Class 9 Chemistry Chapter 2 Electrochemistry Important Questions with Answers

Electrochemistry (Chapter 2) is a core unit in the Class 9 CBSE Chemistry syllabus, consistently featured in board examinations with 8–12 marks allocated across different question formats. The chapter bridges two critical competencies: (1) conceptual understanding of electron transfer at electrodes and (2) quantitative reasoning using equations like the Nernst equation and conductance formulas. Board examiners typically ask MCQs on cell notation and half-reactions, short-answers on cell potentials and efficiency, and long-answers requiring derivations or real-world battery analysis. The 2026–27 pattern emphasizes HOTS (Higher Order Thinking Skills)—scenario-based questions on fuel cells, corrosion prevention, and electroplating. Practising these 18 questions ensures you recognize common question archetypes, master the vocabulary (anode, cathode, EMF, conductivity), and build confidence in numerical problem-solving. Targeted practice also reduces exam anxiety by exposing you to mark-wise difficulty gradation.

**Q1.** In an electrochemical cell, the electrode at which oxidation occurs is called: (a) Cathode (b) Anode (c) Electrolyte (d) Salt bridge **Answer:** (b) Anode **Explanation:** Oxidation (loss of electrons) happens at the anode. In a galvanic cell, the anode is negative; in an electrolytic cell, it is positive. --- **Q2.** The Nernst equation relates cell potential (E) to: (a) Temperature and pressure only (b) Concentration and temperature (c) Mass and volume (d) Density and molarity **Answer:** (b) Concentration and temperature **Explanation:** The Nernst equation is E = E° − (0.0592/n) log Q at 25°C, where Q depends on ion concentrations, and temperature affects the coefficient. --- **Q3.** Which of the following is NOT a primary cell? (a) Dry cell (b) Daniel cell (c) Lead-acid battery (d) Leclanchè cell **Answer:** (c) Lead-acid battery **Explanation:** Lead-acid batteries are secondary (rechargeable) cells. Dry cells, Daniel cells, and Leclanchè cells are primary (non-rechargeable). --- **Q4.** The SI unit of electrical conductance is: (a) Ohm (Ω) (b) Siemens (S) (c) Farad (F) (d) Henry (H) **Answer:** (b) Siemens (S) **Explanation:** Conductance (G) = 1/Resistance; G is measured in Siemens. One Siemens = 1/Ohm. --- **Q5.** A fuel cell generates electricity through: (a) Thermal combustion (b) Electrochemical oxidation-reduction (c) Photosynthesis (d) Hydrolysis **Answer:** (b) Electrochemical oxidation-reduction **Explanation:** Fuel cells (e.g., hydrogen fuel cell) produce electrical energy directly from controlled redox reactions between fuel and oxidant at electrodes, without burning.

**Q1.** Write the half-reactions for a Daniell cell. What is the overall cell reaction? **Answer:** Anode (oxidation): Zn(s) → Zn²⁺(aq) + 2e⁻ Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s) Overall: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) The standard EMF of this cell is approximately 1.1 V at 25°C. --- **Q2.** Define molar conductivity (λₘ) and state its SI unit. **Answer:** Molar conductivity is the conducting power of one mole of an electrolyte in solution. It is defined as: λₘ = κ / c where κ is conductivity (S·cm⁻¹) and c is molar concentration (mol·L⁻¹). **SI unit:** S·m²·mol⁻¹ (or S·cm²·mol⁻¹ in CGS units commonly used in Indian textbooks). --- **Q3.** Why is a salt bridge used in an electrochemical cell? **Answer:** A salt bridge: 1. Completes the internal circuit by allowing anions to migrate to the anode compartment and cations to the cathode compartment. 2. Maintains electrical neutrality in both half-cells. 3. Prevents concentration polarization and maintains steady EMF. 4. Contains inert electrolyte (e.g., KNO₃, KCl) that does not react with the cell components. --- **Q4.** What is the difference between primary and secondary cells? Give one example of each. **Answer:** **Primary Cell:** Non-rechargeable; chemical reaction is irreversible. Once depleted, it cannot be restored. Example: Dry cell (Leclanchè cell). **Secondary Cell:** Rechargeable; chemical reaction is reversible. Can be recharged by passing an external electric current. Example: Lead-acid battery (in vehicles). --- **Q5.** A hydrogen fuel cell operates at 25°C. Write the electrode reactions and the overall cell reaction. **Answer:** Anode (oxidation): H₂(g) − 2e⁻ → 2H⁺(aq) Cathode (reduction): O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l) Overall: 2H₂(g) + O₂(g) → 2H₂O(l) This reaction is spontaneous and generates electrical energy with high efficiency (~60%), making fuel cells ideal for clean power applications.

