A constant current was passed through a solution of AuCl 4 - 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 )

Step-by-step solution

The half-cell reaction is, Au 3 + + 3 e − → Au Given W = 1 . 314 g 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. W E = charge 1 F 1 . 314 197 3 = Q 1 F Q = 2 × 10 - 2 F