Question 25.12: An aqueous solution containing aqueous gold(III) ions was el......

An aqueous solution containing aqueous gold(III) ions was electrolyzed with a current of 0.0381 amperes until 1.47 grams of gold was deposited at the cathode. Determine the duration of the experiment.

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The equation for the reaction at the cathode is

Au^{3+}(aq) + 3\ e^– → Au(s)

Thus, one mole of electrons (96 485 C) deposits one-third of a mole of Au(s). The number of coulombs required to deposit 1.47 grams of Au(s) is given by

Z = (1.47\ g\ Au)\left(\frac{1\ mol\ Au}{197.0\ g\ Au} \right) \left(\frac{3\ mol\ e^–}{1\ mol\ Au} \right) \left(\frac{96\ 485\ C}{1\ mol\ e^-} \right) =2160\ C

The time required at a current of 0.0381 amperes is given by (Equation 25.2)

Z = It                    (25.2)

t =\frac{Z}{I} =\frac{2160\ C}{0.0381\ C·s^{–1}} = 5.67 × 10^{4}\ s = 15.7\ h

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