NCERT Class 12 Chemistry Chapter 3 Electrochemistry

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NCERT Class 12 Chemistry Chapter 3 Electrochemistry

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Also, you can read the NCERT book online in these sections Solutions by Expert Teachers as per Central Board of Secondary Education (CBSE) Book guidelines. NCERT Class 12 Chemistry Chapter 3 Electrochemistry Solutions are part of All Subject Solutions. Here we have given NCERT Class 12 Chemistry Part: I, Part: II Notes. NCERT Class 12 Chemistry Chapter 3 Electrochemistry Notes, NCERT Class 12 Chemistry Textbook Solutions for All Chapters, You can practice these here.

Electrochemistry

Chapter: 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. 

Ans: A metal with stronger reducing power displaces another metal with weaker reducing power from its salt solution.

The correct order of displacement is: Mg > Al > Zn > Fe > Cu. 

This means  that every metal  on the side can displace at the next one (s) from its salt solution.

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: More negative the reduction potential, more easily it is oxidised and hence greater is the reduction power.

To arrange these metals in increasing order of reducing power (weakest to strongest reducing agent):

Ag < Hg < Cr < Mg < K.

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 electrodes is negatively charged? 

(ii) The carriers of the current in the cell. 

(iii) Individual reaction at each electrode. 

Ans: The galvanic cell for the reaction is represented as 

Zn(s) | Zn²⁺ (aq) || 2Ag⁺ (aq) | 2Ag(s)

(i) Zinc electrode (anode) is negatively charged.

(ii) The carriers of current are electrons (through the external circuit) and ions (Zn²⁺ and Ag⁺ in the electrolyte).

(iii) Anode (oxidation):

Zn(s) → Zn²⁺(aq) + 2e⁻

Cathode (reduction):

2Ag⁺(aq) + 2e⁻ → 2Ag(s)

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 DrG o and equilibrium constant of the reactions. 

Ans: (i) 2Cr(s) + 3Cd^{2 +} (aq) → 2Cr³⁺(aq) + 3Cd(s)

(anode is on left and cathode on right)

(Here  n = 6 (as 6e are involved in overall cell reaction i.,e 2Cr + 3Cd²⁺ → 3Cr³⁺ + 3Cd)

= ΔG∘ = − nFEcell∘​

ΔG^∘ = −(6 mol) × (96,500 C/mol) × (0.34 V)

= − (6 × 96,500 × 0.34)

= − (196,860 J/mol)  

E^∘ cell = 0.0591 log ke

Log ke = (0.34 × 6) / 0.0591 

= 34.5177

(ii) 

Here n = 1 as one is involved in overall reaction, i.e, Fe²⁺ + Ag + (Fe²⁺ + Ag )

where n = 1n = 1n = 1 (since 1 mole of electrons is transferred).

ΔG^∘ = − 1 × 96485 C/mol × 0.03 V

= − 2,894.55 J/mol

= − 2.89 kJ/mol

E^∘_cell = 0.0591 / n log kc

= log k_c = nE^∘ _cell / 0.0591 

= 1 × 0.03 / 0.0591 

= 0.5076 .

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) 

Ans: Cell Reaction: Mg(s) + Cu²⁺ (aq) → Mg²⁺ (aq) + Cu(s)

Nernst Equation Calculation: 

= 2.17 – 0.02955 

= 2.68V.

(ii) Fe(s)|Fe²⁺ (0.001M) || H⁺ (1M)|H² (g) (1 bar) | Pt(s) 

Ans: 

= 0.44+ 0.0887

 = 0.5287 V 

Cell Reaction: Fe²⁺(aq) + H² (g) → Fe(s) + 2H⁺ (aq)x

(iii) Sn(s)|Sn²⁺ (0.050 M) || H⁺ (0.020 M)|H₂ (g) (1 bar)|Pt(s) 

Ans: 

= 0.14 = 0.0620 

= 0.08 V.

(iv) Pt(s)|Br^– (0.010 M)|Br₂ (l )||H⁺ (0.030 M)| H₂ (g) (1 bar)|Pt(s). 

Ans: Cell Reaction: Br₂(l) + 2Br⁻(aq) → 2H⁺(aq) + H₂(g)

= 0.08 – 0.0591 / 2 ( 7.0457)

= -1.08 – 0.208 

= – 1.288 V.

Hence Oxidation will occur at hydrogen electron and reduction atBr2 electrode.

6. In the button cells widely used in watches and other devices the following reaction takes place: Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH^– (aq) Determine Δ_r G° and E° for the reaction. 

Ans: From given cell reaction, we find that zinc electrode act as anode as Zn(s) is oxidised to Zn²⁺(aq) and silver electrode act as cathode as Ag⁺ is reduced to Ag(s). 

