Electrochemistry — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Electrochemistry" — 8 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Electrochemistry" — 8 important questions with detailed answers for CBSE board exam…
Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated
NCERT Solutions for Class 12 Chemistry — Electrochemistry. Step-by-step answers to all 8 textbook questions from this chapter, written for CBSE board preparation and free to use.
Electrochemistry — All 8 Questions Solved
Q1. Define electrochemistry. What are electrodes and electrolytes?
Electrochemistry is the study of chemical reactions that produce electrical current or the use of electrical current to drive non-spontaneous chemical reactions.
Electrode:
A conductor (usually metal or graphite) through which current enters or leaves an electrolytic cell or galvanic cell.
Types:
1. Anode: Electrode where oxidation occurs; positive in galvanic cell, negative in electrolytic cell
2. Cathode: Electrode where reduction occurs; negative in galvanic cell, positive in electrolytic cell
Electrolyte:
A substance that conducts electricity when dissolved in water or in molten state due to presence of ions.
Types:
1. Strong electrolytes: Completely ionize (e.g., NaCl, HCl, H₂SO₄)
2. Weak electrolytes: Partially ionize (e.g., acetic acid, NH₃·H₂O)
3. Non-electrolytes: Do not ionize (e.g., glucose, urea)
Example: In a galvanic cell, zinc electrode acts as anode (negative terminal) and copper electrode acts as cathode (positive terminal); electrolyte is the solution containing ions.
Q2. Write the Nernst equation and explain its significance.
The Nernst equation relates cell potential to concentration of reactants and products at non-standard conditions.
Nernst Equation:
E_cell = E°_cell - (RT/nF) × ln(Q)
Alternative form (at 25°C):
E_cell = E°_cell - (0.0592/n) × log(Q)
where:
E_cell = cell potential at non-standard conditions
E°_cell = standard cell potential
R = universal gas constant (8.314 J/mol·K)
T = absolute temperature (K)
n = number of electrons transferred
F = Faraday constant (96485 C/mol)
Q = reaction quotient
Significance:
1. Predicts spontaneity of reaction at any concentration: If E_cell > 0, reaction is spontaneous
2. Determines direction of reaction: Q and E_cell relationship shows which direction reaction proceeds
3. Calculates cell potential at non-standard conditions
4. Shows effect of concentration on cell potential
5. Used in pH measurement: pH = -log[H⁺], related to electrode potential
6. Industrial applications: Electroplating, batteries with varying concentrations
Q3. Calculate the standard cell potential for the reaction: Cu²⁺ + Zn → Cu + Zn²⁺. (Given: E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = -0.76 V)
Given:
Cu²⁺ + Zn → Cu + Zn²⁺
E°(Cu²⁺/Cu) = +0.34 V
E°(Zn²⁺/Zn) = -0.76 V
Step 1: Identify oxidation and reduction
Zn → Zn²⁺ + 2e⁻ (oxidation, occurs at anode)
Cu²⁺ + 2e⁻ → Cu (reduction, occurs at cathode)
Step 2: Calculate standard cell potential
E°_cell = E°(cathode) - E°(anode)
E°_cell = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn)
E°_cell = (+0.34) - (-0.76)
E°_cell = 0.34 + 0.76
E°_cell = +1.10 V
Interpretation:
E°_cell = +1.10 V (positive value)
The reaction is spontaneous under standard conditions.
This is a galvanic cell with Zn as anode and Cu as cathode.
Q4. Explain the working of a galvanic cell. What happens at anode and cathode?
A galvanic cell (voltaic cell) spontaneously converts chemical energy into electrical energy through redox reactions.
Working principle:
Redox reaction occurs spontaneously in two separated compartments (half-cells), driving electron flow through external circuit.
