Question

A current of 1.50A was passed through an electrolytic cell containing AgNO₃​ solution with inert electrodes. The weight of Ag deposited was 1.50g . How long did the current flow? (Given: Ag = 108; N = 14; O = 16; F = 96500 C/mol)

Answer 1

The reaction involved in deposition of Ag on electrode

$Ag^++e-->Ag_s$

1 mol of e− is required for the reduction of 1 mol of Ag⁺, i.e., 1 mol of Ag⁺ require 1 mol of e− (1F)

1F is required to deposit 108kg of Ag.

Deposited Ag=1.5g

So, if 108g Ag→1F=96500C of charge

     

$1.5g Ag-->\frac{96500}{108}×1.5C = 1340.277C$

 We know,

Q=IK

      $t=\frac{Q}{I}=\frac{1340.277}{1.5}= 893.518 sec$