Question

When 4.0 A of current is passed through a 1.0 L, 0.10 M $F{e}^{3+}\left(aq\right)$ solution for 1.0 hour, it is partly reduced to Fe(s) and partly of $F{e}^{2+}\left(aq\right)$. The correct statements(s) is/are :

• A. 0.10 mol 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 $F{e}^{2+}$.

Answer 1

Answer: (A), (B), (C)

0.025 mol of Fe(s) will be deposited.
0.10 mol of electrons are required to convert all $F{e}^{3+}$ to $F{e}^{2+}$.
0.075 mol of iron remains as $F{e}^{2+}$.

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 $F{e}^{2+}$.

$F{e}^{3+}+e^{-}\to F{e}^{2+}$

1 F = 1 mol $F{e}^{3+}$ deposited $\Rightarrow 0.15F\equiv 0.15molF{e}^{3+}$ deposited > $F{e}^{3+}$ available.

Thus, 1 mol $F{e}^{3+}\equiv 1F$ $\Rightarrow 0.1molF{e}^{3+}\equiv 0.1F$ electricity is used $\equiv 0.1molF{e}^{2+}$ produced $\Rightarrow 0.15-0.1=0.05F$ electricity left for the reduction of $F{e}^{2+}$.

$F{e}^{2+}+2e^{-}\to Fe$

2 F = 1 mol $F{e}^{2+}$

$\Rightarrow 0.05F,\equiv \frac{0.05}{2}=0.25molF{e}^{2+}$ reduced $\equiv 0.025$ mol $Fe$ deposited. $\Rightarrow F{e}^{2+}$ left $=0.1-0.025=0.075$ mol