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(C) The reduction reaction at the cathode is: $Ca^{2+} + 2e^{-} \longrightarrow Ca_{(s)}$From the reaction,$2 \ mol$ of electrons are required to depo

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$0.6$

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(C) The reduction reaction at the cathode is: $Ca^{2+} + 2e^{-} \longrightarrow Ca_{(s)}$From the reaction,$2 \ mol$ of electrons are required to deposit $1 \ mol$ of $Ca$.The total charge passed $(Q)$ is given by $Q = I 	imes t = 0.8 \ A 	imes (60 	imes 60 \ s) = 2880 \ C$.The number of moles of electrons passed is $n = \frac{Q}{F} = \frac{2880}{96500} \approx 0.0298 \ mol$.Since $2 \ mol$ of $e^{-}$ deposit $1 \ mol$ of $Ca$,the moles of $Ca$ deposited is $\frac{0.0298}{2} = 0.0149 \ mol$.Mass of $Ca = 	ext{moles} 	imes 	ext{molar mass} = 0.0149 \ mol 	imes 40 \ g \ mol^{-1} \approx 0.597 \ g \approx 0.6 \ g$.

Calculate the mass of $Ca$ deposited at the cathode by passing $0.8 \ A$ current through molten $CaCl_2$ for $60 \ minutes$. [Molar mass of $Ca = 40 \ g \ mol^{-1}$] (in $g$)

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