CBSE Class 12 Chemistry Chapter 3 Important Questions – Free PDF Download
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Important Questions Class 12 Chemistry Chapter 3 – Electrochemistry
CBSE Class 12 Chemistry Important Questions Chapter 3 – Electrochemistry
3 Marks Questions
1.What is the cell potential for the cell at 
;
.
Ans.The cell reaction is
Nernst Equation –
=(-0.44v – (-0.74v) – 
= 
= 0.3V – 0.0394V
= +0.2606 V
2.Calculate 
for the reaction at 
, 
, R = 8.314 J/K.
Ans.The half cell reactions are
Anode: 
Cathode: 
Nernst Equation
= (- 0.403 – (-0.763) – 
= 0.36V – 0.0798V = 0.4398V
= -8488 J mol-1
3.Calculate Equilibrium constant K for the reaction at 
= +0.34v.
Ans.From the reaction, n =2
= + 0.34v – (-0.76v) = 1.10V
At 298k , 
Log kc =
= 
4. For what concentration of 
will the emf of the given cell be zero at
if the concentration of 
is 0.1 M ? 
Ans.
5.Calculate the standard free energy change for the cell- reaction.
How is it related to the equilibrium constant of the reaction?
, 
Ans.
6.How much charge is required for the following reductions:
(i) 1 mol of 
to Al.
(ii) 1 mol of 
to Cu.
(iii) 1 mol of 
to 
.
Ans.(i) 
Therefore, Required charge = 3 F
= 289461 C
(ii) 
Therefore, Required charge = 2 F
= 192974 C
(iii) 
i.e., 
Therefore, Required charge = 5 F
= 482435 C
7. How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten 
.
(ii) 40.0 g of Al from molten 
.
Ans.(i) According to the question,
Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium =
= 1 F
(ii) According to the question,
Electricity required to produce 27 g of Al = 3 F
Therefore, electricity required to produce 40 g of Al = 
= 4.44 F
8. How much electricity is required in coulomb for the oxidation of
(i) 1 mol of 
to 
.
(ii) 1 mol of FeO to 
.
Ans.(i) According to the question,
Now, we can write:
Electricity required for the oxidation of 1 mol of to = 2 F
= 192974 C
(ii) According to the question,
Electricity required for the oxidation of 1 mol of FeO to = 1 F
= 96487 C
9. A solution of 
is electrolysed between platinum electrodes using a current of 5 amperes for 20 minutes. What mass of Ni is deposited at the cathode?
Ans.Given,
Current = 5A
Time = 
= 1200 s
Therefore, 
= 6000 C
According to the reaction,
Nickel deposited by 
= 58.71 g
Therefore, nickel deposited by 6000 C 
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.
10. Depict the galvanic cell in which the reaction takes place. Further show:
(i) Which of the electrode is negatively charged?
(ii) The carriers of the current in the cell.
(iii) Individual reaction at each electrode.
Ans.The galvanic cell in which the given reaction takes place is depicted as:
(i) Zn electrode (anode) is negatively charged.
(ii) Ions are carriers of current in the cell and in the external circuit, current will flow from silver to zinc.
(iii) The reaction taking place at the anode is given by,
The reaction taking place at the cathode is given by,
11. Write the chemistry of recharging the lead storage battery, highlighting all the materials that are involved during recharging.
Ans.A lead storage battery consists of a lead anode, a grid of lead packed with lead oxide 
as the cathode, and a 38% solution of sulphuric acid 
as an electrolyte.
When the battery is in use, the following cell reactions take place:
At anode: 
At cathode: 
The overall cell reaction is given by,
When a battery is charged, the reverse of all these reactions takes place.
Hence, on charging, 
present at the anode and cathode is converted into 
and 
respectively.
12. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire?
Ans.I = 0.5 A
t = 2 hours = 
Thus, Q = It
= 
= 3600 C
We know that 
number of electrons.
Then, 
number of electrons
number of electrons
Hence, 
number of electrons will flow through the wire.
13. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
Ans.For hydrogen electrode, 
, it is given that pH = 10
Therefore, 
= 10 – 10M
Now, using Nernst equation:
= – 0.0591 log 
= – 0.591 V
14. Calculate the emf of the cell in which the following reaction takes place:
Given that 
Ans.Applying Nernst equation we have:
= 1.05 – 0.02955 log 4 
104
= 1.05 – 0.02955 (log 10000 + log 4)
= 1.05 – 0.02955 (4 + 0.6021)
= 0.914 V
15. The cell in which the following reactions occurs:
has 
= 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Ans.Here, n = 2, 
T = 298 K
We know that:
= – 2 × 96487 × 0.236
= – 45541.864 J mol – 1
= – 45.54 kJ mol – 1
Again, 
= 7.981
∴Kc= Antilog (7.981) = 9.57 × 107
16. How would you determine the standard electrode potential of the systemMg2+| Mg?
Ans.The standard electrode potential of 
| Mg can be measured with respect to the standard hydrogen electrode, represented by Pt(s), 
(1 atm) | 
(aq) (1M).
A cell, consisting of Mg | 
(aq 1 M) as the anode and the standard hydrogen electrode as the cathode, is set up.
Then, the emf of the cell is measured and this measured emf is the standard electrode potential of the magnesium electrode.
Here,
for the standard hydrogen electrode is zero.
Therefore, 
=