Electrochemistry 2 Question 42
41. Calculate the equilibrium constant for the reaction, $2 \mathrm{Fe}^{3+}+3 \mathrm{I}^{-} \rightleftharpoons 2 \mathrm{Fe}^{2+}+\mathrm{I}{3}^{-}$. The standard reduction potentials in acidic conditions are $0.77 \mathrm{~V}$ and $0.54 \mathrm{~V}$ respectively for $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ and $\mathrm{I}{3}^{-} / \mathrm{I}^{-}$couples.
(1998, 3M)
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Solution:
- For $2 \mathrm{Fe}^{3+}+3 \mathrm{I}^{-} \rightleftharpoons 2 \mathrm{Fe}^{2+}+\mathrm{I}_{3}^{-}$
$$ \begin{aligned} E^{\circ} & =E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)-E^{\circ}\left(\mathrm{I}_{3}^{-} / \mathrm{I}^{-}\right) \ & =0.77-0.54=0.23 \mathrm{~V} \ \because \quad E^{\circ} & =\frac{0.0592}{2} \log K \quad(n=2) \end{aligned} $$
$$ \begin{array}{rlrl} & & \log K & =\frac{2 E^{\circ}}{0.0592}=\frac{2 \times 0.23}{0.0592}=7.77 \ \Rightarrow \quad K & =5.89 \times 10^{7} \end{array} $$
