Electrochemistry 2 Question 50
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 $0.28 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$.
Write the half-cell reactions and calculate the $\mathrm{pH}$ of the solution at the hydrogen electrode.
$$ E^{\circ}\left(\mathrm{Zn}^{2+} / \mathrm{Zn}\right)=-0.76 \mathrm{~V} E_{\mathrm{H}^{+} / \mathrm{H}_{2}}^{\circ}=0 $$
$(1986,4 M)$
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Solution:
- At anod
$$ \mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+2 e^{-} E^{\circ}=0.76 \mathrm{~V} $$
At cathode $2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}{2}(g) \quad E^{\circ}=0.00 \mathrm{~V}$ $\Rightarrow$ For $\quad \mathrm{Zn}+2 \mathrm{H}^{+} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{H}{2}(g) E^{\circ}=0.76 \mathrm{~V}$
$$ \begin{array}{rlrl} E & =E^{\circ}-\frac{0.0592}{2} \log \frac{\left[\mathrm{Zn}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^{2}} \ \Rightarrow & \frac{2\left(E-E^{\circ}\right)}{0.0592} & =-\log \left[\mathrm{Zn}^{2+}\right]-2 \log \frac{1}{\left[\mathrm{H}^{+}\right]} \ \Rightarrow & -16.2 & =-\log (0.1)-2 \mathrm{pH} \ \Rightarrow & \mathrm{pH} & =\frac{1+16.2}{2}=8.6 \end{array} $$
