NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry plays a pivotal role in the CBSE class 12 Chemistry examination. Chemistry Class 12 solutions chapter 3 is a comprehensive material that has answers to the textbook questions, important questions from previous papers. By studying chemistry class 12 NCERT solutions chapter 3, you will be able to solve different kinds of questions you can expect to appear in the main examination and entrance examinations.
Electrochemistry is the branch of chemistry that deals with the relationship between chemical energy and electrical energy produced in a redox reaction and how they can be converted into each other. The solutions are prepared by the best subject experts. In essence, these solutions can be useful for those students preparing for class 12 board exams and also for JEE advance and other medical entrance exams. Students can successfully answer for numerical problems based on electrochemistry by downloading the free pdf.
Class 12 NCERT Solutions for Electrochemistry
The NCERT solutions for chapter 3 – Electrochemistry has mainly been designed to help the students in preparing well and score good marks in CBSE class 12 Chemistry paper. Further, the solutions consist of well thought or structured questions along with detailed explanations to help students learn and remember concepts easily.
Subtopics for Class 12 Chemistry Chapter 3 – Electrochemistry
 Electrochemical Cells

Galvanic Cells
 Measurement of Electrode Potential

Nernst Equation
 Equilibrium Constant from Nernst Equation
 Electrochemical Cell and Gibbs Energy of Reaction

The conductance of Electrolytic Solutions
 Measurement of the Conductivity of Ionic Solutions
 Variation of Conductivity and Molar Conductivity with Concentration