**Q1.** Using the Nernst equation, calculate the cell potential of a Daniell cell when [Zn²⁺] = 0.01 M and [Cu²⁺] = 1.0 M at 25°C. (Given: E°cell = 1.10 V) **Answer:** The Nernst equation at 25°C is: E = E° − (0.0592/n) log Q For the Daniell cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu n = 2 electrons Q = [Zn²⁺] / [Cu²⁺] = 0.01 / 1.0 = 0.01 E = 1.10 − (0.0592/2) log(0.01) E = 1.10 − 0.0296 × (−2) E = 1.10 + 0.0592 E ≈ 1.16 V The cell potential increases when [Zn²⁺] is lower and [Cu²⁺] is higher, making the reaction more favourable. --- **Q2.** Explain why a lead-acid battery is preferred in automobiles over primary cells. **Answer:** 1. **Rechargeability:** Lead-acid batteries are secondary cells; they can be recharged thousands of times by reversing the internal chemical reaction via an external EMF source. 2. **High Current Capacity:** They deliver large currents (100+ A) required to start engines, whereas primary cells deliver low currents. 3. **Cost Efficiency:** Although initial cost is higher, the cost per use is lower due to reusability; primary cells must be replaced after one use. 4. **Reliability:** Lead-acid batteries have stable voltage (~12 V per cell) and predictable discharge characteristics, crucial for vehicle electrical systems. 5. **Thermal Stability:** They function reliably across temperature ranges typical in vehicles (−10°C to 50°C). --- **Q3.** Define electrical conductivity (κ) and derive the relationship between conductivity, resistance, and cell dimensions. **Answer:** **Electrical Conductivity (κ):** It is the reciprocal of resistivity and measures how easily a substance allows electrons or ions to flow. SI unit: S·m⁻¹ (Siemens per meter). **Derivation:** Resistance R = ρL/A, where ρ is resistivity, L is length, A is cross-section. Since κ = 1/ρ, R = L / (κA) Rearranging: κ = L / (RA) or κ = L × G / A where G is conductance (S). For a conductivity cell, the ratio L/A is called the cell constant (K_cell). Thus: κ = K_cell × G --- **Q4.** A nickel-cadmium battery has the following half-reactions: Anode: Cd(s) + 2OH⁻(aq) → Cd(OH)₂(s) + 2e⁻ Cathode: NiO₂(s) + 2H₂O(l) + 2e⁻ → Ni(OH)₂(s) + 2OH⁻(aq) Write the overall cell reaction and explain why this battery is preferred in portable devices. **Answer:** **Overall Cell Reaction:** Cd(s) + NiO₂(s) + 2H₂O(l) → Cd(OH)₂(s) + Ni(OH)₂(s) or simplified: Cd + NiO₂ + 2H₂O → Cd(OH)₂ + Ni(OH)₂ **Why preferred in portable devices:** 1. **Rechargeable:** Can be cycled 1000+ times, reducing replacement frequency. 2. **High Energy Density:** Delivers consistent power output relative to its mass and volume. 3. **Temperature Tolerance:** Functions well in cold environments (down to −20°C) unlike alkaline batteries. 4. **Robustness:** Resistant to mechanical shock and vibration, ideal for handheld electronics. 5. **Long Shelf Life:** Retains charge for extended periods compared to some primary cells.

