Question

The emf of the cell, $ Pt|C{{e}^{4+}}(90%),C{{e}^{3+}}(10%)| $ normal calomel electrode is 1.464 V at 25?C. Find the value of equilibrium constant of the reaction: $ 2C{{e}^{3+}}+2{{H}^{+}}\xrightarrow[{}]{{}}2C{{e}^{4+}}+{{H}_{2}} $ The electrode potential of the normal calomel electrode is 0.28 V.

• A. $ 2.38\times {{10}^{38}} $
• B. $ 1.08\times {{10}^{42}} $
• C. $ 1.67\times {{10}^{39}} $
• D. $ 3.24\times {{10}^{44}} $

Answer 1

Answer: (A)

$ {{E}_{cell}}={{E}_{c}}-{{E}_{a}}=1.464V $ $ {{E}_{C{{e}^{4+}}/C{{e}^{2+}}}}=E_{C{{e}^{4+}}/C{{e}^{3+}}}^{o}-\frac{0.059}{1}\log \frac{[C{{e}^{3+}}]}{[C{{e}^{4+}}]} $ $ =E_{C{{e}^{4+}}/C{{e}^{3+}}}^{o}-\frac{0.059}{1}\log \frac{10}{90} $ $ {{E}_{cell}}={{E}_{calo}}-{{E}^{o}}C{{e}^{4+}}/C{{e}^{3+}} $ $ 1464={{E}_{calo}}-\left[ \left( {{E}^{o}}_{C{{e}^{4+}}/C{{e}^{3+}}}-0.059\log \frac{1}{8} \right) \right] $ $ 1464=0.28-\left( {{E}^{o}}_{C{{e}^{4+}}/C{{e}^{3+}}}-0.059\log \frac{1}{8} \right) $ $ {{E}^{o}}_{C{{e}^{4+}}/C{{e}^{3+}}}=-1.24V $ Now, $ 2C{{e}^{3+}}+2{{H}^{+}}C{{e}^{4+}}+{{H}_{2}} $ At equilibrium, $ {{E}_{cell}}=0 $ $ \therefore $ $ {{E}^{o}}=\frac{0.059}{2}\log k $ At anode $ 2C{{e}^{3+}}\xrightarrow[{}]{{}}2C{{e}^{4+}}+2{{e}^{-}} $ (oxidation) At cathode $ 2{{H}^{+}}+2{{e}^{-}}\xrightarrow[{}]{{}}{{H}_{2}} $ (reduction) Cell reaction $ 2C{{e}^{3+}}+2{{H}^{+}}\xrightarrow[{}]{{}}2C{{e}^{4+}}+{{H}_{2}} $ $ E_{cell}^{o}={{E}^{o}} $ red(cathode) $ -{{E}^{o}} $ red(anode) $ E_{red({{H}_{2}})}^{o}-E_{red}^{o}(C{{e}^{4+}}/C{{e}^{3+}}) $ $ =0-(-124)=124 $ $ {{E}^{o}}=\frac{0.059}{2}\log k $ $ 1.24=\frac{0.059}{2}\log k $ $ \log k=42.03 $ $ k=1.08\times {{10}^{42}} $