Question

The standard e.m.f. of a cell, involving two electrons transfer is found to be $1.182\, V$ 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

Standard e.m.f. of the cell $\left( E ^{\circ}\right)$
$=\frac{2.303 R T}{n F} \log _{10} K$
$=\frac{0.0591}{n} \log _{10} K$
$\therefore 1.182=\frac{0.0591}{2} \log _{10} K$
$\therefore \frac{1.182 \times 2}{0.0591}=\log _{10} K$
$\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 $40$.