**Q1.** Derive the Nernst equation and explain its significance in electrochemistry. **Complete Solution:** **Derivation:** At any moment during a cell reaction, the spontaneity is determined by ΔG = ΔG°, where ΔG relates to cell potential: ΔG = −nFE (where n = moles of electrons, F = Faraday constant ≈ 96500 C/mol) For a reaction at standard conditions: ΔG° = −nFE° At non-standard conditions, the relationship between ΔG and Q (reaction quotient) is: ΔG = ΔG° + RT ln Q Substituting ΔG = −nFE and ΔG° = −nFE°: −nFE = −nFE° + RT ln Q Dividing by −nF: E = E° − (RT / nF) ln Q At 25°C (298 K), converting to log₁₀ (multiply ln by 2.303): **E = E° − (0.0592 / n) log Q** This is the **Nernst Equation**. **Significance:** 1. **Predicts Cell Behavior:** Determines whether a cell will function (E > 0) at any given ion concentration. 2. **Explains Concentration Effects:** Shows how changes in reactant/product concentrations affect EMF. 3. **Equilibrium Condition:** At equilibrium, Q = K and E = 0, revealing the spontaneity limit. 4. **Electroanalysis:** Used in pH meters, ion-selective electrodes, and potentiometric titrations. 5. **Battery Performance:** Explains why battery voltage drops as the cell discharges (Q increases, E decreases). **Example Calculation:** For Zn | Zn²⁺ (0.1 M) || Cu²⁺ (1 M) | Cu, E° = 1.10 V, n = 2: Q = [Zn²⁺] / [Cu²⁺] = 0.1 / 1 = 0.1 E = 1.10 − (0.0592/2) log(0.1) = 1.10 − 0.0296 × (−1) = 1.13 V --- **Q2.** Compare and contrast galvanic (voltaic) cells and electrolytic cells. Provide labelled diagrams of each (describe in text), electrode designation, and one practical example of each. **Complete Solution:** | **Property** | **Galvanic Cell** | **Electrolytic Cell** | |---|---|---| | **EMF** | Produces electrical energy (E > 0) | Requires external EMF input | | **Spontaneity** | Spontaneous redox reaction (ΔG < 0) | Non-spontaneous; driven by external power | | **Anode** | Negative terminal (oxidation occurs) | Positive terminal (oxidation occurs) | | **Cathode** | Positive terminal (reduction occurs) | Negative terminal (reduction occurs) | | **Electron Flow** | From anode → cathode in external circuit | From power supply cathode → cell cathode | | **Ion Flow** | Anions to anode, cations to cathode (via salt bridge) | Anions to anode, cations to cathode | | **Cell Reaction** | Spontaneous; ΔG° = −nFE° | Non-spontaneous; driven by external energy | **Galvanic Cell Example: Daniell Cell** Zn(s) [anode, negative] | ZnSO₄(aq) || CuSO₄(aq) | Cu(s) [cathode, positive] Anode half-reaction: Zn → Zn²⁺ + 2e⁻ Cathode half-reaction: Cu²⁺ + 2e⁻ → Cu E°cell ≈ 1.10 V; used to power small devices. **Electrolytic Cell Example: Electroplating Copper** Power supply [−] connected to steel object (cathode); [+] connected to copper plate (anode); electrolyte: CuSO₄ solution. Anode (oxidation): Cu(s) → Cu²⁺(aq) + 2e⁻ Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s) Result: Copper deposits on steel, creating a protective coat. Requires external EMF. **Practical Significance:** Galvanic cells power portable devices; electrolytic cells are used in metal refining, water treatment, and electroplating—industries worth billions annually. --- **Q3.** A student measures the conductivity of a 0.01 M NaCl solution using a conductivity cell with a cell constant (K_cell) of 1.0 cm⁻¹. The measured resistance is 1000 Ω. Calculate: (a) Conductivity (κ) of the solution. (b) Molar conductivity (λₘ) of NaCl. (c) Explain why molar conductivity decreases as concentration increases for strong electrolytes. **Complete Solution:** **(a) Conductivity (κ):** Conductance G = 1/R = 1/1000 = 0.001 S = 1 × 10⁻³ S κ = K_cell × G = 1.0 cm⁻¹ × 1 × 10⁻³ S **κ = 1 × 10⁻³ S·cm⁻¹** (or 0.001 S·cm⁻¹) Converting to SI: κ = 0.001 × 100 = 0.1 S·m⁻¹ **(b) Molar Conductivity (λₘ):** λₘ = κ / c where c = 0.01 M = 0.01 mol/L λₘ = (1 × 10⁻³ S·cm⁻¹) / (0.01 mol/L) λₘ = (1 × 10⁻³) / (0.01) S·cm²·mol⁻¹ **λₘ = 0.1 S·cm²·mol⁻¹** (or 100 S·cm²·mol⁻¹ if using reciprocal ohm; check textbook convention) For reference, λₘ of 0.01 M NaCl at 25°C ≈ 118 S·cm²·mol⁻¹. **(c) Why Molar Conductivity Decreases with Concentration:** 1. **Interionic Attractions:** At higher concentrations, ions are closer together and electrostatic attractions between oppositely charged ions increase, slowing ion migration. 