Answer to Question #92169 in Inorganic Chemistry for Ummulkhairi

Question

A constant current was passed through a solution of AuCl4 ions between gold electrolyte .After a period of 10 minutes,the cathode increased in weight by 1.314g.how much charge was passed and what is the current?[Au=196.97]

Answer 1

We take advantage of the united Faraday law:

$$m = {1 \over F} {M \over z} It$$

(where $F = 96485; z = 4; t = 10 min = 600sec$)

$$I = {Fmz \over Mt}$$

$$I = {96485*1.314*4 \over 600*196.97} = 4.291A$$

Find current:

$$q = It$$ $$q = 4.291*600 = 2574.631C$$

Answer: I = 4.291 A; q = 2574.631 C.