NCERT Solutions for Class 12 Chemistry Chapter 2 – Electrochemistry - MyNoteswala

NCERT Solutions for Class 12 Chemistry Chapter 2 – Electrochemistry (English Medium)

⚡ Get NCERT Solutions for Class 12 Chemistry Chapter 2 – Electrochemistry with accurate answers and step-by-step explanations in an exam-friendly format. This chapter covers redox reactions, electrode potential, electrochemical cells (galvanic/voltaic), EMF and cell notation, Nernst equation, conductance (κ, Λ_m), electrolysis, and batteries & corrosion. Perfect for CBSE boards and NEET/JEE practice. ✅

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NCERT Solutions for Class 12 Chemistry Chapter 2 – Electrochemistry

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Q.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

✅ 👉 Displacement (reactivity) order:
📌 Mg > Al > Zn > Fe > Cu

Explanation (Step by Step)

🤔 Rule: A more reactive metal displaces a less reactive metal from the solution of its salt.

📌 So we simply arrange them according to the reactivity series :

Mg is more reactive than Al, Zn, Fe, Cu → displaces all of them from their salt solutions.

Al can displace Zn, Fe, Cu (but not Mg ).

Zn can displace Fe, Cu (but not Mg or Al ).

Fe can displace Cu only.

Cu is least reactive here → cannot displace any of the others .

👉 That’s why the correct order is:
Mg > Al > Zn > Fe > Cu

Did You Know?

✅ Even though Al is very reactive , it often appears “less reactive” in daily life because it forms a thin protective layer of Al₂O₃ on its surface, which prevents further reaction (passivation) .

Q.2: Given the standard electrode potentials:
K⁺/K = −2.93 V, Ag⁺/Ag = +0.80 V, Hg²⁺/Hg = +0.79 V, Mg²⁺/Mg = −2.37 V, Cr³⁺/Cr = −0.74 V.
Arrange these metals in their increasing order of reducing power.

Answer

✅ 👉 Increasing order of reducing power:
📌 Ag < Hg < Cr < Mg < K

Explanation (Step by Step)

🤔 Idea: The metal with more negative standard reduction potential (E°) gets oxidised more easily, so it is a stronger reducing agent .

👉 So, as E° becomes more negative , reducing power increases .

Given E° values (in V)

• Ag⁺/Ag = +0.80 V

• Hg²⁺/Hg = +0.79 V

• Cr³⁺/Cr = −0.74 V

• Mg²⁺/Mg = −2.37 V

• K⁺/K = −2.93 V

📌 Arrange from highest E° (weak reducer) to most negative E° (strongest reducer) :

✅ Ag (+0.80) < Hg (+0.79) < Cr (−0.74) < Mg (−2.37) < K (−2.93)

Did You Know?

✅ Potassium (K) is such a strong reducing agent that it reacts violently with water to form KOH and H₂ gas .

👉 That’s why it is stored under kerosene/oil .

Q.3: Depict the galvanic cell in which the reaction
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
takes place. Further show:
(i) Which electrode is negatively charged?
(ii) The carriers of current in the cell
(iii) Individual reaction at each electrode

Answer

📌 Cell representation (cell notation):
👉 Zn(s) | Zn²⁺(aq) || Ag⁺(aq) | Ag(s)

(i) Negatively charged electrode

👉 Zinc electrode (Zn) = Anode (−)

(ii) Carriers of current in the cell

📌 Electrons (e⁻) in the external wire

📌 Ions in the solutions/salt bridge (cations and anions)

(iii) Half-cell reactions

👉 At anode (oxidation):
Zn(s) → Zn²⁺(aq) + 2e⁻

👉 At cathode (reduction):
2Ag⁺(aq) + 2e⁻ → 2Ag(s)

Explanation (Step by Step)

🤔 In a galvanic cell, the metal that gets oxidised becomes the anode and releases electrons, so it becomes negative .

📌 Here, Zn changes to Zn²⁺ , so Zn is oxidised:
👉 Zn(s) → Zn²⁺(aq) + 2e⁻ ✅ (Anode, −)

📌 Ag⁺ gains electrons and deposits as Ag(s), so Ag is reduced:
👉 Ag⁺(aq) + e⁻ → Ag(s) ✅ (Cathode, +)

📌 Electron flow direction:
👉 Zn (anode) → Ag (cathode) through the external circuit.