Electrolytic Cells and Electrolysis
 Products of Electrolysis

Batteries
 Primary Batteries
 Secondary Batteries
 Fuel Cells
 Corrosion
After studying electrochemistry class 12 important questions and solutions, you will be able to describe an electrochemical cell and differentiate between galvanic and electrolytic cells. You will study the application of the Nernst equation for calculating the emf of a galvanic cell and define the standard potential of the cell.
This chapter has derivations of the relation between the standard potential of the cell, Gibbs energy of cell reaction and its equilibrium constant. This solution will give the definition of resistivity (p), conductivity (K) and molar conductivity ( Am) of ionic solutions; differentiate between ionic (electrolytic) and electronic conductivity; describe the method for measurement of conductivity of electrolytic solutions and calculation of their molar conductivity; justify the variation of conductivity and molar conductivity of solutions with change in their concentration and define Aom (molar conductivity at zero concentration or infinite dilution); enunciate Kohlrausch law and learn its applications; understand quantitative aspects of electrolysis; describe the construction of some primary and secondary batteries and fuel cells and explain corrosion as an electrochemical process.
Class 12 Chemistry NCERT Solutions (Electrochemistry) – Important Questions
Q 3.1:
Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn
Answer:
According to their reactivity, the given metals replace the others from their salt solutions in the said order: Mg, Al, Zn, Fe, Cu.
Mg: Al: Zn: Fe: Cu
Q 3.2:
Given the standard electrode potentials,
K+/K = –2.93V,
Ag+/Ag = 0.80V,
Hg2+/Hg = 0.79V
Mg2+/Mg = –2.37 V, Cr3+/Cr = – 0.74V
Arrange these metals in their increasing order of reducing power.
Ans:
The reducing power increases with the lowering of reduction potential. In order of given standard electrode potential (increasing order) : K^{+}/K < Mg^{2+}/Mg < Cr^{3+}/Cr < Hg^{2+}/Hg < Ag^{+}/Ag
Thus, in the order of reducing power, we can arrange the given metals as Ag< Hg < Cr < Mg < K
Q 3.3 :
Depict the galvanic cell in which the reaction
Zn(s)+2Ag+(aq) →Zn2+(aq)+2Ag(s) 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) The negatively charged electrode is the Zn electrode (anode)
(ii) The current carriers in the cell are ions. Current flows to zinc from silver in the external circuit.
(iii) Reaction at the anode is given by :
Reaction at the anode is given by :
Q 3.4:
Calculate the standard cell potentials of galvanic cell in which the following
reactions take place:
(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd
(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)
Calculate the ∆rGJ 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}
Then,
= −196833.48 CV mol−1
= −196833.48 J mol−1
= −196.83 kJ mol−1
Again,
= 34.496
K = antilog (34.496) = 3.13 × 10^{34}
The galvanic cell of the given reaction is depicted as:
Now, the standard cell potential is
Here, n = 1.
Then,
= −1 × 96487 C mol^{−1} × 0.03 V
= −2894.61 J mol^{−1}
= −2.89 kJ mol^{−1}
Again,
= 0.5073
K = antilog (0.5073)
= 3.2 (approximately)
Q 3.5:
Write the Nernst equation and emf of the following cells at 298 K:
(i) Mg(s)Mg2+(0.001M)Cu2+(0.0001 M)Cu(s)
(ii) Fe(s)Fe2+(0.001M)H+(1M)H2(g)(1bar) Pt(s)
(iii) Sn(s)Sn2+(0.050 M)H+(0.020 M)H2(g) (1 bar)Pt(s)
(iv) Pt(s)Br–(0.010 M)Br2(l )H+(0.030 M) H2(g) (1 bar)Pt(s).
Answer
(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 – ( – 0.14) –
= 0.52865 V
= 0.53 V (approximately)
(iii) For the given reaction, the Nernst equation can be given as:
= 0 – ( – 0.14) –
= 0.14 − 0.0295 × log125
= 0.14 − 0.062
= 0.078 V
= 0.08 V (approximately)
(iv) For the given reaction, the Nernst equation can be given as:
= 0 – 1.09 –
= 1.09 – 0.02955 x
= 1.09 – 0.02955 x
= 1.09 – 0.02955 x
= 1.09 – 0.02955 x (0.0453 + 7)
= 1.09 – 0.208
= 1.298 V
Q 3.6:
In the button cells widely used in watches and other devices the following reaction takes place:
Determine ∆r GJ and EJ for the reaction.
Ans:
We know that,
= −2 × 96487 × 1.04
= −213043.296 J
= −213.04 kJ
Q 3.7:
Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.
Answer
The conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of crosssection 1 sq. cm. Specific conductance is the inverse of resistivity and it is represented by the symbol κ. If ρ is resistivity, then we can write:
At any given concentration, the conductivity of a solution is defined as the unit volume of solution kept between two platinum electrodes with the unit area of crosssection at a distance of unit length.
When concentration decreases there will a decrease in Conductivity. It is applicable for both weak and strong electrolyte. 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 crosssection A and distance of unit length.
Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).
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
Q 3.8:
The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm–1. Calculate its molar conductivity
Ans :
Given, κ = 0.0248 S cm^{−1} c
= 0.20 M
Molar conductivity,
= 124 Scm^{2}mol^{1}
Q 3.9:
