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

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$2$

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

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