Question

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

• A. $0.10$ mole of electrons are required to convert all $F{e}^{3+}$ to $F{e}^{2+}$
• B. $0.025$ mol of $Fe\left(s\right)$ 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)

The correct options are
A $0.10$ mole of electrons are required to convert all $F{e}^{3+}$ to $F{e}^{2+}$
B $0.025$ mol of $Fe\left(s\right)$ will be deposited.
C $0.075$ 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 $F{e}^{2+}$

$F{e}^{3+}+e^{-}⟶F{e}^{2+}$

$1F=$ 1 mol $F{e}^{3+}$ deposited

$\Rightarrow 0.15F=0.15molF{e}^{3+}$ deposited

$\Rightarrow F{e}^{3+}$ available.

Thus, 1 mol $F{e}^{3+}=1F$

$\Rightarrow 0.1molF{e}^{3+}=0.1F$ electricitry is used

$=0.1$ mol $F{e}^{2+}$ is produced

$\Rightarrow$ $0.15-0.1=0.05F$ electricity left for the reduction of $F{e}^{2+}$

$F{e}^{2+}+2e^{-}⟶Fe$

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

$\Rightarrow 0.05F=0.025$ mol $F{e}^{2+}$ reduced $=0.025$ mol $Fe$ deposited

$\Rightarrow F{e}^{2+}$ left $=0.1-0.025=0.075mol$