Hello Students. Are you Searching for NCERT Solutions for Class 12 Chemistry Chapter 3? If yes then you are most welcome to NCERTian. Here we have provided you with the complete Question and Answers of Chapter 3: Electrochemistry. These solutions are written by expert teachers and faculties keeping the need of students in mind.

Chapter | 3. Electrochemistry |

Subject | Chemistry |

Textbook | Chemistry I |

Class | Twelve |

Category | NCERT Solutions for Class 12 |

NCERT Solutions for Class 12 Chemistry is an excellent tool to revise your syllabus quickly. These solutions are made by expert faculties and teachers of Chemistry. It provides you with complete information, making it easier for you to solve the problems. The NCERT textbook includes questions based on concepts and theories. They are important for testing students’ understanding of the topics. On this page, we have provided you with the NCERT Solutions for Class 12 Chemistry Chapter 3: Electrochemistry.

## NCERT Solutions for Class 12 Chemistry Chapter 3

### Electrochemistry Solutions

**Q1) 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) **The following is the order in which the given metals displace each other from the solution of their salts.

Mg, Al, Zn, Fe, Cu

**Q2) 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.**

The lower the reduction potential, the higher is the reducing power. The given standard electrode potentials increase in the order of K^{+/K} < Mg^{2+}/Mg < Cr^{3+}/Cr < Hg^{2+}/Hg < Ag^{+}/Ag.

Hence, the reducing power of the given metals increases in the following order:

Ag < Hg < Cr < Mg < K

**Q3) Depict the galvanic cell in which the reaction Zn(s) + 2Ag**^{+}(aq) → Zn^{2+}(aq) + 2Ag(s) takes place. Further show:

^{+}(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.**

**Answer)** The galvanic cell in which the given reaction takes place is depicted as:

Zn_{(s)} ∣Zn^{2+}_{(aq)} ∣∣ Ag^{+}_{(aq)} ∣ Ag_{(s)}

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,

Zn_{(s)} → Zn^{2+}(aq) + 2e-

The reaction taking place at the cathode is given by,

Ag^{+}(aq) + e^{–} → Ag(s)

**Q4) Calculate the standard cell potentials of galvanic cells 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 Δ _{r}G^{θ }and equilibrium constant of the reactions.**

**Answer i)**

**Answer ii)**

**Q5) 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)|H_{2}(g)(1bar) | Pt(s)**(iii) Sn(s) | Sn**

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

_{2}(l) | Br^{−}(0.010 M) ||^{H+}(0.030 M) | H_{2}(g) (1 bar) | Pt(s).**Answer i)** For the given reaction, the Nernst equation can be given as:

**Answer ii)** For the given reaction, the Nernst equation can be given as:

**Answer iii)** For the given reaction, the Nernst equation can be given as:

**Answer iv)** For the given reaction, the Nernst equation can be given as:

**Q6) In the button cells widely used in watches and other devices the following reaction takes place:**

**Zn(s) + Ag _{2}O(s) + H_{2}O(l) → Zn^{2+}(aq) + 2Ag(s) + 2OH−(aq)**

**Determine ∆ _{r} GJ and EJ for the reaction.**

**Answer)**

**Q7) Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.**

**Answer)** 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 symbolκ. If ρ is resistivity, then we can write:

k = 1 / p

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., G = k(a/l) = k x 1 = k [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.

Λ_{m} = k(A/l)

Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).

A_{m} = kv

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 A_{m} with root c for strong and weak electrolytes is shown in the following plot:

**Q8) The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 Scm**^{−1}. Calculate its molar conductivity.

^{−1}. Calculate its molar conductivity.

**Answer)** Given,

κ = 0.0248 S cm^{−1}

c = 0.20 M

⇒ Molar conductivity, A_{m} = (k × 1000) / c

= 0.0248 × 1000 /0.20

= 124 Scm^{2}mol^{−1}

**Q9) 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 solution at 298 K is 0.146 × 10**^{−3 }S cm^{−1.}

^{−3 }S cm

^{−1.}

**Answer)** Given,

Conductivity, κ = 0.146 × 10^{−3 }S cm^{−1}

Resistance, R = 1500 Ω

⇒ Cell constant = κ × R

= 0.146 × 10^{−3 × 1500}

= 0.219 cm^{−1}

**Q10) 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 }× κ/S m^{−1} | 1.237 | 11.85 | 23.15 | 55.53 | 106.74 |

**Calculate λ _{m} for all concentrations and draw a plot between λ_{m }and c½. Find the value of λº_{m}.**

**Answer)**

Now, we have the following data :

Since the line interrupts λm at 124.0 S cm^{2} mol^{−1, }λº_{m}= 124.0 S cm^{2} mol^{−1}.

**Q11) Conductivity of 0.00241 M acetic acid is 7.896 × 10**^{−5 S} cm^{−1}. Calculate its molar conductivity and if A_{m}º for acetic acid is 390.5 S cm^{2} mol^{−1}, what is its dissociation constant?

^{−5 S}cm

^{−1}. Calculate its molar conductivity and if A

_{m}º for acetic acid is 390.5 S cm

^{2}mol

^{−1}, what is its dissociation constant?

**Answer)**

**Q12) 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+}**

**Answer)**

i) Al^{3+} + 3e^{–} → Al

⇒ Required charge = 3 F

= 3 × 96487 C

= 289461 C

(ii) Cu^{2+} + 2e^{–} → Cu

⇒ Required charge = 2 F

= 2 × 96487 C

= 192974 C

iii) MnO_{4}^{– }→ Mn^{2+}

i.e., Mn^{7+} + 5e^{–} → Mn^{2+}

⇒ Required charge = 5 F

= 5 × 96487 C

= 482435 C

**Q13) 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}.**

**Answer)**

**Q14) 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}.

**Answer)**

i) According to the question,

H_{2}O → H_{2 }+ ½O_{2}.

Now, we can write:

O^{2- }→ ½O_{2 }+ 2e^{–}

Electricity required for the oxidation of 1 mol of H_{2}O to O_{2} = 2 F

= 2 × 96487 C

= 192974 C

ii) According to the question,

Fe^{2+} **→ **Fe^{3+ }+ e^{-1}

Electricity required for the oxidation of 1 mol of FeO to Fe_{2}O_{3} = 1 F

= 96487 C

**Q15) 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?

_{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?

**Answer)** Given,

Current = 5A

Time = 20 × 60 = 1200 s

∵ Charge = current × time

= 5 × 1200

= 6000 C

According to the reaction,

Ni^{2+} +2e^{−} → Ni_{(s)} + *e*^{−}

Nickel deposited by 2 × 96487 C = 58.71 g

Therefore, nickel deposited by 6000 C = (58.71 X 6000) / (2 X 96487) g

= 1.825 g

Hence, 1.825 g of nickel will be deposited at the cathode.

**Q16) Three electrolytic cells A,B,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 deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

_{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 deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were deposited?

**Answer)**

**Q17) 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).**Answer)**

i)

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

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.

**Q18) 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.**Answer)**

i)

ii)

iii)

iv)

**That’s it. These were the solutions of NCERT Class 12 Chemistry Chapter 3 – Electrochemistry. Our team hopes that you have found these solutions helpful for you. If you have any doubt related to this chapter then feel free to comment your doubts below. Our team will try their best to help you with your doubts.**