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 $F{e}^{2+}$ $F{e}^{2+}+2e^{-}\to Fe$ $\text{2F =1mol F}{\text{e}}^{\text{2+}}$ $\Rightarrow$ $\text{0}\text{.05 F =}\frac{\text{0}\text{.05}}{\text{2}}\text{= 0}\text{.025mol F}{\text{e}}^{\text{2+}}$ reduced $\text{= 0}\text{.025 mol Fe}$ deposited $\Rightarrow$ $\text{F}{\text{e}}^{\text{2+}}\text{ left = 0}\text{.1-0}\text{.025 = 0}\text{.075 mol}$