ at $25^{\circ }C$. The equilibrium constant of the reaction is given as $1.0\times 10^X$. The value of ' $X$ ' is________.

Answer 1

Answer: 40

Standard e.m.f. of the cell $\left(E^{\circ }\right)$
$=\frac{2.303RT}{nF}{\log }_{10}K$
$=\frac{0.0591}{n}{\log }_{10}K$
$\therefore 1.182=\frac{0.0591}{2}{\log }_{10}K$
$\therefore \frac{1.182\times 2}{0.0591}={\log }_{10}K$
$\therefore$ Equilibrium constant $(K)$
$=10^{\left(\frac{1.182\times 2}{0.0591}\right)}=1.0\times 10^{40}$
Hence, the value of ' $X$ ' is $40$.