Class 12th Chemistry 2017 Set3 Delhi Board Paper Solution
Calculate the mass of Ag deposited at cathode when a current of 2A was passed through a solution of AgNO3 for 15 min.
(Given : Molar mass of Ag = 108 g mol−1 , 1F = 96500 C mol−1).
As Given:
Molar Mass of Ag = 108 g/mol
1F = 96500 C mol−1
Reaction at cathode = Ag + e- → Ag(s)
w = Zlt
Where, w = Mass deposited at cathode
Z = electrochemical constant
I = current
t = time
Now I = 2amp
t = 15 mnts = 15x60 = 900 seconds
Z = Eq. wt of substance / 96500 = 108/96500
So,
w = 108/96500 x 900 x2 = 2.015g
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