On passing 1 F of electricity through electr | vedclass

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(A) According to Faraday's first law of electrolysis,the mass of a substance deposited is given by $W = Z 	imes Q$,where $Z$ is the electrochemical e

Vedclass · Accepted Answer

$108 \ g$,$29.5 \ g$,$17.3 \ g$

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(A) According to Faraday's first law of electrolysis,the mass of a substance deposited is given by $W = Z 	imes Q$,where $Z$ is the electrochemical equivalent and $Q$ is the charge in Faradays.For $1 \ F$ of electricity,the mass deposited is equal to the equivalent weight of the metal.Equivalent weight $= \frac{	ext{Atomic weight}}{	ext{Valency factor (n-factor)}}$.For $Ag^{+}$ $(n=1)$: $Eq. wt. = \frac{108}{1} = 108 \ g$.For $Ni^{2+}$ $(n=2)$: $Eq. wt. = \frac{59}{2} = 29.5 \ g$.For $Cr^{3+}$ $(n=3)$: $Eq. wt. = \frac{52}{3} \approx 17.33 \ g$.Thus,the masses deposited are $108 \ g$,$29.5 \ g$,and $17.3 \ g$ respectively.

On passing $1 \ F$ of electricity through electrolytic cells containing $Ag^{+}$,$Ni^{2+}$,and $Cr^{3+}$ ions,the mass of $Ag$ $(At. wt. = 108)$,$Ni$ $(At. wt. = 59)$,and $Cr$ $(At. wt. = 52)$ deposited is:

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