The amount of metal deposited at the cathode | vedclass

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Question

(B) The reaction at the cathode for the reduction of ferric ions $(Fe^{3+})$ is: $Fe^{3+} + 3e^- \rightarrow Fe(s)$.According to Faraday's law,the mas

Vedclass · Accepted Answer

$0.261$

Vedclass · Answer

(B) The reaction at the cathode for the reduction of ferric ions $(Fe^{3+})$ is: $Fe^{3+} + 3e^- ightarrow Fe(s)$.According to Faraday's law,the mass deposited $(w)$ is given by $w = \frac{Z 	imes I 	imes t}{96500}$,where $Z = \frac{	ext{Equivalent weight}}{96500}$.Equivalent weight of $Fe = \frac{	ext{Atomic weight}}{	ext{Valency factor}} = \frac{56}{3} \approx 18.67 \ g/eq$.Given: $I = 0.75 \ A$,$t = 30 \ min = 30 	imes 60 \ s = 1800 \ s$.$w = \frac{56 	imes 0.75 	imes 1800}{3 	imes 96500} = \frac{56 	imes 1350}{289500} = \frac{75600}{289500} \approx 0.261 \ g$.Thus,the correct option is $B$.

The amount of metal deposited at the cathode on passing an electric current of $0.75 \ A$ in an aqueous ferric sulphate solution for $30 \ min$ will be - ........... $g$ (atomic weight of $Fe = 56$).

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