In the Hall process,aluminium is produced by | vedclass

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

Question

(B) Total mass of $Al$ required = $24 	imes 5.0 \ g = 120 \ g$. Using Faraday's law: $w = \frac{M}{n 	imes F} 	imes I 	imes t 	imes 	ext{efficie

Vedclass · Accepted Answer

$148.14$

Vedclass · Answer

(B) Total mass of $Al$ required = $24 	imes 5.0 \ g = 120 \ g$. Using Faraday's law: $w = \frac{M}{n 	imes F} 	imes I 	imes t 	imes 	ext{efficiency}$. Here,$w = 120 \ g$,$M = 27 \ g/mol$,$n = 3$ (for $Al^{3+} + 3e^- ightarrow Al$),$F = 96500 \ C/mol$,$I = 9650 \ A$,and efficiency = $0.90$. $120 = \frac{27}{3 	imes 96500} 	imes 9650 	imes t 	imes 0.90$. $120 = \frac{27}{30} 	imes t 	imes 0.90$. $120 = 0.9 	imes 0.9 	imes t$. $120 = 0.81 	imes t$. $t = \frac{120}{0.81} \approx 148.14 \ s$.

Similar Questions

On passing $96500 \ C$ of charge through a $CuSO_4$ solution,the amount of copper liberated is:

The $SI$ unit of electrochemical equivalent is

Calculate the mass of $Ca$ deposited at the cathode by passing $0.8 \ A$ current through molten $CaCl_2$ for $60 \ minutes$. [Molar mass of $Ca = 40 \ g \ mol^{-1}$] (in $g$)

Faraday's laws of electrolysis are related to

In the electrolysis of an aqueous $NiI_2$ solution using $Ni$ electrodes,what happens to the mass of the cathode upon passing $1$ equivalent of charge? (Given: atomic mass of $Ni = 59$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module