A constant current was passed through a solution of AuCl _ 4 ^ -...

The half-cell reaction is,

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

Given W = $1 . 314g$

Atomic mass of $Au = 197$

When one Faraday of charge is passed through a circuit, then one equivalent of charge is deposited at the respective electrode.

$\dfrac{W}{E} = \dfrac{charge}{1F}$

$\dfrac{1 . 314}{\dfrac{197}{3}} = \dfrac{Q}{1F}$

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