(B) The reduction reaction at the cathode is: $Fe^{2+} + 2e^- \to Fe(s)$.From the stoichiometry,$2 \ mol$ of electrons (i.e.,$2 \ F$ of electricity) a

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(B) The reduction reaction at the cathode is: $Fe^{2+} + 2e^- \to Fe(s)$.From the stoichiometry,$2 \ mol$ of electrons (i.e.,$2 \ F$ of electricity) are required to deposit $1 \ mol$ of $Fe$ $(56 \ g)$.The equivalent weight of $Fe$ is $E_{Fe} = \frac{56}{2} = 28 \ g/eq$.Using Faraday's law,the weight of metal deposited is $W = E_{Fe} \times \text{Number of Faradays}$.$W = 28 \times 3 = 84 \ g$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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Three faradays of electricity were passed through an aqueous solution of iron $(II)$ bromide. The weight of iron metal (atomic weight $= 56$) deposited at the cathode (in $g$) is

A
$56$
B
$84$
C
$112$
D
$168$

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Electrochemistry

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Faraday’s law of electrolysis

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Three faradays of electricity were passed through an aqueous solution of iron $(II)$ bromide. The weight of iron metal (atomic weight $= 56$) deposited at the cathode (in $g$) is

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