(B) The number of equivalents of $AgNO_3$ present is $N \times V(L) = 0.1 \times 0.2 = 0.02 \ \text{equivalents}$.We need to remove half of the silver

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(B) The number of equivalents of $AgNO_3$ present is $N \times V(L) = 0.1 \times 0.2 = 0.02 \ \text{equivalents}$.We need to remove half of the silver,which is $0.02 / 2 = 0.01 \ \text{equivalents}$.According to Faraday's law,the number of equivalents deposited is given by $W = \frac{I \times t}{96500}$.Substituting the values: $0.01 = \frac{0.1 \times t}{96500}$.Solving for $t$: $t = 0.01 \times 96500 / 0.1 = 9650 \ \text{seconds}$.

Class 12 Chemistry Electrochemistry Faraday’s law of electrolysis

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Silver is removed electrically from $200 \ mL$ of a $0.1 \ N$ solution of $AgNO_3$ by a current of $0.1 \ A$. How long will it take to remove half of the silver from the solution? $(F = 96500 \ C \ mol^{-1})$

A
$16$
B
$9650$
C
$100$
D
$10$

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Faraday’s law of electrolysis

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Silver is removed electrically from $200 \ mL$ of a $0.1 \ N$ solution of $AgNO_3$ by a current of $0.1 \ A$. How long will it take to remove half of the silver from the solution? $(F = 96500 \ C \ mol^{-1})$

Similar Questions

$A$ current of $2 \ A$ passing for $5 \ hours$ through molten tin salt deposited $22.2 \ g$ of tin. The oxidation state of tin in the salt is . . . . . . . (Atomic mass of $Sn = 119$)

The $E.C.E.$ of $Cu$ and $Ag$ are $7 \times 10^{-6} \ g/C$ and $1.2 \times 10^{-6} \ g/C$ respectively. If a certain current deposits $14 \ g$ of $Cu$,the amount of $Ag$ deposited is .............. $g$.

What current strength in $amp$ will be required to liberate $10 \ g$ of chlorine from $NaCl$ solution in one hour?

The quantity of charge required to obtain $2$ moles of aluminium from $Al_2O_3$ is: (in $F$)

When the same quantity of electricity is passed through solutions of different electrolytes connected in series,the amounts of substances deposited at the electrodes are in the ratio of their ...?

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