How long will it take for a uniform current of 6.00 A to deposit 78 g of gold from a solution of $A u C l_{4}^{-}$ ? What mass of chlorine gas will be formed simultaneously at anode of the cell?
(Atomic mass of Au = 197)

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TestHub · Accepted Answer

Reactions take place at electrodes are
At cathode $A u C l_{4}^{-} + 3 e^{-} \to A u + 4 C l^{-}$
At cathode $C l^{-} \to \frac{1}{2} C l_{2} + e^{-}$
For the deposition of 197 g (1 mole) of Au = 3F of charge is required thus, for the deposition of 78 g of Au, charge required
$= \frac{3}{197} \times 78 = 1.19 F$
$= 1.19 \times 96500$ coulumbs
From $Q = I \times t$
$t = \frac{Q}{I} = \frac{1.19 \times 96500}{6}$
$= 19139.16$ sec
By 1 F charge, 35.5 g of $C l_{2}$ gas is formed, thus from 1.19 F, mass of chlorine gas formed $= 35.5 \times 1.19 = 42.24$ g

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