(A) The reaction at the cathode is: $Au^{3+} + 3e^- \rightarrow Au(s)$.The equivalent mass of $Au$ is $E = \frac{\text{Atomic mass}}{n-factor} = \frac

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(A) The reaction at the cathode is: $Au^{3+} + 3e^- \rightarrow Au(s)$.The equivalent mass of $Au$ is $E = \frac{\text{Atomic mass}}{n-factor} = \frac{197}{3} \ \text{g/eq}$.According to Faraday's law of electrolysis,the mass deposited $W$ is given by $W = \frac{Q \times E}{1 \ \text{F}}$,where $Q$ is the charge in Faradays.Substituting the values: $1.314 = \frac{Q \times 197}{3}$.$Q = \frac{1.314 \times 3}{197} \ \text{F}$.$Q = \frac{3.942}{197} \ \text{F} = 0.02 \ \text{F}$.$Q = 2 \times 10^{-2} \ \text{F}$.Thus,the total charge passed is $2 \times 10^{-2} \ \text{F}$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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$A$ constant current was passed through a solution of $AuCl_4^-$ ion between gold electrodes. After a period of $10.0 \ \text{minutes}$,the increase in mass of cathode was $1.314 \ \text{g}$. The total charge passed through the solution is . . . . . . $\times 10^{-2} \ \text{F}$. (Given atomic mass of $Au = 197$)

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DifficultMHT CET 2019

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For electroplating,$1.5 \ A$ current is passed for $250 \ s$ through $250 \ mL$ of $0.15 \ M$ solution of $MSO_4$. Only $85 \ \%$ of the current was utilised for electrolysis. The molarity of $MSO_4$ solution after electrolysis is closest to $.......$ [Assume that the volume of the solution remained constant] (in $M$)

MediumKVPY 2016

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$A$ current of $10.0 \, A$ flows for $2.00 \, h$ through an electrolytic cell containing a molten salt of metal $X$. This results in the decomposition of $0.250 \, mol$ of metal $X$ at the cathode. The oxidation state of $X$ in the molten salt is $(F = 96,500 \, C \, mol^{-1})$ (in $+$)

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 $= x \ g/mol$)

Two electrolytic cells are connected in series containing $CuSO_4$ solution and molten $AlCl_3$. If in electrolysis $0.4$ moles of $Cu$ are deposited on the cathode of the first cell,the number of moles of $Al$ deposited on the cathode of the second cell is: (in $moles$)

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