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Question

(B) The reduction reactions at the electrodes are as follows:$Al^{3+} + 3e^- \rightarrow Al$$Ag^+ + 1e^- \rightarrow Ag$$Mg^{2+} + 2e^- \rightarrow Mg

Vedclass · Accepted Answer

$2: 6: 3$

Vedclass · Answer

(B) The reduction reactions at the electrodes are as follows:$Al^{3+} + 3e^- ightarrow Al$$Ag^+ + 1e^- ightarrow Ag$$Mg^{2+} + 2e^- ightarrow Mg$According to Faraday's laws of electrolysis,the number of moles of metal deposited is given by $n = \frac{	ext{moles of electrons}}{	ext{n-factor}}$.For $Al$: $n_{Al} = \frac{1}{3} 	ext{ mole}$.For $Ag$: $n_{Ag} = \frac{1}{1} = 1 	ext{ mole}$.For $Mg$: $n_{Mg} = \frac{1}{2} 	ext{ mole}$.The ratio of moles deposited is $\frac{1}{3} : 1 : \frac{1}{2}$.Multiplying by $6$ to simplify the ratio: $2 : 6 : 3$.Thus,the correct option is $B$.

If $1$ mole of electrons is passed through the solutions of $AlCl_3$,$AgNO_3$,and $MgSO_4$,in what ratio will $Al$,$Ag$,and $Mg$ be deposited at the electrodes?

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