## NCERT Solutions for Class 12 Chemistry Chapter 3 – Free PDF Download

**NCERT Solutions for Class 12 Chemistry Chapter 3 Electrochemistry** play a pivotal role in the CBSE Class 12 Chemistry board examination. NCERT Solutions for Class 12 Chemistry are comprehensive materials that have answers to the exercise present in the NCERT Textbook. These solutions are developed by subject experts at BYJU’S, following the latest CBSE Syllabus for 2022-23 and its guidelines.

By studying these **NCERT Solutions for Class 12 Chemistry,** students will be able to solve different kinds of questions that might appear in the board examination and entrance examinations. These solutions are presented in a clear and step-wise format for ease of understanding. To download the NCERT Solutions for Class 12 Chemistry Chapter 3 PDF, click the link given below.

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### Class 12 Chemistry NCERT Solutions Chapter 3 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, and Cu.

Mg: Al: Zn: Fe: Cu

*Q 3.2:*

**Given the standard electrode potentials.****K ^{+}/K = –2.93V**

**Ag ^{+}/Ag = 0.80V**

**Hg ^{2+}/Hg = 0.79V**

**Mg ^{2+}/Mg = –2.37 V**

**Cr ^{3+}/Cr = – 0.74V**

**Arrange these metals in their increasing order of reducing power.**

*Ans:*

The reducing power increases with the lowering of the 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) →Zn^{2+}(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 the galvanic cell in which the following****reactions take place.****(i) 2Cr(s) + 3Cd ^{2+}(aq) → 2Cr^{3+}(aq) + 3Cd**

**(ii) Fe**

^{2+}(aq) + Ag^{+}(aq) → Fe^{3+}(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, ^{−1} × 0.34 V

= −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)|Mg ^{2+}(0.001M)||Cu^{2+}(0.0001 M)|Cu(s)**

**(ii) Fe(s)|Fe**

^{2+}(0.001M)||H^{+}(1M)|H2(g)(1bar)| Pt(s)**(iii) Sn(s)|Sn**

^{2+}(0.050 M)||H^{+}(0.020 M)|H2(g) (1 bar)|Pt(s)**(iv) Pt(s)|Br–(0.010 M)|Br**

_{2}(l )||H^{+}(0.030 M)| H_{2}(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 cross-section 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 the cross-section at a distance of unit length.

When concentration decreases, there will be a decrease in Conductivity. It is applicable for both 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 – **

The 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)

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 the 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

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 the conductivity of 0.001M KCl solution at 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 ^{2} mol^{−1}, ^{2} mol^{−1}

*Q 3.11:*

**The 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 cm**

^{2}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}

^{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 Al ^{3+} to Al**

**(ii) 1 mol of Cu**

^{2+}to Cu**(iii) 1 mol of MnO**

_{4}^{–}to Mn^{2+}*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 CaCl _{2}?
(ii) 40.0 g of Al from molten Al_{2}O_{3}?*

*Ans:*

(i) From the 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 the 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 H _{2}O to O_{2}?**

**(ii) 1 mol of FeO to Fe**

_{2}O_{3}?*Ans :*

(i) From the 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 the 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(NO _{3})_{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, and C, containing solutions of ZnSO _{4}, AgNO_{3} and CuSO_{4}, respectively, are connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver was 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) Fe ^{3+}(aq) and I^{–}(aq)**

**(ii) Ag**

^{+}(aq) and Cu(s)**(iii) Fe**

^{3+}(aq) and Br^{–}(aq)**(iv) Ag(s) and Fe**

^{3+}(aq)**(v) Br**

_{2}(aq) and Fe^{2+}(aq)*Ans :*

(i)

(ii)

E^{0 } is positive; hence, the reaction is feasible.

(iii)

E^{0 } is negative; hence, the reaction is not feasible.

(iv)

E^{0 } is negative; hence, the reaction is not feasible.

(v)

E^{0 } is positive; hence, the reaction is feasible.

*Q 3.18:*

**Predict the products of electrolysis in each of the following.****(i) An aqueous solution of AgNO _{3} with silver electrodes**

**(ii) An aqueous solution of AgNO**

_{3}with platinum electrodes**(iii) A dilute solution of H**

_{2}SO_{4}with platinum electrodes**(iv) An aqueous solution of CuCl**

_{2}with platinum electrodes*Ans:*

(i) At the cathode,

The following reduction reactions compete to take place at the cathode.

^{0 }= 0.80 V

^{0 }= 0.00 V

The reaction with a higher value of E^{0} takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.

At the anode,

The Ag anode is attacked by ^{+}.

(ii) At the cathode,

The following reduction reactions compete to take place at the cathode.

^{0 }= 0.80 V

^{0 }= 0.00 V

The reaction with a higher value of E^{0} takes place at the cathode. Therefore, the deposition of silver will take place at the cathode.

At the anode,

Since Pt electrodes are inert, the anode is not attacked by ^{−} or ^{−} ions have a lower discharge potential and get preference and decompose to liberate O_{2}.

(iii) At the cathode, the following reduction reaction occurs to produce H_{2} gas.

At the anode, the following processes are possible.

^{0}= +1.23 V —–(i)

^{0}= +1.96 V —–(ii)

For dilute sulphuric acid, reaction (i) is preferred to produce O_{2} gas. But for concentrated sulphuric acid, reaction (ii) occurs.

(iv) At the cathode,

The following reduction reactions compete to take place at the cathode.

^{0 }= 0.34 V

^{0 }= 0.00 V

The reaction with a higher value takes place at the cathode. Therefore, the deposition of copper will take place at the cathode.

At the anode,

The following oxidation reactions are possible at the anode.

^{0 }= 1.36 V

^{0 }= +1.23 V

At the anode, the reaction with a lower value of E^{0} is preferred. But due to the overpotential of oxygen, Cl^{−} gets oxidised 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 |

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. **NCERT Solutions for Class 12** are prepared by the best subject experts. In essence, these solutions can be useful for those students preparing for Class 12 exams, JEE Advance and other medical entrance exams. Students can successfully answer the numerical problems based on electrochemistry by downloading the free PDF.

## Class 12 NCERT Solutions for Chemistry Chapter 3 Electrochemistry

Chapter 3 Electrochemistry of Class 12 Chemistry is prepared as per the CBSE Syllabus for 2022-23. Class 12 Chemistry **NCERT Solutions** for Chapter 3 – Electrochemistry have been designed to help the students prepare well and score good marks in the CBSE Class 12 Chemistry exam. Further, the solutions consist of well thought and 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 textbook questions solutions, students will be able to describe an electrochemical cell and differentiate between galvanic and electrolytic cells. They will also 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.

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With these **NCERT Solutions for Class 12**, students can easily customise how they learn. Apart from these NCERT Solutions, BYJU’S also has the best subject experts who can guide students to learn the concepts in a more simple and precise manner. Further, if students come across any doubts or queries while going through the **NCERT Class 12 Chemistry Solutions**, they can always approach BYJU’S responsive support team to clear all their doubts. Besides, BYJU’S keeps track of all the progress that students make and offers feedback, as well as counselling via periodic assessments. Moreover, students can bring in all their queries regarding Chemistry, and other subjects, including Physics, Biology and Maths.