📌 Ion flow (inside cell):
👉 In salt bridge, anions move towards anode (to balance Zn²⁺) and cations move towards cathode (to balance loss of Ag⁺).

Did You Know?

✅ The double line “||” in cell notation shows a salt bridge , which is like a “charge balancer” .

👉 Without a salt bridge, charge builds up quickly, and the cell stops working even though reactants are still present.

Q.4: Calculate the standard cell potential (E°_cell) of the galvanic cell for the reactions:
(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)
(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)
Also calculate Δ_rG° and equilibrium constant (K_c).

Answer

(i) 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s)

📌 Given standard reduction potentials:
• E°(Cr³⁺/Cr) = −0.74 V

• E°(Cd²⁺/Cd) = −0.40 V

✅ E°_cell = +0.34 V

✅ Δ_rG° = −196.86 kJ mol⁻¹

✅ K_c ≈ 3.2 × 10³⁴

(ii) Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)

📌 Given standard reduction potentials:
• E°(Ag⁺/Ag) = +0.80 V

• E°(Fe³⁺/Fe²⁺) = +0.77 V

✅ E°_cell = +0.03 V

✅ Δ_rG° = −2.895 kJ mol⁻¹

✅ K_c ≈ 3.22

Explanation (Step by Step)

Common formulas (use in both) 📌

👉 E°_cell = E°_cathode − E°_anode (in V)

👉 Δ_rG° = −n F E°_cell (in J mol⁻¹)

👉 Δ_rG° = −RT ln K_c ⇒ ln K_c = −Δ_rG° / (RT)

Constants:
• F = 96500 C mol⁻¹

• R = 8.314 J mol⁻¹ K⁻¹

• T = 298 K

(i) Calculation 🤔

Step 1: Identify cathode and anode
• Cd²⁺ → Cd happens at cathode (reduction), E° = −0.40 V

• Cr → Cr³⁺ happens at anode (oxidation), E°(Cr³⁺/Cr) = −0.74 V

Step 2: E°_cell
👉 E°_cell = (−0.40) − (−0.74) = +0.34 V

Step 3: n (electrons transferred)
Cr(s) → Cr³⁺ + 3e⁻
For 2 Cr: electrons = 2 × 3 = 6, so n = 6

Step 4: Δ_rG°
👉 Δ_rG° = −nFE°_cell
= −(6 mol) × (96500 C mol⁻¹) × (0.34 V)
= −196860 J mol⁻¹ = −196.86 kJ mol⁻¹

Step 5: K_c
👉 ln K_c = −Δ_rG°/(RT)
= 196860 J mol⁻¹ / (8.314 × 298)
≈ 196860 / 2477.6 ≈ 79.45

So,
👉 K_c = e^79.45 ≈ 3.2 × 10³⁴ (dimensionless)

(ii) Calculation 🤔

Step 1: Cathode and anode
• Ag⁺ + e⁻ → Ag(s) is cathode, E° = +0.80 V

• Fe²⁺ → Fe³⁺ + e⁻ is anode (reverse of Fe³⁺/Fe²⁺), E°(Fe³⁺/Fe²⁺) = +0.77 V

Step 2: E°_cell
👉 E°_cell = 0.80 − 0.77 = 0.03 V

Step 3: n
Only 1 electron is transferred, so n = 1

Step 4: Δ_rG°
👉 Δ_rG° = −(1) × 96500 × 0.03
= −2895 J mol⁻¹ = −2.895 kJ mol⁻¹

Step 5: K_c
👉 ln K_c = 2895 / (8.314 × 298)
= 2895 / 2477.6 ≈ 1.168

So,
👉 K_c = e^1.168 ≈ 3.22

Did You Know?

✅ If E°_cell is positive , then Δ_rG° is negative , and the reaction is spontaneous under standard conditions.

✅ A very large K_c (like 10³⁴ ) means the reaction goes almost completely to products, so it’s practically “one-way” in the lab.

