Class 12th Chemistry 2017 Set2 Delhi Board Paper Solution

Question 16

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). 

Answer

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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