A constant current was passed through a solution of AuCl₄⁻ 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 _______ × 10⁻² F.
(Given atomic mass of Au=197)

Solution

Chemistry ma’am · 3 steps

Gold is +3 in AuCl₄⁻, so every mole of Au deposited takes 3 faradays.

1. Charge on gold. In AuCl₄⁻, the four chloride ions give -4 and the whole ion carries -1, so gold is +3. At the cathode Au³⁺ + 3e⁻ → Au, so each mole of Au deposited needs 3 F.
2. Moles deposited. The cathode gains 1.314 g of gold. Moles of Au = 1.314/197 = 6.67 × 10⁻³ mol.

   Q = n × z = 6.67 × 10⁻³ × 3 = 2.0 × 10⁻² F

3. So. The total charge is 2 × 10⁻² F, so the answer is 2. You do not need the 10 minutes, because the question asks for the charge in faradays and not for the current.

The common slip: 0.67. Treating gold as Au⁺, with one electron per atom, gives 0.67 × 10⁻² F instead of 2. In AuCl₄⁻ gold is +3.

The idea to keep. Charge in faradays equals moles of metal deposited times the electrons each ion takes. Always get that electron count from the oxidation state in the given ion.

Answer from the NTA final answer key. Solution by PrepWiser.