QuestionElectrochemistry / transport number
During electrolysis of silver nitrate solution, loss in the anode compartment was 0.589 g. In a Voltameter connected in series, 1.259 g of silver was deposited. What would be the transport number of two ions?
- (a) t₊ = 0.468 and t– = 0.532
- (b) t₊ = 0.532 and t– = 0.468
- (c) t₊ = 0.670 and t– = 0.330
- (d) t₊ = 0.330 and t– = 0.670
Answer: (a) t₊ = 0.468 and t– = 0.532
Hittorf method with Ag electrodes: loss of Ag⁺ in the anode compartment = Ag⁺ carried away = t₊ × total. t₊ = 0.589/1.259 = 0.468; t₋ = 1 − 0.468 = 0.532 → (a).