When X amperes of current is passed through m | vedclass

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Question

(A) The reaction for the deposition of aluminium is: $Al^{3+} + 3e^- \rightarrow Al(s)$.According to Faraday's law of electrolysis,the mass deposited

Vedclass · Accepted Answer

$10 \ 	ext{ampere}$

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(A) The reaction for the deposition of aluminium is: $Al^{3+} + 3e^- ightarrow Al(s)$.According to Faraday's law of electrolysis,the mass deposited $w = \frac{M 	imes i 	imes t}{n 	imes F}$,where $M$ is the molar mass $(27 \ g/mol)$,$n$ is the number of electrons $(3)$,$i$ is the current in amperes,$t$ is the time in seconds $(96.5 \ s)$,and $F$ is Faraday's constant $(96500 \ C/mol)$.Substituting the values: $0.09 = \frac{27 	imes i 	imes 96.5}{3 	imes 96500}$.$0.09 = \frac{9 	imes i 	imes 96.5}{96500}$.$0.09 = \frac{9 	imes i}{1000}$.$i = \frac{0.09 	imes 1000}{9} = 10 \ 	ext{ampere}$.Therefore,the value of $X$ is $10$.

When $X$ amperes of current is passed through molten $AlCl_3$ for $96.5 \ s$,$0.09 \ g$ of aluminium is deposited. What is the value of $X$?

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