1. Tardigrade
  2. Question
  3. Chemistry
  4. The EMF of a cell corresponding to the reaction: Zn(s)+2H+(aq) → Zn2+(0.1 M)+H2(g)(1 atm) is 0.28 volt at 15°C. The pH of the solution at the hydrogen electrode is (Given: E°Zn2+/Zn=-0.76 volt; E°H+/H2=0 volt)

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

Electrochemistry

Solution:

The half-cell reactions are
$Zn \rightarrow Zn^{2+} +2e^-$
$2H^+ +2e^- \rightarrow H_2$
We know that
$E_{Zn/Zn^{2+}}=E^{\circ}_{Zn/Zn^{2+}}-\frac{RT}{nF} ln \frac{\left[Zn^{2+}\right]}{\left[Zn\right]}$
$\therefore E_{Zn/Zn^{2+}}=0.76-\frac{2.303\times8.314\times298}{2\times96500}log \frac{0.1}{1}$
$=0.76-\left(-0.03\right)=0.79\,V$
Also
$E_{H^{+}/H_{2}}=E^{\circ}_{H^{+}/H_{2}}-\frac{RT}{nF}-\frac{RT}{nF} ln \frac{\left[H_{2}\right]}{\left[H^{+}\right]^{2}}$
$=0-\frac{2.303\times8.314\times298}{2\times96500}log \frac{1}{\left[H^{+}\right]^{2}}$
$=0.0591\,log \left[H^{+}\right]$
$=-0.0591\,pH$
Since $E_{cell}=E_{Zn/Zn^{2+}}+E_{H^{+}/H_{2}}$
or $0.28=0.79-0.0591\,pH$
or $pH=\frac{0.79-0.28}{0.0591}=\frac{0.51}{0.0591}$
$=8.62$