Two electrolytic cells containing molten solu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes connected in seri

Vedclass · Accepted Answer

$58.5$

Vedclass · Answer

(B) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through different electrolytes connected in series,the masses of the substances deposited are proportional to their equivalent masses.$\frac{	ext{Mass of } Ni}{	ext{Mass of } Al} = \frac{	ext{Equivalent mass of } Ni}{	ext{Equivalent mass of } Al}$Equivalent mass of $Ni = \frac{	ext{Atomic mass of } Ni}{	ext{Valency factor}} = \frac{58.5}{2} = 29.25 \ g/eq$Equivalent mass of $Al = \frac{	ext{Atomic mass of } Al}{	ext{Valency factor}} = \frac{27}{3} = 9 \ g/eq$Let the mass of $Ni$ be $w$.$\frac{w}{18} = \frac{29.25}{9}$$w = 18 	imes \frac{29.25}{9} = 2 	imes 29.25 = 58.5 \ g$Therefore,the weight of nickel obtained is $58.5 \ g$.

Similar Questions

Calculate the amount of electricity required in coulombs to convert $0.08 \ mol$ of $MnO_4^{-}$ to $Mn^{2+}$. (in $C$)

The number of Faradays required to completely deposit magnesium from $1 \ L$ of $0.1 \ M$ $MgCl_2$ aqueous solution is

How much charge in Coulombs is required for the oxidation of $1 \ mol$ of $H_2O$ to $O_2$?

When a current of $9.65 \ A$ is passed for $10 \ \text{minutes}$,$3.0 \ g$ of metal is deposited. What is the equivalent weight of the metal?

State Faraday's laws of electrolysis and mention their uses.

Vedclass Test Series

Exam Paper Generator

Online Exam Module