Question

When $ \text{4A} $ of current is passed through a 1.0 L, 0.10 M $ \text{F}{{\text{e}}^{3+}}(aq) $ ) solution for 1 h, it is partly reduced to Fe(s) and partly of. $ F{{e}^{2+}}(aq) $ Identify the incorrect statement.

• A. 0.10 mole of electrons are required to convert all $ F{{e}^{3+}} $ to $ F{{e}^{2+}} $
• B. 0.025 mol of Fe(s) will be deposited
• C. 0.075 mol of iron remains as $ \text{F}{{\text{e}}^{2+}} $
• D. 0.050 mol of iron remains as $ \text{F}{{\text{e}}^{2+}} $

Answer 1

Answer: (D)

Number of Faradays $ =\frac{4\times 1\times 3600}{96500}=0.15 $ Initially, moles of $ F{{e}^{3+}}=0.1\times 1=0.1 $ First, $ F{{e}^{3+}} $ will get reduced to $ F{{e}^{2+}} $ $ F{{e}^{3+}}+{{e}^{-}}\to F{{e}^{2+}} $ $ \text{1F =1mol F}{{\text{e}}^{\text{3+}}} $ deposited $ \Rightarrow $ $ \text{0}\text{.15 F = 0}\text{.15 mol F}{{\text{e}}^{\text{3+}}} $ $ \text{F}{{\text{e}}^{3+}} $ available. Thus, $ \text{1 mol F}{{\text{e}}^{\text{3+}}}\text{=1F} $ $ \Rightarrow $ $ \text{0}\text{.1 mol F}{{\text{e}}^{\text{3+}}}\text{= 0}\text{.1F} $ electricity is used $ \text{= 0}\text{.1mol F}{{\text{e}}^{2+}} $ is produced $ \Rightarrow $ $ \text{0}\text{.15-0}\text{.1= 0}\text{.05 F} $ electricity left for the reduction of $ F{{e}^{2+}} $ $ F{{e}^{2+}}+2{{e}^{-}}\to Fe $ $ \text{2F =1mol F}{{\text{e}}^{\text{2+}}} $ $ \Rightarrow $ $ \text{0}\text{.05 F =}\frac{\text{0}\text{.05}}{\text{2}}\text{= 0}\text{.025mol F}{{\text{e}}^{\text{2+}}} $ reduced $ \text{= 0}\text{.025 mol Fe} $ deposited $ \Rightarrow $ $ \text{F}{{\text{e}}^{\text{2+}}}\text{ left = 0}\text{.1-0}\text{.025 = 0}\text{.075 mol} $