Electrochemistry Result Question 49
49. The emf of a cell corresponding to the reaction.
$ \mathrm{Zn}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Zn}^{2+}(0.1 \mathrm{M})+\mathrm{H}_2,(g, 1 \mathrm{~atm}) $
is 0.28 V at $25^{\circ} \mathrm{C}$.
Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode.
$ E^{\circ}\left(\mathrm{Zn}^{2+} / \mathrm{Zn}\right)=-0.76 \mathrm{~V} E_{\mathrm{H}^{+} / \mathrm{H}_2}^{\circ}=0 $
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Answer:
Correct Answer: 49. $(8.6)$
Solution:
- At anode: $\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 e^{-} \quad E^{\circ}=0.76 \mathrm{~V}$
At cathode: $2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_2 (\mathrm{~g}) \quad E^{\circ}=0.00 \mathrm{~V}$
$\Rightarrow \quad$ For $\quad \mathrm{Zn}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{H}_2(g) \quad E^{\circ}=0.76 \mathrm{~V}$
$
\begin{aligned}
& E=E^{\circ}-\frac{0.0592}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^2} \\
& \Rightarrow \quad \frac{2\left(E-E^{\circ}\right)}{0.0592}=-\log \left[\mathrm{Zn}^{2+}\right]-2 \log \frac{1}{\left[\mathrm{H}^{+}\right]} \\
& \Rightarrow \quad-16.2=-\log (0.1)-2 \mathrm{pH} \\
& \Rightarrow \quad \mathrm{pH}=\frac{1+16.2}{2}=8.6
\end{aligned}
$
BETA
