Answer to Question #319042 in Physical Chemistry for Sir mencho

Question

A solution of AgNO3 containing 12.14g of Ag in 50ml of solution was electrolyzed between pt electrodes . After electrolysis 50ml of the anode solution was found to contain 11.55g of Ag ,while 1.25g of metallic Ag was deposited on the cathode . calculate the transport number of Ag+

Answer 1

M(Ag) = 108 g/mol;

W(Ag⁺)_{a, before} = 12.14 g;

W(Ag⁺)_{a, after} = 11.55 g;

$\Delta$ W(Ag)_k = 1.25;

$\Delta$ (Ag⁺)_k= $\Delta$ W(Ag⁺)_k /M(Ag) = 1.25/108 = 0.0116 mol;

Fall in concentration due to migration of Ag⁺: $\Delta$ W(Ag⁺)_a = W(Ag⁺)_{a, before} - W(Ag⁺)_{a, after} = 12.14 - 11.55 = 0.59 g;

$\Delta$ (Ag⁺)_a= $\Delta$ W(Ag⁺)_a /M(Ag) = 0.59/108 = 0.0055 mol;

Transport number of Ag⁺ : t(Ag⁺) = $\Delta$ (Ag⁺)_a/$\Delta$ (Ag⁺)_k = 0.0055/0.0116 = 0.4741.

Answer: t(Ag⁺) = 0.4741.