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 \[F{{e}^{2+}}\]
D) 0.050 mol of iron remains as \[F{{e}^{2+}}\]
Correct Answer: D
Solution :
Number of Faraday \[=\frac{4\times 1\times 3600}{96500}=0.15\] Initially, moles of \[F{{e}^{3+}}=0.1\times 1=0.1\] First, \[F{{e}^{3+}}\] will get reduced to \[F{{e}^{2+}}\] \[F{{e}^{3+}}+{{e}^{-}}\xrightarrow{{}}F{{e}^{2+}}\] \[1F=1\,mol\,F{{e}^{3+}}\] deposited \[\Rightarrow \] \[0.15\,F=0.15\,mol\,F{{e}^{3+}}\] deposited \[<\,F{{e}^{3+}}\] available. Thus, \[1\,mol\,F{{e}^{3+}}=1\,F\] \[\Rightarrow \] \[0.1\,mol\,F{{e}^{3+}}=0.1\,F\] electricity is used \[=0.1\,mol\,F{{e}^{2+}}\] is produced \[\Rightarrow \] 0.15 - 0.1 = 0.05 F electricity left for the reduction of \[F{{e}^{2+}}\] \[F{{e}^{2+}}+2{{e}^{-}}\xrightarrow{{}}Fe\] \[2F=1\,\,mol\,\,F{{e}^{2+}}\] \[\Rightarrow \] \[0.05\,F=\frac{0.05}{2}=0.025\,mol\,F{{e}^{2+}}\] reduced \[=0.025\,mol\,Fe\] deposited \[\Rightarrow \] \[F{{e}^{2+}}\] left \[=0.1-0.025=0.075\]
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