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
5 Marks Questions
1. Explain construction and working of standard Hydrogen electrode?
Ans. Construction :
SHE consists of a platinum electrode coated with platinum black. The electrode is dipped in an acidic solution and pure Hydrogen gas is bubbled through it. The concentration of both the reduced and oxidized. Forms of Hydrogen is maintained at unity i.e) pressure of 
gas is 1 bar and concentration of Hydrogen ions in the solution is 1 molar.
Working – The reaction taking place in SHE is 
At 298 K , the emf of the cell constructed by taking SHE as anode and other half cell as cathode, gives the reduction potential of the other half cell where as for a cell constructed by taking SHE as anode gives the oxidation potential of other half cell as conventionally the electrode potential of SHE is zero.
2. The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S 
. Calculate its degree of dissociation and dissociation constant. Given 
= 349.6 Sand 
Ans. C = 0.025 
= 349.6 + 54.6 = 
Now, degree of dissociation:
= 0.114 (approximately)
Thus, dissociation constant:
3. Explain how rusting of iron is envisaged as setting up of an electrochemical cell.
Ans. In the process of corrosion, due to the presence of air and moisture, oxidation takes place at a particular spot of an object made of iron. That spot behaves as the anode. The reaction at the anode is given by, 
Electrons released at the anodic spot move through the metallic object and go to another spot of the object.
There, in the presence of 
ions, the electrons reduce oxygen. This spot behaves as the cathode. These H+ ions come either from
, which are formed due to the dissolution of carbon dioxide from air into water or from the dissolution of other acidic oxides from the atmosphere in water.
The reaction corresponding at the cathode is given by, 
The overall reaction is: 
Also, ferrous ions are further oxidized by atmospheric oxygen to ferric ions. These ferric ions combine with moisture, present in the surroundings, to form hydrated ferric oxide 
i.e., rust.
Hence, the rusting of iron is envisaged as the setting up of an electrochemical cell.
4. Calculate the standard cell potentials of galvanic cells in which the following reactions take place:
(i) 
(ii) 
Calculate the 
¸ and equilibrium constant of the reactions
Ans. (i)
The galvanic cell of the given reaction is depicted as: 
Now, the standard cell potential is 
= 0.40 – (–0.74)
= +0.34 V
In the given equation,
n = 6
F = 96487 C mol – 1
= +0.34 V
Then, 
= –196833.48 CV 
= –196833.48 J
= –196.83 kJ v
Again, 
= 34.496
Therefore, K = antilog (34.496)
(ii) 
The galvanic cell of the given reaction is depicted as: 
Now, the standard cell potential is
= 0.80 – 0.77
= 0.03 V
Here, n = 1.
Then,
= 
= –2894.61 J
= –2.89 kJ
Again, 
= 0.5073
Therefore, K = antilog (0.5073)
= 3.2 (approximately)
5. Write the Nernst equation and emf of the following cells at 298 K:
(i) 
(ii) 
(iii) 
(iv) 
Ans. (i) For the given reaction, the Nernst equation can be given as:
= 2.7 – 0.02955
= 2.67 V (approximately)
(ii) For the given reaction, the Nernst equation can be given as:
= 0.44–0.02955(–3)
= 0.52865 V
= 0.53 V (approximately)
(iii) For the given reaction, the Nernst equation can be given as:
= 
= 0.14–0.062
= 0.078 V
= 0.08 V (approximately)
(iv) For the given reaction, the Nernst equation can be given as:
= –1.09–0.02955(0.0453+7)
= –1.09–0.208
= –1.298 V
6. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Ans. Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol K. If 
is resistivity, then we can write: 
The conductivity of a solution at any given concentration is the conductance (G) of one unit volume of solution kept between two platinum electrodes with the unit area of cross-section and at a distance of unit length.
i.e., 
(Since a = 1, l = 1)
Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.
Molar conductivity:
Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length. 
Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).
Therefore, 
Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.
The variation of 
with 
for strong and weak electrolytes is shown in the following plot:
7. The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below: Concentration/M 0.001 0.010 0.020 0.050 0.100 
Calculate for all concentrations and draw a plot betweenand 
. Find the value of 
.
Ans. Given, 
Then, 
Therefore, 
Given,
Then, 
Therefore,
= 
Given,
Then, 
Therefore,
Given,
Then, 
Therefore, 
= 
Given,
Then, 
Therefore,
Now, we have the following data:
 | 0.0316 | 0.1 | 0.1414 | 0.2236 | 0.3162 |
 | 123.7 | 118.5 | 115.8 | 111.1 | 106.74 |
Since the line interrupts at 
, 
8. Conductivity of 0.00241 M acetic acid is 
. Calculate its molar conductivity and if for acetic acid is 
, what is its dissociation constant?
Ans. Given, 
Then, molar conductivity,
Again, 
Now, 
= 0.084
Therefore, Dissociation constant, 
9. Three electrolytic cells A,B,C containing solutions of 
and 
respectively are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?
Ans. According to the reaction:
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by = 
= 1295.43 C
Given,
Current = 1.5 A
Therefore, Time 
= 863.6 s
= 864 s
= 14.40 min
Again,
i.e., 
C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit = 
= 0.426 g of Cu
i.e., C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit =
= 0.439 g of Zn
10. Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) 
and
(ii) 
and Cu(s)
(iii) and 
(iv) Ag(s) and
(v) 
and 
.
Ans.
Since
for the overall reaction is positive, the reaction between and is feasible.
Since for the overall reaction is positive, the reaction between 
and Cu(s) is feasible.
Sincefor the overall reaction is negative, the reaction between and is not feasible.
Since E for the overall reaction is negative, the reaction between Ag(s) and is not feasible.
Sincefor the overall reaction is positive, the reaction between and 
is feasible.
11. Predict the products of electrolysis in each of the following:
(i) An aqueous solution of 
with silver electrodes.
(ii) An aqueous solution of with platinum electrodes.
(iii) A dilute solution of 
with platinum electrodes.
(iv) An aqueous solution of 
with platinum electrodes.
Ans. (i) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
The Ag anode is attacked by 
ions. Therefore, the silver electrode at the anode dissolves in the solution to form 
.
(ii) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
Since Pt electrodes are inert, the anode is not attacked by ions. Therefore, 
or ions can be oxidized at the anode. But ions having a lower discharge potential and get preference and decompose to liberate
.
(iii) At the cathode, the following reduction reaction occurs to produce 
gas.
At the anode, the following processes are possible.
……………(1)
……………(2)
For dilute sulphuric acid, reaction (i) is preferred to produce gas. But for concentrated sulphuric acid, reaction (ii) occurs.
(iv) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of 
takes place at the cathode. Therefore, deposition of copper will take place at the cathode.
Atanode:
The following oxidation reactions are possible at the anode.
At the anode, the reaction with a lower value of is preferred. But due to the over-potential of oxygen, 
gets oxidized at the anode to produce 
gas.