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CBSE Class 12 Chapter Test: ELECTROCHEMISTRY
General Instructions:
- This question paper consists of 15 questions divided into 5 sections: A, B, C, D, and E.
- Section A comprises 5 MCQs and 2 Assertion-Reasoning questions of 1 mark each.
- Section B comprises 2 Very Short Answer questions of 2 marks each.
- Section C comprises 3 Short Answer questions of 3 marks each.
- Section D comprises 1 Case-Based question of 4 marks.
- Section E comprises 2 Long Answer questions of 5 marks each.
- Use of log tables is permitted. Faraday's constant (F) = 96500 C mol-1.
Directions for Q6 & Q7: These questions consist of two statements, each printed as Assertion (A) and Reason (R). Select the correct answer from the codes (a), (b), (c), and (d) as given below:
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.
Reason (R): The standard Gibbs free energy change (ΔG°) is related to standard cell potential as ΔG° = -nFE°cell.
Reason (R): Zinc is more reactive than copper and has a lower standard reduction potential than copper.
(Molar mass of Cu = 63.5 g mol-1, 1 F = 96500 C)
Mg(s) | Mg2+ (0.1 M) || Cu2+ (0.01 M) | Cu(s)
[Given: E°Mg2+/Mg = -2.36 V, E°Cu2+/Cu = +0.34 V]
(i) An aqueous solution of AgNO3 with silver electrodes.
(ii) An aqueous solution of NaCl with platinum electrodes.
(iii) A dilute aqueous solution of H2SO4 with platinum electrodes.
Galvanic cells that are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy are called fuel cells. One of the most successful fuel cells uses the reaction of hydrogen with oxygen to form water. The cell was used for providing electrical power in the Apollo space programme. The water vapors produced during the reaction were condensed and added to the drinking water supply for the astronauts. In the cell, hydrogen and oxygen are bubbled through porous carbon electrodes into concentrated aqueous sodium hydroxide solution.
(a) Write the half-cell reaction taking place at the anode of the H2-O2 fuel cell. [1 Mark]
(b) State two distinct advantages of fuel cells over ordinary thermal power plants. [1 Mark]
(c) Unlike primary batteries (like dry cells) which become dead over time, fuel cells do not become dead. Explain why. [2 Marks]
(b) Why does the specific conductivity of a solution decrease with dilution? [1 Mark]
(c) The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2 mol-1. Calculate its degree of dissociation (α) and dissociation constant (Ka).
[Given: λ°(H+) = 349.6 S cm2 mol-1 and λ°(HCOO-) = 54.6 S cm2 mol-1] [3 Marks]
(b) Calculate the standard Gibbs free energy change (ΔG°) and the equilibrium constant (Kc) for the following reaction at 298 K:
Zn(s) + Cu2+(aq) ⇔ Zn2+(aq) + Cu(s)
[Given: E°cell = 1.10 V, F = 96500 C mol-1, R = 8.314 J K-1 mol-1] [3 Marks]
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SOLUTIONS & MARKING SCHEME
1. Answer: (c) 3 F
Reasoning: In Al2O3, aluminium is in the +3 oxidation state (Al3+). The reduction reaction is Al3+ + 3e- → Al. Since 3 moles of electrons are required to reduce 1 mole of Al3+, the charge required is 3 Faradays (3 F).
2. Answer: (c) The pH of the electrolyte increases
Reasoning: During the discharging of a lead storage battery, sulphuric acid (H2SO4) is consumed to form PbSO4 and water. Because the concentration of H+ ions decreases and water is produced (diluting the acid), the pH of the electrolyte increases.
3. Answer: (c) Λ°m(NH4Cl) + Λ°m(NaOH) - Λ°m(NaCl)
Reasoning: According to Kohlrausch's law, Λ°m(NH4OH) = λ°(NH4+) + λ°(OH-). We can obtain this by adding the conductivities of NH4Cl and NaOH, and subtracting the conductivity of NaCl to cancel out the extra Na+ and Cl- ions.
4. Answer: (b) κ decreases but Λm increases
Reasoning: Specific conductivity (κ) is the conductance of 1 cm3 of the solution. On dilution, the number of current-carrying ions per unit volume decreases, so κ decreases. However, molar conductivity (Λm = κ × V) increases because the total volume (V) containing one mole of electrolyte increases much faster than the decrease in κ.
5. Answer: (b) Decrease
Reasoning: According to the Nernst equation: Ecell = E°cell - (0.059/n) log([Zn2+]/[Cu2+]). Increasing the concentration of the product ion (Zn2+) increases the log term, which is subtracted from E°cell. Hence, the overall EMF of the cell decreases.
6. Answer: (a) Both A and R are true and R is the correct explanation of A.
Reasoning: For a reaction to be spontaneous, ΔG° must be negative. Since ΔG° = -nFE°cell, the term E°cell must be positive so that the overall ΔG° value becomes negative.
7. Answer: (d) A is false but R is true.
Reasoning: Zinc is more reactive than copper (lower reduction potential). If aqueous copper sulphate is stored in a zinc vessel, the zinc will displace copper from the solution (Zn + CuSO4 → ZnSO4 + Cu), creating holes in the vessel. Thus, Assertion is false.
8. Solution:
Molar Conductivity (Λm): It is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in a given volume of solution.
Λm = κ × 1000 / M (Unit: S cm2 mol-1).
Graph description (1 Mark): A graph of Λm vs √c shows a straight line with a small negative slope for strong electrolytes (e.g., KCl), which slowly approaches a finite value at infinite dilution. For weak electrolytes (e.g., CH3COOH), the curve rises steeply at low concentrations and does not intersect the y-axis (infinite dilution).
