Electrochemistry Result Question 41
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 V and 0.54 V respectively for $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ and $\mathrm{I}_3^{-} / \mathrm{I}^{-}$couples.
(1998, 3M)
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Answer:
Correct Answer: 41. $(5.89 \times 10^7)$
Solution:
- For
$ \quad \begin{aligned} 2 \mathrm{Fe}^{3+} & +3 \mathrm{I}^{-} \rightleftharpoons 2 \mathrm{Fe}^{2+}+\mathrm{I}_3^{-}\\ 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{aligned} \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{aligned} $
BETA