At Anode (negative terminal in galvanic cell):
1. Oxidation occurs: Loss of electrons
2. Anode material is oxidized
3. Electrons are released here and flow to cathode through external circuit
4. Anions from salt bridge move toward anode to maintain charge neutrality
5. Example: Zn → Zn²⁺ + 2e⁻
At Cathode (positive terminal in galvanic cell):
1. Reduction occurs: Gain of electrons
2. Cations from electrolyte are reduced
3. Electrons from anode arrive here through external circuit
4. Cations from salt bridge move toward cathode
5. Example: Cu²⁺ + 2e⁻ → Cu
Electrolyte and Salt Bridge:
1. Salt bridge contains inert salt (e.g., KNO₃, KCl)
2. Prevents accumulation of charge in compartments
3. Allows ion flow without mixing solutions
4. Maintains electrical neutrality
Overall:
Electrons flow from anode to cathode through external circuit (opposite to conventional current direction).
Cell potential (E°_cell) = E°_cathode - E°_anode
Q5. Calculate moles of Cu deposited when 2 A of current is passed for 1.5 hours through CuSO₄ solution.
Given:
Current (I) = 2 A
Time (t) = 1.5 hours = 1.5 × 3600 = 5400 s
Reaction: Cu²⁺ + 2e⁻ → Cu
Step 1: Calculate total charge passed
Charge (Q) = I × t = 2 × 5400 = 10,800 C
Step 2: Determine electrons required per Cu atom
Cu²⁺ + 2e⁻ → Cu
For 1 Cu atom, 2 electrons are required
Step 3: Calculate moles of electrons
Moles of electrons = Q / F
where F = Faraday constant = 96,500 C/mol
Moles of electrons = 10,800 / 96,500 = 0.1119 mol
Step 4: Calculate moles of Cu deposited
From equation: 2 mol e⁻ produces 1 mol Cu
Moles of Cu = 0.1119 / 2 = 0.05595 mol ≈ 0.056 mol
Alternatively:
Using Faraday's law: W = (M × I × t) / (n × F)
Moles = (I × t) / (n × F) = (2 × 5400) / (2 × 96,500) = 0.056 mol
Answer: Moles of Cu deposited = 0.056 mol
Q6. What is electrolysis? Explain the electrolysis of molten NaCl.
Electrolysis is the process of using electrical current to drive non-spontaneous redox reactions in an electrolytic cell.
Electrolysis of molten NaCl:
Setup:
1. Molten NaCl contains Na⁺ and Cl⁻ ions
2. Inert electrodes (typically graphite or platinum) used
3. External source provides electric current
At Cathode (negative electrode):
Reduction occurs: Na⁺ + e⁻ → Na
Sodium metal deposits on cathode.
At Anode (positive electrode):
Oxidation occurs: 2Cl⁻ → Cl₂ + 2e⁻
Chlorine gas is produced at anode.
Overall reaction:
2NaCl(molten) → 2Na + Cl₂
Products:
1. Sodium metal (Na): Produced at cathode, highly reactive metal used in manufacturing
2. Chlorine gas (Cl₂): Produced at anode, used in disinfection and chemical synthesis
This is the chlor-alkali process used industrially.
Products are valuable: Na for various chemical processes, Cl₂ for bleach and water disinfection.
Key point: Molten NaCl is required because solid NaCl doesn't conduct electricity (ions are fixed in lattice).
Q7. State and explain Faraday's Law of Electrolysis.
Faraday's Laws of Electrolysis describe the relationship between the amount of substance produced and the electrical charge passed.
Faraday's First Law:
The amount of chemical change produced by electrolysis is proportional to the quantity of electricity passed through the electrolyte.
Mathematical form:
Mass of substance = (Equivalent weight × Charge) / Faraday constant
m = (E × Q) / F or m = (M × I × t) / (n × F)
where:
m = mass of substance deposited/dissolved
E = equivalent weight
Q = total charge (coulombs)
F = Faraday constant = 96,500 C/mol
M = molar mass
I = current (A)
t = time (s)
n = number of electrons transferred
Faraday's Second Law:
When the same quantity of electricity is passed through different electrolytes, the amounts of different substances deposited are proportional to their equivalent weights.
Mathematical form:
m₁/m₂ = E₁/E₂
where m₁, m₂ are masses and E₁, E₂ are equivalent weights.
