Electrochemistry Result Question 42
42. Calculate the equilibrium constant for the reaction
$ \mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \rightleftharpoons \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+} $
Given, $E^{\circ}\left(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\right)=1.44 \mathrm{V}, E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.68 \mathrm{V}$
(1997, 2M)
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Answer:
Correct Answer: $(6.88 \times 10^{12})$
Solution:
$ \mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \rightleftharpoons \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+} $
$E^{\circ} =E^{\circ}\left(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\right)-E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right) $
$ =1.44-0.68=0.76 \mathrm{~V} $
$\because \quad E^{\circ} =0.0592 $ $\log$ $ K $
$\Rightarrow \log K =\frac{E^{\circ}}{0.0592}=\frac{0.76}{0.0592}=12.83 $
$\Rightarrow K =6.88 \times 10^{12} $
BETA
