(D) According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = \frac{M \times I \times t}{n \times F}$.Here,$M = x \ g/mol$,$I

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(D) According to Faraday's law of electrolysis,the mass $W$ deposited is given by $W = \frac{M \times I \times t}{n \times F}$.Here,$M = x \ g/mol$,$I = 5 \ A$,$t = 100 \times 60 \ s = 6000 \ s$,$n = 2$ (for a divalent metal),and $F \approx 96500 \ C/mol$.Substituting the values: $W = \frac{x \times 5 \times 6000}{2 \times 96500}$.$W = \frac{30000 x}{193000} = \frac{30}{193} x \ g$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

Medium

Calculate the mass of a divalent metal produced at the cathode by passing $5 \ A$ current through its salt solution for $100 \ minutes$. (Molar mass of metal $= x \ g/mol$)

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A
$\frac{193}{30} x \ g$
B
$\frac{193}{15} x \ g$
C
$\frac{15}{193} x \ g$
D
$\frac{30}{193} x \ g$

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Electrochemistry

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Faraday’s law of electrolysis

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Calculate the mass of a divalent metal produced at the cathode by passing $5 \ A$ current through its salt solution for $100 \ minutes$. (Molar mass of metal $= x \ g/mol$)

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