Question

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

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

Answer 1

Answer: (B)

The half-cell reactions are
$Zn\to Z{n}^{2+}+2e^{-}$
$2H^{+}+2e^{-}\to H_2$
We know that
$E_{Zn/Z{n}^{2+}}=E_{Zn/Z{n}^{2+}}^{\circ }-\frac{RT}{nF}ln\frac{\left[Z{n}^{2+}\right]}{\left[Zn\right]}$
$\therefore E_{Zn/Z{n}^{2+}}=0.76-\frac{2.303\times 8.314\times 298}{2\times 96500}log\frac{0.1}{1}$
$=0.76-\left(-0.03\right)=0.79V$
Also
$E_{H^{+}/H_2}=E_{H^{+}/H_2}^{\circ }-\frac{RT}{nF}-\frac{RT}{nF}ln\frac{\left[H_2\right]}{{\left[H^{+}\right]}^2}$
$=0-\frac{2.303\times 8.314\times 298}{2\times 96500}log\frac{1}{{\left[H^{+}\right]}^2}$
$=0.0591log\left[H^{+}\right]$
$=-0.0591pH$
Since $E_{cell}=E_{Zn/Z{n}^{2+}}+E_{H^{+}/H_2}$
or $0.28=0.79-0.0591pH$
or $pH=\frac{0.79-0.28}{0.0591}=\frac{0.51}{0.0591}$
$=8.62$