2. **Increased Viscosity:** Denser solution increases frictional resistance, reducing ion mobility. 3. **Atmosphere Formation:** Cations are surrounded by anion "atmospheres" and vice versa (ionic atmosphere), impeding movement. 4. **Reduced Mean Free Path:** Ions collide more frequently, losing directional momentum. 5. **Mathematical Expression:** For strong electrolytes, Kohlrausch's law states: λₘ = λ°ₘ − A√c where λ°ₘ is molar conductivity at infinite dilution (no ion-ion interactions), A is a constant, and c is concentration. As c ↑, the −A√c term becomes larger (more negative), so λₘ decreases. At infinite dilution (c → 0), λₘ → λ°ₘ (maximum value). --- **Bonus Context for Board Success:** These calculations appear frequently in board exams. Ensure you can: - Convert between μS·cm⁻¹, S·cm⁻¹, and S·m⁻¹. - Use K_cell values (often given in problems). - Apply Kohlrausch's law when dilute and concentrated solutions are compared. Start a 3-day free trial at cbsetutor.ai to practice conductivity problems with instant feedback and step-by-step explanations tailored to your learning pace.

**Case Study: Fuel Cell vs. Traditional Battery—A Sustainability Challenge** **Scenario:** A renewable energy startup is evaluating power solutions for a backup electricity system in a remote hospital. Two options are proposed: **Option A:** Hydrogen fuel cell (H₂ + ½O₂ → H₂O) - Efficiency: ~60% - Byproduct: Water only - Refueling time: 5 minutes - Lifespan: 5000+ operating hours - Cost: ₹ 3 lakh initially, low maintenance **Option B:** Lead-acid battery bank (12 V cells in series) - Efficiency: ~80% round-trip - Byproduct: Lead oxide, sulfuric acid (hazardous) - Recharge time: 8–10 hours via solar panel - Lifespan: 500–800 charge cycles (~3 years) - Cost: ₹ 1.5 lakh initially, high replacement cost **Questions:** 1. **Electrochemistry Analysis (3 marks):** Write the overall cell reaction for the hydrogen fuel cell. Calculate the standard EMF if E°cathode = +0.401 V (O₂ reduction) and E°anode = 0.00 V (H₂ oxidation). Is the reaction spontaneous? Justify. 2. **Nernst Application (2 marks):** If the fuel cell operates at [H⁺] = 0.1 M and P(O₂) = 1 atm at 25°C, use the Nernst equation to determine if the cell potential differs significantly from E°. 3. **Efficiency & Conductance Logic (2 marks):** Why does the hydrogen fuel cell maintain ~60% electrical efficiency while the lead-acid battery achieves only ~80% round-trip efficiency? Consider internal resistance and ion conductivity in your answer. 4. **Real-World Decision (3 marks):** Despite lower initial cost, why might the hospital ultimately choose the fuel cell? Discuss in terms of sustainability, byproducts, long-term electrochemistry principles, and operational reliability. --- **Full Solution:** **Q1: Electrochemistry Analysis** Overall reaction: 2H₂(g) + O₂(g) → 2H₂O(l) Anode (oxidation): H₂ − 2e⁻ → 2H⁺ [E° = 0.00 V] Cathode (reduction): O₂ + 4H⁺ + 4e⁻ → 2H₂O [E° = +0.401 V] Balancing electrons: multiply anode by 2: 2H₂ − 4e⁻ → 4H⁺ E°cell = E°cathode − E°anode = 0.401 − 0.00 = **0.401 V** (per half-reaction pair; commonly reported as ~1.23 V for the full reaction in literature, depending on half-cell definitions). **Spontaneity Check:** E° > 0 ✓ → ΔG° < 0 ✓ → Reaction is **spontaneous** under standard conditions. The positive cell potential drives