Question

The EMF of a cell corresponding to the reaction:
$Z{n}_{\left(s\right)}+2H_{\left(aq\right)}^{+}\to Z{n}^{2+}\left(0.1M\right)+H_{2\left(g\right)}$($1$ atm) is $0.28$ volt at ${15}^o$C.
What is the pH of the solution at the hydrogen electrode is?(Given: $E_{Z{n}^{2+}/Zn}^o=-0.76$volt; $E_{H^{+}/H_2}^o=0$volt)

• A. $7.05$
• B. $8.62$
• C. $8.75$
• D. $9.57$

Answer 1

The correct option is B $8.62$

$Zn+2H^{+}⟶Z{n}^{2+}+H_2$

$E_{cell}^o=E_{H^{+}/H_2}-E_{Z{n}^{2+}/Zn}^o=0.76$ $V$

From nernst equation,

$E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log \left[\frac{\left(Z{n}^{2+}\right)}{(H^{+}{)}^2}\right]$

$⟹0.28=0.76-\frac{0.0591}{2}\log \left[\frac{0.1}{(H^{+}{)}^2}\right]$

$⟹\frac{0.48\times 2}{0.0591}=\log \left[\frac{0.1}{(H^{+}{)}^2}\right]$

$⟹1.86\times {10}^{16}=\frac{0.1}{(H^{+}{)}^2}$

$⟹1.86\times {10}^{16}=\frac{0.1}{(H^{+}{)}^2}$

$⟹(H^{+}{)}^{+2}=5.37\times {10}^{-18}$

$⟹\left(H^{+}\right)=2.31\times {10}^{-9}$

$⟹pH=8.62$