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Electrochemistry

Electrochemistry

Question
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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 1F = 96500 C mol–1)

Answer

Given,
Molar mass of Ag = 108 g mol–1 
1F = 96500 C mol–1

Q=It
Q= 2×15×60=1800 C
W(weight of substance deposited)= ZQ
Z= M/nF
n-factor here is 1
So Z = 108/1×96500


fraction numerator 108 space straight x space 2 space straight x space 15 space straight x 60 over denominator 1 space straight x 96500 end fraction =2.01 g
The mass of Ag deposited at the cathode is2.01 g

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