Electrochemistry Result Question 44
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
$ 2 \mathrm{Hg}+2 \mathrm{Fe}^{3+} \longrightarrow \mathrm{Hg}_2^{2+}+2 \mathrm{Fe}^{2+} $
$ \text{Given}, E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.77 $ $\mathrm{V}$
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Answer:
Correct Answer: 44. $(0.7926 \mathrm{V})$
Solution:
- For reaction,
$ 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 $
$ \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} $
BETA
