Question

10nA current passed through AgNO₃ solution for 80 minutes. Both electrodes were made of gold. The cathode solution (total weight 40.28 g) was titrated with 86 ml of 0.02 M KSCN. Anode solution of same weight was titrated with 112 ml of the same solution of KSCN. What is thesilver ion transport number?

Answer 1

At first let's apply Faraday's first law of electrolysis for both cathode and anode(AgNO₃ in Gold electrodes). At Cathode : Ag⁺ + e Ag

At Anode : 4 OH^- 2H₂O + O₂ + 4e

Faraday's first law applied at cathode gives : m = E x I x t /F = 107.9 x 10 x 10^-9 x 80 x 60 / 96485 grams, This gives, mass of silver to be deposited = 53.68 x 10^-9 grams

Molar amount of AgNO₃ present at the cathode prior to electrolysis = c(cathode) x V(cathode) = 86/1000 L x 0.02 mol/L = 1.72 mmol

Now, n_AgNO3,tot = -I x t / v_e x F = -10 x 10^-9 x 80 x 60 / 96485 = 0.497 x 10^-9 moles

n(anode) = - I x t / v_e x F = -10 x 10^-9 x 80 x 60 / 96485 x 4 = - 0.124 x 10^-9 moles

For anode , c(anode) x V(anode) = 0.02 x 112 mmol = 2.24 mmol

n(anode) = -t₊ x n = -t₊ x 2.24 x 10^-3 = - 0.124 x 10^-9

Therefore, t₊ = 55 x 10^-9