Q.5: Write the Nernst equation and calculate emf (E_cell) of the following cells at 298 K:
(i) Mg(s) | Mg²⁺(0.001 M) || Cu²⁺(0.0001 M) | Cu(s)
(ii) Fe(s) | Fe²⁺(0.001 M) || H⁺(1 M) | H₂(g, 1 bar) | 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)

Given (E° values):
E°(Mg²⁺/Mg) = −2.37 V, E°(Cu²⁺/Cu) = +0.34 V, E°(Fe²⁺/Fe) = −0.44 V,
E°(Sn²⁺/Sn) = −0.14 V, E°(Br₂/Br⁻) = +1.08 V

Answer

Nernst equation at 298 K (general) 📌

👉 E_cell = E°_cell − (0.0591/n) log₁₀ Q

where n = electrons transferred , and Q = reaction quotient .

✅ Final emf values:

(i) 2.68 V

(ii) 0.529 V (≈ 0.53 V)

(iii) 0.078 V (≈ 0.08 V)

(iv) −1.288 V (negative)

Explanation

(i) Mg(s)|Mg²⁺(10⁻³ M) || Cu²⁺(10⁻⁴ M)|Cu(s)

Cell reaction: Mg(s) + Cu²⁺(aq) → Mg²⁺(aq) + Cu(s) , n = 2

• E°_cell = E°_cathode − E°_anode
= (+0.34) − (−2.37) = 2.71 V

• Q = [Mg²⁺]/[Cu²⁺] = 10⁻³/10⁻⁴ = 10

👉 E_cell = 2.71 − (0.0591/2) log(10)
= 2.71 − 0.02955 × 1
= 2.68 V

✅ emf = 2.68 V

(ii) Fe(s)|Fe²⁺(10⁻³ M) || H⁺(1 M)|H₂(1 bar)|Pt(s)

Cell reaction: Fe(s) + 2H⁺(aq) → Fe²⁺(aq) + H₂(g) , n = 2

(Here, a(H₂) = 1 because pressure = 1 bar)

• E°_cell = E°(H⁺/H₂) − E°(Fe²⁺/Fe)
= 0.00 − (−0.44) = 0.44 V

• Q = [Fe²⁺] / [H⁺]² = 10⁻³ / (1)² = 10⁻³

👉 E_cell = 0.44 − (0.0591/2) log(10⁻³)
= 0.44 − 0.02955 × (−3)
= 0.529 V

✅ emf = 0.529 V (≈ 0.53 V)

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

Cell reaction: Sn(s) + 2H⁺(aq) → Sn²⁺(aq) + H₂(g) , n = 2

• E°_cell = 0.00 − (−0.14) = 0.14 V

• Q = [Sn²⁺] / [H⁺]² = 0.050 / (0.020)²
= 0.050 / 0.0004 = 125

👉 E_cell = 0.14 − (0.0591/2) log(125)
= 0.14 − 0.02955 × 2.0969
= 0.0785 V

✅ emf = 0.078 V (≈ 0.08 V)

(iv) Pt|Br⁻(0.010 M)|Br₂(l) || H⁺(0.030 M)|H₂(1 bar)|Pt

As written cell reaction: 2Br⁻(aq) + 2H⁺(aq) → Br₂(l) + H₂(g) , n = 2

• E°_cell = E°(H⁺/H₂) − E°(Br₂/Br⁻)
= 0.00 − 1.08 = −1.08 V

• Q = 1 / ( [Br⁻]² [H⁺]² )
= 1 / (0.010² × 0.030²)
= 1 / (10⁻⁴ × 9×10⁻⁴) = 1 / (9×10⁻⁸)
= 1.11 × 10⁷

👉 E_cell = −1.08 − (0.0591/2) log(1.11×10⁷)
= −1.08 − 0.02955 × 7.0457
= −1.288 V

✅ emf = −1.288 V (negative)

Did You Know?

✅ If E_cell comes out negative , it means the cell reaction as written is not spontaneous under those conditions.

👉 In such a case, simply reverse the cell notation . The reaction will proceed in the reverse direction, and the emf becomes positive.