E^∘ _cell​ = E^∘_cathode​ − E^∘_anode

​ E^∘_cell​ = E^∘_Ag​ − E^∘_Zn​ 

E^∘_cell = 0.344V − (−0.76V)

= 1.104V

Δr​G° = − nFE°cell​

= − 2 × 96500 × 1.104 

= − 213072

= – 2.13 × 10⁵ J.

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

Ans: Conductivity is the reciprocal of resistance of an electrolyte in aqueous solution is known as its conductivity. It is equal to 1/R.

Molar conductivity is conductivity per mole of electrolyte, reflecting ion mobility. As concentration increases, conductivity rises but molar conductivity decreases due to ion interactions and reduced mobility. At very high concentrations, molar conductivity approaches zero.

8. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm^–1. Calculate its molar conductivity. 

Ans: Molarity of solution, M = 0.20 conducti-vity, i.e., specific conductivity = k = 0.248 s cm-1¹ 

= 2.48 × 10-2 ohm-¹ cm^-1,

The formula for molar conductivity is:

λ_m = (1000 ​× κ)/C​ 

λ_m​ = (1000 × 0.0248S cm⁻¹​)/ 0.20

= 124.0s cm² mol⁻¹

∴ conductivity is 124 S cm² mol⁻¹

9. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 W. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10^–3 S cm^–1.

Ans: Given: 

Resistance (R) = 1500 Ω

Conductivity (κ) = 0.146 × 10−3 S cm⁻¹

A is cell constant

Specific conductivity,

K = A × R

The cell constant.

1/A = K.R

= 0.146 × 10⁻³ × 1500 cm⁻¹

= 0.219cm^-1

10. The conductivity of sodium chloride at 298 K has been determined at different concentrations and the results are given below:

Calculate Λ_m for all concentrations and draw a plot between Λ_m and c^½. Find the value of Λ°_m. 

Ans: 

11. Conductivity of 0.00241 M acetic acid is 7.896 × 10^–5 S cm^–1. Calculate its molar conductivity. If Λ°_m for acetic acid is 390.5 S cm² mol^–1, what is its dissociation constant? 

Ans: Calculation of molar conductivity(Λ°_m )

Here:

Λ_m​ is the molar conductivity (in S cm² mol⁻¹),

Conductivity κ = 7.896 × 10⁻⁵cm⁻¹,

Concentration C = 0.00241M.

= 0.00241 / 10³ mol cm^-3

= 241 × 10^-8 cm^-3

= 32.76Scm²mo

Calculation of Degree of Dissociation (α)

= 0.084 = 8.4 × 10⁻²

Calculation of Dissociation Constant (K_a)

= 0.000185 mol-¹

= 1.85 × 10^-5 mol L^-1.

Molar Conductivity (Λm​) = 32.76Scm²mo,

The degree of dissociation of acetic acid is 8.4 × 10⁻²

Dissociation Constant (Ka​) = 1.85 × 10^-5 mol L^-1.

12. How much charge is required for the following reductions: 

(i) 1 mol of Al³⁺ to Al?

Ans: 

The reduction will require 3 Faraday of electricity

3 × 96500 = 2.895 × 10⁵C

(ii) 1 mol of Cu²⁺ to Cu? 

Ans: 

The reduction will require 2 Faraday of electricity or 2 × 96500 = 1.93 × 105C

(iii) 1 mol of MnO^–₄ to Mn²⁺? 

Ans: 

The reduction will require 5 Faraday of electricity

5 × 96500 = 4.825 × 105C.

13. How much electricity in terms of Faraday is required to produce

(i) 20.0 g of Ca from molten CaCl₂? 

Ans: Ca²⁺ (aq) + 2e⁻ 2F​Ca (s)

Since the molar mass of calcium (Ca) is 40 g/mol, it takes 2 Faradays of electricity to produce 40 g of Ca.

Faraday required = (20.0 / 40.0) × 2

= 1 Faraday.

(ii) 40.0 g of Al from molten Al₂O₃? 

Ans: Molar mass of Al (Aluminum) = 26.98 g/mol Electrons involved (n) = 3 (since Al³⁺ needs 3 electrons to reduce to Al)

Faraday required = (40.0 / 26.98) × 3

= 4.448F

14. How much electricity is required in coulomb for the oxidation of: 

(i) 1 mol of H₂O to O₂? 

Ans: 

For 1 mole of H₂O, 2 moles of electrons (2e⁻) are needed. Therefore, the amount of electricity required is:

Q = 2 × 96500

= 193000C 

= 1.93 × 10⁵C.

(ii) 1 mol of FeO to Fe₂O₃? 

Ans:

For 1 mole of Fe₂+, 1 mole of electrons is required. Therefore, the electricity required is:

Q = 1 × 96500

= 96500C

= 9.65 × 10⁴C.

