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 \[\text{F}{{\text{e}}^{2+}}\]
Correct Answer: D
Solution :
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 \[\Rightarrow \] \[\text{F}{{\text{e}}^{\text{2+}}}\text{ left = 0}\text{.1-0}\text{.025 = 0}\text{.075 mol}\]
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