Applications:
1. Electroplating: Controlling metal deposition thickness
2. Electro-refining: Purifying metals
3. Calculations in electrochemistry problems
4. Industrial electrochemical processes
Example: If 0.5 A current is passed for 1 hour, same charge through AgNO₃ deposits more Ag than through CuSO₄ (Ag equivalent weight is higher)
Q8. Define conductance and conductivity. How do they differ?
Conductance (G):
The ability of a substance to conduct electricity.
Formula: G = 1/R = κA/l
where κ = conductivity, A = cross-sectional area, l = length of conductor
Unit: Siemens (S) or ohm⁻¹ (Ω⁻¹)
Also expressed as: Mho (℧) = ohm⁻¹
Conductivity (κ or σ):
The ability of 1 cm length and 1 cm² cross-section of material to conduct electricity.
Formula: κ = G × l / A
Unit: S·cm⁻¹ or ohm⁻¹·cm⁻¹
Molar Conductance (Λ_m):
Conductance of solution containing 1 mole of solute.
Formula: Λ_m = κ × V
where V = molar volume in cm³/mol
Alternatively: Λ_m = (κ × 1000) / C
where C = concentration in mol/L
Unit: S·cm²·mol⁻¹
Key Differences:
1. Conductance: Property of conductor; depends on geometry (length, area)
2. Conductivity: Intrinsic property; independent of geometry
3. Conductance: Specific to sample, changes with dimensions
4. Conductivity: Material property, remains constant for same material
5. Conductance: Used for specific cells or wires
6. Conductivity: Used to compare materials
Relationship: Conductivity is conductance normalized for geometry
Electrochemistry — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. Which of the following statements is true for a galvanic cell?
- A. It converts electrical energy into chemical energy.
- B. It converts chemical energy into electrical energy.
- C. It requires an external power source to function.
- D. Its overall cell reaction is non-spontaneous.
Answer: B. It converts chemical energy into electrical energy. — Galvanic cells are electrochemical cells that convert the chemical energy of a spontaneous redox reaction into electrical energy.
MCQ 2. For the reaction 2Cr(s) + 3Cd²⁺(aq) → 2Cr³⁺(aq) + 3Cd(s), if E°cell = +0.34 V, what will be the cell potential (Ecell) when [Cd²⁺] = 0.01 M and [Cr³⁺] = 0.1 M at 298 K? (Use 0.059 V for 2.303RT/F)
- A. 0.34 V
- B. 0.30 V
- C. 0.38 V
- D. 0.26 V
Answer: B. 0.30 V — Using the Nernst equation, Ecell = E°cell - (0.059/n) log([Cr³⁺]² / [Cd²⁺]³). Here n=6 and Q = (0.1)² / (0.01)³ = 0.01 / 0.000001 = 10⁴. Ecell = 0.34 - (0.059/6) * log(10⁴) = 0.34 - (0.059/6) * 4 = 0.34 - 0.0393 ≈ 0.30 V.
MCQ 3. What is the SI unit of specific conductivity (κ)?
- A. Ohm cm⁻¹
- B. S cm⁻¹
- C. S m⁻¹
- D. Ohm⁻¹ m⁻²
Answer: C. S m⁻¹ — Specific conductivity, also known as conductivity, is the reciprocal of resistivity. Its SI unit is Siemens per meter (S m⁻¹), though S cm⁻¹ is also commonly used.
MCQ 4. How much charge in Coulombs is required for the reduction of 1 mole of Al³⁺ to Al?
- A. 96487 C
- B. 2 * 96487 C
- C. 3 * 96487 C
- D. 6 * 96487 C
Answer: C. 3 * 96487 C — The reduction of 1 mole of Al³⁺ to Al (Al³⁺ + 3e⁻ → Al) requires 3 moles of electrons. Since 1 mole of electrons carries 1 Faraday (F) or 96487 Coulombs of charge, 3 moles of electrons will carry 3 * 96487 C.
MCQ 5. Given the limiting molar conductivities (Λ°m) at 298 K: Λ°m(HCl) = 425 S cm² mol⁻¹, Λ°m(NaCl) = 126 S cm² mol⁻¹, Λ°m(CH₃COONa) = 91 S cm² mol⁻¹. Calculate Λ°m for CH₃COOH.
