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(C) The chemical reaction for the reduction of $Mg^{2+}$ to $Mg$ is: $Mg^{2+} + 2e^- ightarrow Mg(s)$.From the stoichiometry, $1 	ext{ mole}$ of $M

Vedclass · Accepted Answer

$0.666$

Vedclass · Answer

(C) The chemical reaction for the reduction of $Mg^{2+}$ to $Mg$ is: $Mg^{2+} + 2e^- ightarrow Mg(s)$.From the stoichiometry, $1 	ext{ mole}$ of $Mg$ requires $2 	ext{ moles}$ of electrons, which is equal to $2 	ext{ F}$ of electricity.Molar mass of $Mg = 24 	ext{ g mol}^{-1}$.Number of moles of $Mg$ produced = $\frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{8 	ext{ g}}{24 	ext{ g mol}^{-1}} = \frac{1}{3} 	ext{ mol}$.Since $1 	ext{ mole}$ of $Mg$ requires $2 	ext{ F}$, then $\frac{1}{3} 	ext{ mole}$ of $Mg$ requires $\frac{1}{3} 	imes 2 	ext{ F} = 0.666 	ext{ F}$.Therefore, the correct option is $C$.

What is the quantity of electricity in Faradays required to produce $8 \text{ g}$ of $Mg$ (molar mass = $24 \text{ g mol}^{-1}$) from $MgCl_2$ solution (in $\text{ F}$)?

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