The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 Ω. What is the cell constant if conductivity of 0.001M KCl solutionat 298 K is 0.146 × 10–3 S cm–1
Answer
Given,
Conductivity, k = 0.146 × 10^{−3} S cm−1
Resistance, R = 1500 Ω
Cell constant = k × R
= 0.146 × 10^{−3} × 1500
= 0.219 cm^{−1}
Q 3.10:
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
10^{2} × k/S m^{−1} 1.237 11.85 23.15 55.53 106.74
Calculate Λm for all concentrations and draw a plot between Λm and c½. Find the value of 0 Λ m.
Ans:
Given,
κ = 1.237 × 10^{−2} S m−1, c = 0.001 M
Then, κ = 1.237 × 10^{−4} S cm^{−1}, c^{1⁄2} = 0.0316 M^{1/2}
= 123.7 S cm^{2} mol^{−1}
Given,
κ = 11.85 × 10^{−2} S m^{−1}, c = 0.010M
Then, κ = 11.85 × 10^{−4} S cm^{−1}, c^{1⁄2} = 0.1 M^{1/2}
= 118.5 S cm^{2} mol^{−1}
Given,
κ = 23.15 × 10^{−2} S m^{−1}, c = 0.020 M
Then, κ = 23.15 × 10^{−4} S cm^{−1}, c^{1/2} = 0.1414 M^{1/2}
= 115.8 S cm^{2} mol^{−1 }
Given,
κ = 55.53 × 10^{−2} S m^{−1}, c = 0.050 M
Then, κ = 55.53 × 10^{−4} S cm^{−1}, c^{1/2} = 0.2236 M^{1/2}
= 111.1 1 S cm^{2} mol^{−1}
Given,
κ = 106.74 × 10^{−2} S m^{−1}, c = 0.100 M
Then, κ = 106.74 × 10^{−4} S cm^{−1}, c^{1/2} = 0.3162 M^{1/2}
= 106.74 S cm^{2} mol^{−1}
Now, we have the following data :
Since the line interrupts
Q 3.11:
Conductivity of 0.00241 M acetic acid is 7.896 × 10–5 S cm–1. Calculate its molar conductivity. If 0 Λ m for acetic acid is 390.5 S cm2 mol–1, what is its dissociation constant?
Ans:
Given, κ = 7.896 × 10^{−5} S m^{−1} c
= 0.00241 mol L^{−1}
Then, molar conductivity,
=
= 32.76S cm^{2} mol^{−1}
Again,
=
Now,
= 0.084
Dissociation constant,
=
= 1.86 × 10^{−5} mol L^{−1}
Q 3.12:
How much charge is required for the following reductions:
(i) 1 mol of Al3+ to Al?
(ii) 1 mol of Cu2+ to Cu?
(iii) 1 mol of MnO4– to Mn2+?
Ans :
(i)
Required charge = 3 F
= 3 × 96487 C
= 289461 C
(ii)
Required charge = 2 F
= 2 × 96487 C
= 192974 C
(iii)
i.e
Required charge = 5 F
= 5 × 96487 C
= 482435 C
Q 3.13:
How much electricity in terms of Faraday is required to produce
(i) 20.0 g of Ca from molten CaCl2?
(ii) 40.0 g of Al from molten Al2O3?’
Ans:
(i) From given data,
Electricity required to produce 40 g of calcium = 2 F
Therefore, electricity required to produce 20 g of calcium = (2 x 20 )/ 40 F
= 1 F
(ii) From given data,
Electricity required to produce 27 g of Al = 3 F
Therefore, electricity required to produce 40 g of Al = ( 3 x 40 )/27 F
= 4.44 F
Q 3.14:
How much electricity is required in coulomb for the oxidation of
(i) 1 mol of H2O to O2?
(ii) 1 mol of FeO to Fe2O3?
Ans :
(i) From given data,
We can say that :
Electricity required for the oxidation of 1 mol of H_{2}O to O_{2} = 2 F
= 2 × 96487 C
= 192974 C
(ii) From given data,
Electricity required for the oxidation of 1 mol of FeO to Fe_{2}O_{3} = 1 F
= 96487 C
Q 3.15:
A solution of Ni(NO3)2 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 = 20 × 60 = 1200 s
Charge = current × time
= 5 × 1200
= 6000 C
According to the reaction,
Nickel deposited by 2 × 96487 C = 58.71 g
Therefore, nickel deposited by 6000 C =
= 1.825 g
Hence, 1.825 g of nickel will be deposited at the cathode.
Q 3.16:
Three electrolytic cells A,B,C containing solutions of ZnSO4, AgNO3 and CuSO4, 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
Time = 1295.43/ 1.5 s
= 863.6 s
= 864 s
= 14.40 min
Again,
i.e., 2 × 96487 C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit
= 0.426 g of Cu
i.e., 2 × 96487 C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit
= 0.439 g of Zn
Q 3.17:
Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:
(i) Fe3+(aq) and I–(aq)
(ii) Ag+ (aq) and Cu(s)
(iii) Fe3+ (aq) and Br– (aq)
(iv) Ag(s) and Fe 3+ (aq)
(v) Br2 (aq) and Fe2+ (aq).
Ans :
(i)
(ii)
E^{0 } is positive, hence reaction is feasible.
(iii)
E^{0 } is negative, hence reaction is not feasible.
(iv)
E^{0 } is negative, hence reaction is not feasible.
(v)
E^{0 } is positive, hence reaction is feasible.
Q 3.18:
Predict the products of electrolysis in each of the following:
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of AgNO3 with platinum electrodes.
(iii) A dilute solution of H2SO4 with platinum electrodes.
(iv) An aqueous solution of CuCl2 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 E^{0} takes place at the cathode. Therefore, deposition of silver will take place at the cathode.
At anode:
The Ag anode is attacked by
(ii) At cathode:
The following reduction reactions compete to take place at the cathode.
The reaction with a higher value of E^{0} 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
(iii) At the cathode, the following reduction reaction occurs to produce H_{2} gas.
At the anode, the following processes are possible.
For dilute sulphuric acid, reaction (i) is preferred to produce O_{2} 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.
At anode:
The following oxidation reactions are possible at the anode.
At the anode, the reaction with a lower value of E^{0} is preferred. But due to the over potential of oxygen, Cl^{−} gets oxidized at the anode to produce Cl_{2} gas.
Also Access 
NCERT Exemplar for class 12 Chemistry Chapter 3 
CBSE Notes for class 12 Chemistry Chapter 3 
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