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Question
Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes.
(Given : Molar mass of Ag = 108 g mol−1 lF = 96500 C mol−1)
Solution
t = 900 s
Charge = Current × Time = 2 A × 900 s = 1800 C
According to the reaction:
Ag+(aq) + e– = Ag(s)
We require 1F to deposit 1 mol or 108 g of Ag.
For 1800 C, the mass of Ag deposited = (108 g mol–1 × 1800 C)/(1 × 96500 C mol–1) = 2.0145 g.
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