- A. 642 S cm² mol⁻¹
- B. 390 S cm² mol⁻¹
- C. 208 S cm² mol⁻¹
- D. 540 S cm² mol⁻¹
Answer: B. 390 S cm² mol⁻¹ — According to Kohlrausch's law, Λ°m(CH₃COOH) = Λ°m(CH₃COONa) + Λ°m(HCl) - Λ°m(NaCl). Substituting the given values, Λ°m(CH₃COOH) = 91 + 425 - 126 = 390 S cm² mol⁻¹.
MCQ 6. In an electrolytic cell, which process occurs at the anode?
- A. Reduction
- B. Oxidation
- C. Neutralization
- D. Precipitation
Answer: B. Oxidation — In any electrochemical cell (galvanic or electrolytic), oxidation always occurs at the anode, and reduction always occurs at the cathode.
MCQ 7. For a spontaneous reaction in an electrochemical cell, which of the following conditions is true?
- A. ΔG > 0 and Ecell > 0
- B. ΔG < 0 and Ecell > 0
- C. ΔG > 0 and Ecell < 0
- D. ΔG < 0 and Ecell < 0
Answer: B. ΔG < 0 and Ecell > 0 — For a spontaneous reaction, the Gibbs free energy change (ΔG) must be negative. Since ΔG = -nFEcell, a negative ΔG implies a positive cell potential (Ecell).
MCQ 8. A conductivity cell with 0.01 M KCl solution has a resistance of 100 Ω. If the conductivity of 0.01 M KCl is 0.141 S m⁻¹, what is the cell constant?
- A. 1.41 m⁻¹
- B. 0.0141 m⁻¹
- C. 14.1 m⁻¹
- D. 141 m⁻¹
Answer: C. 14.1 m⁻¹ — The cell constant (G*) is given by G* = conductivity (κ) × resistance (R). Substituting the given values, G* = 0.141 S m⁻¹ × 100 Ω = 14.1 m⁻¹.
MCQ 9. What is the standard electrode potential of the Standard Hydrogen Electrode (SHE)?
- A. +1.00 V
- B. -1.00 V
- C. 0.00 V
- D. +0.76 V
Answer: C. 0.00 V — The standard electrode potential of the Standard Hydrogen Electrode (SHE) is conventionally taken as exactly 0.00 V at all temperatures.
MCQ 10. Which of the following is a primary battery?
- A. Lead-acid battery
- B. Nickel-cadmium battery
- C. Dry cell (Leclanché cell)
- D. Lithium-ion battery
Answer: C. Dry cell (Leclanché cell) — A primary battery is one that cannot be recharged and reused, meaning the reactions are irreversible. A dry cell (Leclanché cell) is a common example of a primary battery.
MCQ 11. If the standard Gibbs free energy change (ΔG°) for a reaction is -287 kJ/mol and the standard cell potential (E°cell) is +1.48 V, what is the number of electrons (n) involved in the reaction? (Given: F = 96487 C/mol)
- A. 1
- B. 2
- C. 3
- D. 4
Answer: B. 2 — Using the relationship ΔG° = -nFE°cell, where ΔG° = -287000 J/mol, F = 96487 C/mol, and E°cell = +1.48 V. We calculate n = -ΔG° / (F × E°cell) = -(-287000) / (96487 × 1.48) ≈ 2.
MCQ 12. Which of the following methods is used for cathodic protection against corrosion?
- A. Painting the surface
- B. Greasing
- C. Sacrificial protection
- D. Alloying with noble metals
Answer: C. Sacrificial protection — Sacrificial protection is a method of cathodic protection where a more reactive metal (sacrificial anode) is connected to the metal to be protected, corroding preferentially.
MCQ 13. Molar conductivity (Λm) is defined as the conductivity of the solution containing:
- A. One gram of electrolyte.
- B. One gram equivalent of electrolyte.
- C. One mole of electrolyte.
- D. A fixed volume of electrolyte.
Answer: C. One mole of electrolyte. — Molar conductivity (Λm) is the conductivity of a solution containing one mole of electrolyte, placed between two electrodes separated by unit distance and having a large enough area to contain the entire volume.
