A constant current was passed through a solution of ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed through the solution is

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Solution

Q.
A constant current was passed through a solution of ${AuCl}_{4}^{-}$ ion between gold electrodes. After a period of 10.0 minutes, the increase in mass of cathode was 1.314 g. The total charge passed through the solution is _________ $\times 10^{- 2}$ F. (Given atomic mass of Au = 197) [2024]

Ans.
(2)

   $Moles of Au (n_{Au}) = \frac{Mass deposited}{Molar mass} = \frac{1.314 g}{197 {g mol}^{- 1}}$

$= \frac{1.314}{197} mol$

  Reaction at cathode is:

   ${Au}^{3 +} + 3 e^{-} \to Au$

By stoichiometry:

   $\frac{n_{e^{-}}}{3} = \frac{n_{Au}}{1}$

$\frac{n_{e^{-}}}{3} = \frac{1.314}{197}$

   $n_{e^{-}} = 3 \times \frac{1.314}{197} mol$

Charge passed (Q) = $n_{e^{-}} \times$ charge on 1 mole

   $= 3 \times \frac{1.314}{197} mol \times 1 F {mol}^{- 1}$

   $= 0.02 F = 2 \times 10^{- 2} F$

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