Question

The amount of electricity which releases $2.0g$ of gold from a gold salt is the same as that which dissolves $0.967g$ of copper anode during the electrolysis of copper sulphate solution. What is the oxidation number of gold in the gold state? (At. Wt. of $Cu$ = 63.5; $Au$ = 197)

Answer 1

The number of moles of copper is $\frac{0.967}{63.5}=0.015$ moles.

The number of moles of gold is $\frac{2.0}{197}=0.01$ moles.

The oxidation number of copper is 2. Let x be the oxidation number of gold.

$0.015\times 2=0.01\times x$

$x=\frac{0.030}{0.010}=3$.

Hence, the oxidation number of gold is 3.