One ampere of current is passed for 9650 s t | vedclass

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

Vedclass · Accepted Answer

$0.9$

Vedclass · Answer

(A) The reaction at the cathode is: $Al^{3+} + 3e^- ightarrow Al(s)$.According to Faraday's law of electrolysis,the mass of substance deposited is given by $w = \frac{M 	imes I 	imes t}{n 	imes F}$,where $M$ is the molar mass,$I$ is the current,$t$ is the time,$n$ is the number of electrons involved,and $F$ is Faraday's constant $(96500 \ C/mol)$.Given: $M = 27 \ g/mol$,$I = 1 \ A$,$t = 9650 \ s$,$n = 3$,and $F = 96500 \ C/mol$.Substituting the values: $w = \frac{27 	imes 1 	imes 9650}{3 	imes 96500}$.$w = \frac{27 	imes 9650}{289500} = \frac{27}{30} = 0.9 \ g$.

One ampere of current is passed for $9650 \ s$ through molten $AlCl_3$. What is the weight in grams of $Al$ deposited at the cathode? (Atomic weight of $Al = 27$)

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