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(A) The dissociation of $MgCl_2$ is: $MgCl_2 \rightarrow Mg^{2+} + 2Cl^-$. The reduction reaction at the cathode is: $Mg^{2+} + 2e^- \rightarrow Mg_{(

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$0.2$

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(A) The dissociation of $MgCl_2$ is: $MgCl_2 ightarrow Mg^{2+} + 2Cl^-$. The reduction reaction at the cathode is: $Mg^{2+} + 2e^- ightarrow Mg_{(s)}$. From the stoichiometry,$2 \ moles$ of electrons are required to deposit $1 \ mole$ of $Mg_{(s)}$. Since $1 \ mole$ of electrons is equivalent to $1 \ Faraday$ $(F)$,$2 \ moles$ of electrons correspond to $2 \ F$. The number of moles of $MgCl_2$ is calculated as: $n = M 	imes V = 0.1 \ M 	imes 1 \ L = 0.1 \ mol$. Thus,$n(Mg^{2+}) = 0.1 \ mol$. The number of Faradays required to deposit $0.1 \ mol$ of $Mg$ is: $2 	imes 0.1 \ mol = 0.2 \ F$.

The number of Faradays required to completely deposit magnesium from $1 \ L$ of $0.1 \ M$ $MgCl_2$ aqueous solution is

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