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: Quantity of electricity passed:

20 × 60 seconds = 1200s

Q = 5 A × 1200s = 6000 Coulombs

Where: 

I = Current = 5 A

t = Time = 20 minutes 

The molar mass of nickel (Ni) = 58.7 g/mol:

2 × 96500 colombs 58.7 gNi

Therefore, 6,000 coulombs produce 

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

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: Cathode reaction in Cell B (AgNO₃):

Current (I) = 1.5 A

Mass of silver deposited (m_Ag) = 1.45 g

Molar mass of silver (Ag) = 107.87 g/mol

1 mole of Ag (108 g) requires 96500 C.

On time for which current is passed 

= 1295.6​ / 1.5 

= 863.7s

= 14.40 min.

Cathode reaction in Cell B (ZnSO₄):

Molar mass of Zn = 65.38 g/mol

2 moles of electrons (193000 C) deposit 1 mole of Zn (65.38 g).

Charge used = 1295.6 C

Cathode reaction in Cell C (CuSO₄)

= 0.44g.  

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)

Ans: Reduction Potentials:

Fe³⁺ + 3e⁻ → Fe E° = + 0.77V

I₂​ + 2e⁻ → 2I⁻ E° = + 0.54 V

= 0.77V – 0.54v 

= + 0.23v

As the E°_cell is , the reaction between Fe³⁺ (aq) and I(- (aq)) occurs as indicated by possible reaction given above.

(ii) Ag⁺ (aq) and Cu(s) 

Ans: Reduction Potentials:

Ag⁺ + e⁻ → Ag E° = + 0.80 V

Cu²⁺ + 2e⁻ → Cu E° = + 0.34 

= 0.80 V – 0.34 

= 0.46V.

As the E°cell is positive, the reaction between Ag⁺ (aq) and Cu(s) occurs as indicated by possible reaction given above.

(iii) Fe³⁺ (aq) and Br^– (aq) 

Ans: Reduction Potentials: 

Fe³⁺ + 3e⁻ → Fe E° = + 0.77 V

Br₂​ + 2e⁻ → 2Br⁻ E° = + 1.09 V

= 0.77 V – 1.09 V

= -0.31V.

As E°_cell is negative, no reaction occurs between Fe³⁺ (aq) and Ag(s)

(iv) Ag(s) and Fe³⁺ (aq)

Ans: Reduction Potentials:

Ag⁺ + e⁻ → Ag E° = + 0.80 V

Fe³⁺ + 3e⁻ → Fe E° = + 0.77 V

= 0.80 V – 0.77 V

= -0.3V.

As E°_cell  is negative, no reaction will occur between Fe (aq) and Br(aq)

(v) Br₂ (aq) and Fe²⁺ (aq). 

Ans: Reduction Potentials:

Br₂​ + 2e⁻ → 2Br⁻ E° = + 1.09 V

Fe²⁺ + 2e⁻ → Fe E° = + 0.77 V

= 1.09 V- 0.77 V

= 0.31 V 

As the E°cell is positive, the reaction is feasible, i.e., reaction between Br₂(aq) and Fe²⁺ (aq) occurs as indicated by the possible reaction given above.

18. Predict the products of electrolysis in each of the following: 

(i) An aqueous solution of AgNO₃ with silver electrodes. 

Ans: In an aqueous solution of AgNO₃ with silver electrodes, silver ions are preferentially discharged at the cathode due to their lower discharge potential compared to hydrogen ions, resulting in silver deposition.

At anode: Silver anode will dissolve to form silver ions in the solution.

Ag → Ag⁺ + e⁻

(ii) An aqueous solution of AgNO₃ with platinum electrodes.

Ans: Silver ions (Ag⁺) are preferentially reduced to form silver metal, as they have a lower discharge potential compared to hydrogen ions.

Ag⁺ + e⁻ → Ag(s)

Hydroxide ions (OH⁻) are discharged over sulphate ions (SO₄²⁻​) due to their lower discharge potential. Hydroxide ions decompose to produce oxygen gas and water

4OH⁻ → 2H₂​O + O₂​(g) + 4e⁻

(iii) A dilute solution of H₂SO₄ with platinum electrodes.

Ans: A dilute solution of H₂SO₄ with platinum electrodes.At cathode:

2H⁺ + e⁻ → H₂(g)

4OH⁻(aq) → H₂O(l) + O₂(g) + 4e⁻

At the anode, hydroxide ions are preferentially discharged over sulphate ions due to their lower discharge potential. 

Hydroxide ions decompose to produce oxygen gas.

(iv) An aqueous solution of CuCl₂ with platinum electrodes

Ans: At the Anode: Copper ions (Cu²⁺) are reduced to form copper metal. This occurs because copper ions have a lower reduction potential compared to hydrogen ions in the solution.

Cu²⁺+ 2e− → Cu(s)

At the Anode: Chloride ions (Cl⁻) are preferentially oxidised to form chlorine gas. Although hydroxide ions (OH⁻) can also be oxidised, chloride ions are usually oxidised in preference to water in this scenario.

2Cl⁻ → Cl₂(g) + 2e^–