the irreversible combustion of hydrogen to form water, releasing significant energy (~286 kJ/mol). --- **Q2: Nernst Equation Application** At 25°C, the Nernst equation for the fuel cell: E = E° − (0.0592/n) log Q For 2H₂ + O₂ → 2H₂O, n = 4 electrons transferred. Reaction quotient: Q = 1 / ([H⁺]⁴ × P(O₂)) (Note: gases and pure liquids have activity ≈ 1; focusing on H⁺ concentration.) Q = 1 / (0.1⁴ × 1) = 1 / 0.0001 = 10,000 E = 0.401 − (0.0592/4) log(10,000) E = 0.401 − 0.0148 × 4 E = 0.401 − 0.0592 E ≈ **0.342 V** Comparison: E drops by ~0.06 V (15%) from standard conditions due to high [H⁺]. This is **significant** because lower potential = lower available electrical energy. In practical cells, pH is controlled to optimize output. --- **Q3: Efficiency & Conductance Comparison** **Hydrogen Fuel Cell (~60% efficiency):** - Theoretical max: ~80–85% (from thermodynamic Gibbs free energy vs. enthalpy). - Actual ~60% due to: - Overpotential at electrodes (kinetic barriers to electron transfer) - Internal resistance from ion transport in the electrolyte membrane (conductivity κ must be very high to minimize losses) - Parasitic reactions and gas crossover - **Key:** Efficiency is limited by how fast H⁺ ions can conduct through the proton exchange membrane (PEM). **Lead-Acid Battery (~80% round-trip efficiency):** - Appears higher because "round-trip" includes charging efficiency (~95%) and discharge (~95%), giving 0.95 × 0.95 ≈ 90%. - Actual discharge efficiency is ~60–70% due to: - Charge acceptance inefficiency during recharge (some energy → heat) - Internal resistance of lead plates and sulfuric acid electrolyte - The sulfation process progressively reduces conductivity with cycling. - **Paradox resolved:** Lead-acid reports round-trip; fuel cells report continuous discharge. On a per-cycle basis, fuel cells are more efficient for continuous power. --- **Q4: Real-World Decision for the Hospital** **Why choose the hydrogen fuel cell despite higher upfront cost:** 1. **Environmental & Regulatory Compliance (Sustainability):** - Zero emissions of CO₂, NOx, or particulates → meets international green hospital standards. - No hazardous waste disposal (lead-acid requires recycling, a cost and environmental burden). - Byproduct is pure water—can be used for hospital needs (humidification, cooling). 2. **Long-Term Cost of Ownership:** - Lead-acid bank requires replacement every 3 years = ₹1.5 lakh × 3 cycles over 9 years = ₹4.5 lakh. - Fuel cell: ₹3 lakh upfront + minimal maintenance = ₹3.5 lakh over 9 years. - **Breakeven at ~4.5 years; fuel cell becomes cheaper in year 5.** 3. **Operational Reliability:** - Fuel cell: 5-minute refueling → critical for hospital; batteries need 8–10 hours, risking outage. - Medical backup must be rapid; lead-acid cannot meet emergency restart requirements in remote locations. 4. **Electrochemistry Advantage:** - Fuel cell conductivity doesn't degrade with use (unlike sulfuric acid in lead-acid cells, which loses conductivity as sulfation progresses). - Nernst equation shows fuel cell potential remains stable across a wider operating range. - No "memory effect" or cycle-life limitation. 5. **Future-Proofing:** - Green hydrogen production is scaling globally; fuel cell infrastructure will improve. - Lead-acid technology is stagnant; no performance gains expected. - Hospital alignment with climate commitments (SDG 7—Affordable and Clean Energy). **Conclusion:** Despite lead-acid's lower initial cost, the hydrogen fuel cell is the **optimal choice** for a remote hospital because it prioritizes reliability (critical for human life), environmental stewardship, and long-term economic sustainability—all aligned with CBSE's emphasis on real-world problem-solving using electrochemistry principles.

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