CBSE Class 12 Chemistry 2020 Delhi Set 1 Solved Paper

Question : 29

Total: 37

When a steady current of 2A was passed through two electrolytic cells A and B containing electrolytes ZnSO4 and CuSO4 connected in series, 2g of Cu were deposited at the cathode of cell B. How long did the current flow? What mass of Zn was deposited at cathode of cell A?
[Atomic mass: Cu=63.5gmol−1,Zn=65gmol−1 ; 1F=96500Cmol−1]

Solution:

‌Zn2+(aq)+‌

2e−

2‌mol

→‌

Zn( s

1‌mol

)
‌Cu2++‌

2e−

2‌mol

→‌

Cu(s)

1‌mol

( 2 gm given)
The charge Q on a mole of electrons, Q=nF
Calculation of time for the flow of current:
‌n=1‌mol
‌Q=1×96500Cmol−1=96500C
Molar mass of Cu=63.5‌gm‌mol−1
‌∵‌‌63.5‌gm‌ of ‌Cu‌ is deposited by electric charge ‌
‌=96500C
‌∴‌‌2‌gm‌ of ‌Cu‌ is deposited by electric charge ‌
‌=‌

96500

63.5

×2=3039.37C
Let 2A of current be passed for time t, quantity of electricity used =2A×t=3039.37C
‌‌ or, ‌t=‌

3039.37C

=1519.68 s
‌=25‌min‌.33‌ s
Calculation of mass of Zn deposited:
‌

‌=‌

=‌

‌ Mass of ‌Zn

‌ Mass of ‌Cu

‌=‌

‌ Molar mass of ‌Zn∕‌ Charge on ‌Cu

‌ Molar mass of ‌Cu∕‌ Charge on ‌Cu

Amount of Zn deposited:
=2×‌

‌

635

=2.0472‌gm

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