MCQ 14. When aqueous NaCl solution is electrolysed using inert electrodes, what are the products formed at the anode and cathode, respectively?
- A. Anode: O₂, Cathode: H₂
- B. Anode: Cl₂, Cathode: Na
- C. Anode: Cl₂, Cathode: H₂
- D. Anode: O₂, Cathode: Na
Answer: C. Anode: Cl₂, Cathode: H₂ — At the anode, Cl⁻ ions are preferentially oxidized over water to produce Cl₂ gas, especially at high concentrations, due to overpotential for O₂. At the cathode, water is reduced to H₂ gas and OH⁻ ions, as Na⁺ is difficult to reduce in aqueous solution.
MCQ 15. How does molar conductivity (Λm) vary with dilution for a strong electrolyte?
- A. Decreases slightly
- B. Increases slightly
- C. Remains constant
- D. First increases then decreases
Answer: B. Increases slightly — For a strong electrolyte, molar conductivity increases slightly with dilution because interionic attractions decrease, allowing ions to move more freely.
MCQ 16. Which of the following is an advantage of a fuel cell over a traditional battery?
- A. It requires frequent recharging.
- B. It produces harmful byproducts.
- C. It operates continuously as long as reactants are supplied.
- D. Its efficiency is very low.
Answer: C. It operates continuously as long as reactants are supplied. — Fuel cells continuously convert the chemical energy of a fuel (e.g., H₂) and an oxidant (e.g., O₂) into electrical energy, as long as the reactants are supplied, unlike batteries which have a finite charge.
MCQ 17. A current of 0.5 A is passed through a solution of CuSO₄ for 30 minutes. Calculate the mass of copper deposited at the cathode. (Atomic mass of Cu = 63.5 g/mol, F = 96487 C/mol)
- A. 0.148 g
- B. 0.296 g
- C. 0.444 g
- D. 0.592 g
Answer: B. 0.296 g — First, calculate the total charge Q = I × t = 0.5 A × (30 × 60 s) = 900 C. For Cu²⁺ + 2e⁻ → Cu, 2 Faradays are needed per mole of Cu. Moles of Cu = (Q / 2F) = 900 / (2 × 96487) = 0.00466 mol. Mass of Cu = moles × molar mass = 0.00466 × 63.5 = 0.296 g.
MCQ 18. What is the primary function of a salt bridge in a galvanic cell?
- A. To complete the external circuit.
- B. To prevent the mixing of electrolytes.
- C. To maintain electrical neutrality in the half-cells.
- D. To supply electrons for the redox reaction.
Answer: C. To maintain electrical neutrality in the half-cells. — The salt bridge maintains electrical neutrality in the half-cell solutions by allowing the migration of ions, preventing charge build-up that would stop the cell reaction.
MCQ 19. Which of the following equations correctly relates molar conductivity (Λm) with specific conductivity (κ)? (Assume concentration C is in mol L⁻¹ and κ in S cm⁻¹)
- A. Λm = κ * C
- B. Λm = κ / C
- C. Λm = (κ * 1000) / C
- D. Λm = C / κ
Answer: C. Λm = (κ * 1000) / C — Molar conductivity is related to specific conductivity by the formula Λm = (κ × 1000) / C, where κ is in S cm⁻¹, C is in mol L⁻¹, and Λm is in S cm² mol⁻¹. The factor of 1000 converts L to cm³.
MCQ 20. Given the standard electrode potentials: Zn²⁺/Zn = -0.76 V, Cu²⁺/Cu = +0.34 V. Which of the following reactions is spontaneous under standard conditions?
- A. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s)
- B. Cu(s) + Zn²⁺(aq) → Cu²⁺(aq) + Zn(s)
- C. Zn²⁺(aq) + Cu²⁺(aq) → Zn(s) + Cu(s)
- D. Zn(s) + Cu(s) → Zn²⁺(aq) + Cu²⁺(aq)
Answer: A. Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) — A reaction is spontaneous if its standard cell potential (E°cell) is positive. For Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), E°cell = E°(Cu²⁺/Cu) - E°(Zn²⁺/Zn) = 0.34 V - (-0.76 V) = +1.10 V, which is positive.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.