Question

The emf of a cell corresponding to the reaction $Zn\text{(s)}+2H^{+}\text{(aq)}\to Z{n}^{2+}\left(0.1\text{M}\right)+H_2(\text{g},1\text{bar})$ is $0.28\text{V}$ at ${25}^{\circ }\text{C}$. Calculate the pH of the solution at the hydrogen electrode. Given $E_{Z{n}^{2+}|Zn}^{\circ }=-0.763\text{V}$

• A. $8.67$
• B. $6$
• C. $7.20$
• D. $10.35$

Answer 1

The correct option is A $8.67$

The half-cell reactions are:
$\text{Anodic reaction}:Zn\to Z{n}^{2+}+2e^{-}$
$\text{Cathodic reaction}:2H^{+}+2e^{-}\to H_2$
The Nernst equation for the cell reaction is
$E=E^{\circ }-\frac{RT}{2F}ln\frac{\left[Z{n}^{2+}p\right(H_2)}{[H^{+}{]}^2}$
Where, $E^{\circ }=E_{H^{+}|H_2|Pt}^{\circ }-E_{Z{n}^{2+}|Zn}^{\circ }=0-(-0.763\text{V})=0.763\text{V}$
Substituting given data in Nernst equation, we get
$0.28\text{V}=0.763\text{V}-\frac{0.05915\text{V}}{2}log\frac{\left(0.1\right)\left(1\right)}{[H^{+}{]}^2}$
$\Rightarrow 0.28\text{V}=0.763\text{V}-\frac{0.05915\text{V}}{2}\left(log\right(0.1)-log[H^{+}{]}^2)$
$\Rightarrow 0.28\text{V}=0.763\text{V}-\frac{0.05915\text{V}}{2}\left(log\right(0.1)-2log[H^{+}\left]\right)$
$\Rightarrow -log\left[H^{+}\right]=\frac{1}{2}[\frac{2(0.763-0.28)}{0.05915}-log(0.1\left)\right]=\frac{1}{2}[16.33+1]$
$\Rightarrow -log\left[H^{+}\right]=\text{pH}=\frac{17.33}{2}=8.67$
Thus, the pH of the solution is $8.67$.