Question

. 1.08 g of pure silver was converted into silver nitrate and its solution was taken in a beaker. It was electrolysed using platinum cathode and silver anode. 0.01 faraday of electricity was passed using 0.15 volt above the decomposition potential of silver. The silver content of the beaker after the above shall be

Answer 1

What will be the molar ratio of the cations deposited at the cathodes if three Faraday of electricity is passed through the solution of AgNO3, CuSo4 and AuCl3.

This may be a simple question but I am not getting the fundamentals right. Please help

Answer 2

Answer:

Explanation:

1F = 1mole of e =96500

Ag+ +e- = Ag

1F = 108g

0.01 F = 1.08g Ag

Ag left = 1.08 - 1.08 = 0