9. Solution:
Reaction at cathode: Cu2+ + 2e- → Cu. Here, n = 2.
Mass deposited (m) = 1.27 g. Current (I) = 2.0 A.
Using Faraday's First Law: m = Z × I × t = (M / nF) × I × t
1.27 = (63.5 / (2 × 96500)) × 2 × t
1.27 = (63.5 / 96500) × t
t = (1.27 × 96500) / 63.5 = 1930 seconds (or 32.16 minutes).
10. Solution:
Step 1: Cell reaction and E°cell
Anode: Mg → Mg2+ + 2e-
Cathode: Cu2+ + 2e- → Cu
Overall: Mg + Cu2+ → Mg2+ + Cu (n = 2)
E°cell = E°Cathode - E°Anode = 0.34 V - (-2.36 V) = 2.70 V
Step 2: Nernst Equation
Ecell = E°cell - (0.0591 / n) log ([Mg2+] / [Cu2+])
Ecell = 2.70 - (0.0591 / 2) log (0.1 / 0.01)
Ecell = 2.70 - 0.02955 × log(10)
Ecell = 2.70 - 0.02955 = 2.67 V
11. Solution:
Given: M = 0.05 M, d = 1 cm ⇒ r = 0.5 cm, l = 50 cm, R = 5.55 × 103 Ω.
Area (A) = πr2 = 3.14 × (0.5)2 = 0.785 cm2.
(i) Resistivity (ρ):
R = ρ(l/A) ⇒ ρ = R × A / l = (5.55 × 103 × 0.785) / 50 = 87.135 Ω cm.
(ii) Conductivity (κ):
κ = 1 / ρ = 1 / 87.135 = 0.01148 S cm-1.
(iii) Molar Conductivity (Λm):
Λm = κ × 1000 / M = (0.01148 × 1000) / 0.05 = 229.6 S cm2 mol-1.
12. Solution:
(i) AgNO3 with Ag electrodes: Since the anode is reactive (Ag), it will undergo oxidation.
Anode: Ag(s) → Ag+(aq) + e-
Cathode: Ag+(aq) + e- → Ag(s).
(ii) NaCl(aq) with Pt electrodes: Cl- is oxidized at anode (due to overpotential of O2), and H2O is reduced at cathode (Na+ is highly stable).
Anode: 2Cl- → Cl2(g) + 2e-
Cathode: 2H2O + 2e- → H2(g) + 2OH-.
(iii) Dilute H2SO4 with Pt electrodes: Water is oxidized and reduced.
Anode: 2H2O → O2(g) + 4H+ + 4e-
Cathode: 2H+ + 2e- → H2(g).
13. Solution (Case-Based):
(a) At the anode, hydrogen gas is oxidized in the alkaline medium.
Anode Reaction: 2H2(g) + 4OH-(aq) → 4H2O(l) + 4e-
(b) Advantages of fuel cells:
1. Highly Efficient: They operate at an efficiency of around 70% compared to thermal plants (~40%).
2. Pollution Free: The only byproduct in the H2-O2 cell is water, producing no toxic emissions.
(c) Primary batteries become dead when the active chemicals (reactants) stored inside them are completely consumed. Fuel cells do not become dead because the reactants (fuel and oxidant) are continuously supplied from an external source. As long as the supply of H2 and O2 is maintained, the cell will keep producing electricity.
14. Solution:
(a) Kohlrausch's Law: It states that the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.
(b) Specific conductivity is the conductance of one unit volume (1 cm3) of the solution. Upon dilution, the concentration of ions decreases, meaning the number of current-carrying ions per unit volume decreases, resulting in a drop in specific conductivity.
(c) Numerical:
Given: C = 0.025 M, Λm = 46.1 S cm2 mol-1.
Step 1: Calculate Λ°m(HCOOH) using Kohlrausch's law.
Λ°m = λ°(H+) + λ°(HCOO-) = 349.6 + 54.6 = 404.2 S cm2 mol-1.
Step 2: Calculate Degree of dissociation (α).
α = Λm / Λ°m = 46.1 / 404.2 = 0.114.
Step 3: Calculate Dissociation constant (Ka).
Ka = Cα2 / (1 - α) = (0.025 × (0.114)2) / (1 - 0.114)
Ka = (0.025 × 0.012996) / 0.886 = 3.67 × 10-4.
15. Solution:
(a) Derivation:
At equilibrium, the cell potential becomes zero (Ecell = 0), and the reaction quotient (Q) becomes equal to the equilibrium constant (Kc).
Using Nernst Equation: Ecell = E°cell - (2.303RT / nF) log Q
Substituting Ecell = 0 and Q = Kc:
0 = E°cell - (2.303RT / nF) log Kc
E°cell = (2.303RT / nF) log Kc
At 298 K, E°cell = (0.0591 / n) log Kc.
(b) Numerical Calculation:
For the reaction: Zn + Cu2+ ⇔ Zn2+ + Cu, the number of electrons transferred (n) = 2.
Given E°cell = 1.10 V.
Step 1: Calculate ΔG°
ΔG° = -nFE°cell
ΔG° = -2 × 96500 C mol-1 × 1.10 V
ΔG° = -212300 J mol-1 = -212.3 kJ mol-1.
Step 2: Calculate Kc
Using E°cell = (0.0591 / n) log Kc
1.10 = (0.0591 / 2) log Kc
1.10 = 0.02955 × log Kc
log Kc = 1.10 / 0.02955 = 37.22
Kc = Antilog(37.22) = 1.66 × 1037.
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