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(D) The reduction reaction at the cathode is: $Ca^{2+} + 2e^{-} \longrightarrow Ca$. From the stoichiometry,$1 \ mol$ of $Ca$ requires $2 \ F$ of char

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) The reduction reaction at the cathode is: $Ca^{2+} + 2e^{-} \longrightarrow Ca$. From the stoichiometry,$1 \ mol$ of $Ca$ requires $2 \ F$ of charge. Number of moles of $Ca$ $(n)$ = $\frac{	ext{mass}}{	ext{molar mass}} = \frac{30 \ g}{40 \ g/mol} = 0.75 \ mol$. Total charge required $(Q)$ = $n 	imes 2 \ F = 0.75 	imes 2 	imes 96500 \ C = 144750 \ C$. Using the relation $Q = i 	imes t$,where $i = 5 \ A$: $t = \frac{Q}{i} = \frac{144750 \ C}{5 \ A} = 28950 \ s$. To convert time into hours: $t = \frac{28950}{3600} \ h \approx 8.04 \ h$. Thus,the approximate time is $8 \ h$.

The approximate time duration in hours to electroplate $30 \ g$ of calcium from molten calcium chloride using a current of $5 \ A$ is [At. mass of $Ca=40$]

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