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 first law of electrolysis,the mass $w$ deposited is give

Vedclass · Accepted Answer

$0.9$

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(A) The reaction at the cathode is: $Al^{3+} + 3e^- ightarrow Al(s)$.According to Faraday's first law of electrolysis,the mass $w$ deposited is given by $w = \frac{M \cdot I \cdot t}{n \cdot F}$.Here,$M = 27 \ g/mol$,$I = 1 \ A$,$t = 9650 \ s$,$n = 3$,and $F = 96500 \ C/mol$.Substituting the values: $w = \frac{27 	imes 1 	imes 9650}{3 	imes 96500}$.$w = \frac{27 	imes 9650}{289500} = \frac{27}{10} = 0.9 \ g$.

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

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