Electrochemistry 2 Question 45
44. An excess of liquid mercury is added to an acidified solution of $1.0 \times 10^{-3} \mathrm{M} \mathrm{Fe}^{3+}$. It is found that $5 %$ of $\mathrm{Fe}^{3+}$ remains at equilibrium at $25^{\circ} \mathrm{C}$. Calculate $E^{\circ}\left(\mathrm{Hg}^{2+} / \mathrm{Hg}\right)$ assuming that the only reaction that occurs is
$$ \begin{aligned} & 2 \mathrm{Hg}+2 \mathrm{Fe}^{3+} \longrightarrow \mathrm{Hg}_{2}^{2+}+2 \mathrm{Fe}^{2+} \ & \text { Given, } \quad E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.77 \mathrm{~V} \end{aligned} $$
$(1995,4 \mathrm{M})$
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Solution:
- For reaction,
$2 \mathrm{Hg}+2 \mathrm{Fe}^{3+} \rightleftharpoons \mathrm{Hg}_{2}^{2+}+2 \mathrm{Fe}^{2+}$
Initial : $\quad 10^{-3} \mathrm{M} \quad 0 \quad 0$
Equilibrium : $\quad 5 \times 10^{-5} \quad 4.75 \times 10^{-4} 9.5 \times 10^{-4}$
$$ \begin{aligned} K & =\frac{\left[\mathrm{Fe}^{2+}\right]^{2}\left[\mathrm{Hg}_{2}^{2+}\right]}{\left[\mathrm{Fe}^{3+}\right]^{2}} \ & =\frac{\left(9.5 \times 10^{-4}\right)^{2}\left(4.75 \times 10^{-4}\right)}{\left(5 \times 10^{-5}\right)^{2}}=0.17 \end{aligned} $$
$\because \quad E^{\circ}=\frac{0.0592}{2} \log K=-0.0226 \mathrm{~V}$
$$ =E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)-E^{\circ}\left(\mathrm{Hg}_{2}^{2+} / \mathrm{Hg}\right) $$
$\Rightarrow E^{\circ}\left(\mathrm{Hg}_{2}^{2+} / \mathrm{Hg}\right)=0.77+0.0226=0.7926 \mathrm{~V}$
