The number of Faradays required to completely deposit magnesium from $1 \mathrm{~L}$ of $0.1 \mathrm{M}…
The number of Faradays required to completely deposit magnesium from $1 \mathrm{~L}$ of $0.1 \mathrm{M} \mathrm{MgCl}_2$ aq. solution is
- $0.2$
- $2$
- $0.1$
- $0.4$
Solution
$\begin{aligned}
& \text {} \mathrm{MgCl}_2 \rightarrow \mathrm{Mg}^{2+}+2 \mathrm{Cl}^{-} \\
& \mathrm{Mg}^{2+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mg} \text { (s) }
\end{aligned}$
2 moles of $\mathrm{e}^{-}$are required to deposit 1 mole of $\mathrm{Mg}(\mathrm{s})$.
$\begin{aligned}
& \Rightarrow 1 \text { mole of } \mathrm{e}^{-}=1 \mathrm{~F} \\
& \Rightarrow 2 \text { moles of } \mathrm{e}^{-}=2 \mathrm{~F} \\
& \mathrm{n}\left(\mathrm{MgCl}_2\right)=0.1 \mathrm{M} \times 1 \mathrm{~L}=0.1 \mathrm{~mol} . \\
& \mathrm{n}\left(\mathrm{Mg}^{2+}\right)=0.1 \mathrm{~mol} .
\end{aligned}$
Therefore, the number of moles of electrons/faradays required to deposit $0.1 \mathrm{~mol}$ of $\mathrm{Mg}$ will be $2 \times 0.1=\mathbf{0 . 2} \mathbf{m o l}$.
Asked in: AP EAMCET 2023 (15 May Shift 2)
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