Question

The EMF of the cell $M|M^{n^{+}}\left(0.02\text{M}\right)||H^{+}\left(1\text{M}\right)|H_2\text{(g)}\left(1\text{atm}\right),Pt$ at ${25}^{\circ }\text{C}$ is $0.81\text{V}$. Calculate the valency of the metal if the standard oxidation potential of the metal is $0.76\text{V}$.
(Given, $\log 4=0.6$)

• A. $4$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is B $2$

$M\to M^{n+}+n{e}^{-}$ ... (1)
$2e^{-}+2H^{+}\to H_2$ ... (2)
If we multiply equation $1$ by $2$ and $2$ by $n$ and add both equation then we get
$2M+2n{H}^{+}\to 2M^{n+}+n{H}_2$
$E=E^{\circ }-\frac{0.059}{n}\log \frac{[M^{n+}{]}^2}{[H^{+}{]}^{2n}}$
$0.81=(0.76+0)-\frac{0.059}{2n}\log \frac{[0.02{]}^2}{[1{]}^{2n}}$
$n=\frac{0.059}{2\times 0.05}\log 4\times {10}^{-4}$
$n=-0.59(-4+0.6)$
$n=2$
Therefore, valency of metal is $2$.