Electrochemistry 2 Question 43
42. Calculate the equilibrium constant for the reaction
$$ \begin{gathered} \mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \rightleftharpoons \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+} \ \text { Given, } E^{\circ}\left(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\right)=1.44 \mathrm{~V}, E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right)=0.68 \mathrm{~V} \end{gathered} $$
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Solution:
$$ \begin{array}{rlrl} & \mathrm{Fe}^{2+}+\mathrm{Ce}^{4+} \rightleftharpoons \mathrm{Fe}^{3+}+\mathrm{Ce}^{3+} \ E^{\circ} & =E^{\circ}\left(\mathrm{Ce}^{4+} / \mathrm{Ce}^{3+}\right)-E^{\circ}\left(\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}\right) \ & & =1.44-0.68=0.76 \mathrm{~V} \ & & E^{\circ} & =0.0592 \log K \ \Rightarrow & & \log K & =\frac{E^{\circ}}{0.0592}=\frac{0.76}{0.0592}=12.83 \ \Rightarrow & & K & =6.88 \times 10^{12} \end{array} $$
