Question #319042

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+

1
Expert's answer
2022-03-28T17:13:04-0400

M(Ag) = 108 g/mol;

W(Ag+)a, before = 12.14 g;

W(Ag+)a, after = 11.55 g;

Δ\Delta W(Ag)k = 1.25;

Δ\Delta (Ag+)k= Δ\Delta W(Ag+)k /M(Ag) = 1.25/108 = 0.0116 mol;

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

Δ\Delta (Ag+)a= Δ\Delta W(Ag+)a /M(Ag) = 0.59/108 = 0.0055 mol;

Transport number of Ag+ : t(Ag+) = Δ\Delta (Ag+)a/Δ\Delta (Ag+)k = 0.0055/0.0116 = 0.4741.

Answer: t(Ag+) = 0.4741.


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