📌 For part (iv), to get a positive emf, write the cell as:
👉 Pt | H₂(1 bar) | H⁺(0.030 M) || Br⁻(0.010 M) | Br₂(l) | Pt

Q.6: In button cells (watches etc.) the reaction is:
Zn(s) + Ag₂O(s) + H₂O(l) → Zn²⁺(aq) + 2Ag(s) + 2OH⁻(aq)

Determine E° and Δ_rG° for the reaction.
[Given, E°(Zn/Zn²⁺) = 0.76 V and E°(Ag⁺/Ag) = +0.34 V]

Answer

✅ 👉 Standard cell potential (E°_cell) = 1.10 V

✅ 👉 Standard Gibbs free energy change (Δ_rG°) = −2.12 × 10⁵ J mol⁻¹ = −212.3 kJ mol⁻¹

Explanation (Step by Step)

Half reactions (standard conditions) 📌

Anode (oxidation)

Zn(s) → Zn²⁺(aq) + 2e⁻

E°(Zn²⁺/Zn) = −0.76 V

Cathode (reduction)

Ag₂O(s) + H₂O(l) + 2e⁻ → 2Ag(s) + 2OH⁻(aq)

E°(Ag₂O/Ag) = +0.34 V

✅ Step 1: Calculate E°_cell

👉 E°_cell = E°_cathode − E°_anode
= (+0.34) − (−0.76)
= +1.10 V

✅ Step 2: Calculate Δ_rG°

🤔 Electrons transferred n = 2

👉 Δ_rG° = −n F E°_cell
= −(2 mol) × (96500 C mol⁻¹) × (1.10 V)
= −212300 J mol⁻¹

= −212.3 kJ mol⁻¹

Did You Know?

✅ Button cells using Ag₂O–Zn are popular because they give a nearly constant voltage during most of their life.

👉 That stable output is why they’re perfect for watches, calculators, hearing aids, and small medical devices.

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

Answer

(i) Conductivity (κ)

📌 Conductivity (κ) is the ability of an electrolyte solution to conduct electricity.
It is the reciprocal of resistivity (ρ).

👉 κ = 1/ρ

Also,
👉 κ = (1/R) × (l/A) = G × (l/A)

Where:
• R = resistance (unit: Ω )

• G = 1/R = conductance (unit: S or Ω⁻¹ )

• l = distance between electrodes (unit: cm )

• A = area of electrodes (unit: cm² )

📌 If l = 1 cm and A = 1 cm² , then κ = G .

✅ Unit of κ: S cm⁻¹ (or Ω⁻¹ cm⁻¹ )

(ii) Molar conductivity (Λ_m)

📌 Molar conductivity (Λ_m) is the conductance of the solution containing 1 mole of electrolyte .

👉 Λ_m = (κ × 1000) / C

Where:
• κ = conductivity (S cm⁻¹)

• C = concentration in mol L⁻¹

• 1000 converts L to cm³ (since 1 L = 1000 cm³ )

✅ Unit of Λ_m: S cm² mol⁻¹ (or Ω⁻¹ cm² mol⁻¹ )

Explanation (Variation with Concentration)

1) Variation of Conductivity (κ) with concentration 🤔

📌 When a solution is diluted, the number of ions per unit volume decreases.
👉 So, conductivity generally decreases on dilution (for both strong and weak electrolytes).

✅ On dilution (concentration ↓): κ ↓

2) Variation of Molar Conductivity (Λ_m) with concentration 📌

📌 On dilution, the volume of solution containing 1 mole of electrolyte becomes larger.
👉 Also, ions get more space to move, and in weak electrolytes, ionisation increases.

✅ On dilution (concentration ↓): Λ_m ↑

Reason (simple):

• κ goes down because fewer ions exist per cm³

• Λ_m goes up because it counts conduction for the whole solution volume that contains 1 mole .

Did You Know?

✅ For strong electrolytes (like NaCl, HCl), Λ_m increases slowly on dilution because they are already almost fully ionised.

✅ For weak electrolytes (like CH₃COOH), Λ_m increases sharply on dilution because ionisation increases a lot as concentration decreases.