A constant current (0.5 , A) is passed for 1 | vedclass

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(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes,the mass of the s

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$6M_{Ag} : 3M_{Cu} : 2M_{Al}$

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(C) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes,the mass of the substances deposited is directly proportional to their equivalent weights.Mass of metal deposited $\propto$ Equivalent weight.Equivalent weight $= \frac{	ext{Molar mass}}{	ext{n-factor (number of electrons transferred)}}$.$(i)$ For $AgNO_3$: $Ag^+ + e^- ightarrow Ag(s)$,n-factor $= 1$. Mass $\propto \frac{M_{Ag}}{1}$.$(ii)$ For $CuSO_4$: $Cu^{2+} + 2e^- ightarrow Cu(s)$,n-factor $= 2$. Mass $\propto \frac{M_{Cu}}{2}$.$(iii)$ For $AlF_3$: $Al^{3+} + 3e^- ightarrow Al(s)$,n-factor $= 3$. Mass $\propto \frac{M_{Al}}{3}$.Ratio of masses $= \frac{M_{Ag}}{1} : \frac{M_{Cu}}{2} : \frac{M_{Al}}{3}$.Multiplying by $6$ to clear the denominators,we get $6M_{Ag} : 3M_{Cu} : 2M_{Al}$.

$A$ constant current $(0.5 \, A)$ is passed for $1 \, hour$ through $(i)$ aqueous $AgNO_3$,$(ii)$ aqueous $CuSO_4$ and $(iii)$ molten $AlF_3$,separately. The ratio of the mass of the metals deposited on the cathode is $[M_{Ag}, M_{Cu}, M_{Al}$ are molar masses of the respective metals.]

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