CBSE Class 12 Chemistry 2019 Delhi Set 1 Solved Paper

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Question : 25
Total: 27
SECTION - D

E°‌cell ‌ for the given redox reaction is 2.71V.
Mg+Cu2+‌(0.01M)→Mg2+‌(0.001M)+Cu
Calculate E‌cell ‌ for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) Less than 2.71V
(ii) Greater than 2.71V
OR
(a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4 and ZnSO4 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass
‌‌ (Molar mass: Fe ‌=56gmol−1,Zn =65.3gmol−1,1F
‌=96500Cmol−1)
(b) In the plot of molar conductivity Λm vs. square root of concentration (C1∕2), following curve obtained for two electrolytes A and B:
Answer the following :
(i) Predict the nature of electrolytes A and B:
(ii) What happens on extrapolation of Λm to concentration approaching zero for electrolytes A and B ?
Solution:  
‌E‌cell ‌=E°‌cell ‌−‌
0.059
n
‌log
‌
[Mg2+]
[Cu2+]

‌E‌cell ‌=2.71−‌
0.059
2
‌log
‌
[0.001]
[0.01]

‌E‌cell ‌=2.71−(−0.0295)=2.74V
(i) When external opposite applied voltage is less than 2.71, it is less than E°‌cell ‌, therefore, the electrons will flow from the anode to the cathode, and current will flow from cathode (copper electrode) to anode (magnesium electrode).

(ii) When external opposite applied potential is greater than 2.71, it is greater than E°‌cell ‌, therefore, the reaction will be reversed, and the current will flow from anode to cathode.
OR
(a) Charge required to deposite 2.8g‌Fe :

mol‌ ‌Fe=‌
‌ mass ‌
‌ molar mass ‌
=‌
2.8g
56g⋅mol−1
=0.05‌mol

2F charge is required to discharge 1‌mol of Fe2+ ions as Fe, therefore deposition of 0.05‌mol‌Fe will need
0.05×2=0.1F=0.1F ×‌
96500C
F
=9650C

The quantity of charge is related to current as
Q=It
Therefore, the time needed to deposit 2.8g Fe is:

t=‌
Q
I
=‌
9650C
2A
=4825 s

So, the current flowed through the cells for 4825 seconds.

The amount of Zn deposited in cell Y can be calculated using Faraday's second law:

‌‌
‌ mass of ‌Zn
‌ mass of ‌Fe
=‌
‌ Eq.wt of ‌Zn
‌ Eq wt of ‌Fe

‌‌‌=‌
‌ molar mass of ‌Zn∕‌ chargeon zincion ‌
‌ molarmass of ‌Fe∕‌ chargeon iron ion ‌

‌‌ mass of ‌Zn=2.8g×‌
65.3g∕2
56g∕2
=3.265g≈3.3g
Therefore, the mass of Zn deposited in cell Y in the same time is 3.3g.

(b) (i) Molar conductivity of strong electrolytes increases linearly as the square root of the concentration decreases; therefore, electrolyte A is a strong electrolyte. Molar conductivity of weak electrolytes increases non-linearly as square root of concentration decreases; therefore, electrolyte B is a weak electrolyte.
(ii) As concentration of strong electrolyte approaches zero, the molar conductivity of the plot intercepts the molar conductivity axis, giving the limiting value of molar conductivity Em0. The plot of molar conductivity of weak electrolyte tends to infinity as its concentration approaches zero; it does not intersect the molar conductivity axis.
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