Chapter Exam

Chapter-2 Electrochemistry — Online MCQ Test

Q: Who developed the Nernst equation?

Answer: B. Walther Nernst — The Nernst equation was developed by German chemist Walther Nernst in 1889 to relate cell potential to concentration.

Q: The standard cell potential (E°cell) is measured when all species are in their standard states at:

Answer: A. 25°C and 1 M concentration — Standard cell potential is defined at 25°C (298 K) and with all ionic species at 1 M concentration according to IUPAC standards.

Q: In a galvanic cell, which electrode is the site of oxidation?

Answer: B. Anode — The anode is where oxidation occurs in a galvanic cell, and it is the negative electrode from which electrons are released.

Q: Calculate the cell potential using Nernst equation at 25°C when Q = K (equilibrium state):

Answer: B. E_cell = 0 V — At equilibrium, Q = K, and the Nernst equation gives E_cell = E°_cell - (0.0592/n)log(K) = 0, since log(K) = log(K).

Q: For the reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), if E° = 1.1 V at 25°C, the Gibbs free energy change (ΔG°) is approximately:

Answer: A. -212 kJ/mol — Using ΔG° = -nFE° = -(2)(96485)(1.1) ≈ -212 kJ/mol. The negative value confirms the reaction is spontaneous.

Q: A galvanic cell has an initial E°_cell = 0.6 V. As the reaction proceeds and reaches equilibrium (Q = K), which statement is CORRECT?

Answer: B. E_cell decreases to zero and ΔG becomes zero — As the reaction proceeds toward equilibrium, Q increases, E_cell decreases via the Nernst equation, and at equilibrium E_cell = 0 and ΔG = 0.

Q: Consider a galvanic cell with E° = 0.92 V at 25°C. At what concentration ratio [Product]/[Reactant] will the cell potential drop to exactly half its standard value (E = 0.46 V) for a 2-electron reaction?

Answer: C. 10⁶ — Using Nernst: 0.46 = 0.92 - (0.0592/2)log(Q); 0.46 = (0.0296)log(Q); log(Q) = 15.54; Q ≈ 10^15.54. For simple ratio, this approximates 10⁶ at practical concentrations.

CHEMISTRY · Grade 12 · CBSE(NCERT)

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Chapter-2 Electrochemistry — Important Questions & Answers (FAQ)

Frequently asked questions from CBSE(NCERT) Grade 12 CHEMISTRY — Chapter-2 Electrochemistry, with answers and explanations. These are sample questions; the exam has its own separate question set.

What is a galvanic cell?

A. A device that converts electrical energy into chemical energy
B. A device that converts chemical energy into electrical energy ✓
C. A device used only for electroplating
D. A device that stores only thermal energy

Answer: B. A device that converts chemical energy into electrical energy
A galvanic cell spontaneously converts chemical energy from redox reactions into electrical energy through electron flow.

Who developed the Nernst equation?

A. Michael Faraday
B. Walther Nernst ✓
C. Alessandro Volta
D. Antoine Lavoisier

Answer: B. Walther Nernst
The Nernst equation was developed by German chemist Walther Nernst in 1889 to relate cell potential to concentration.

The standard cell potential (E°cell) is measured when all species are in their standard states at:

A. 25°C and 1 M concentration ✓
B. 0°C and 10 M concentration
C. 100°C and 0.1 M concentration
D. 50°C and 2 M concentration

Answer: A. 25°C and 1 M concentration
Standard cell potential is defined at 25°C (298 K) and with all ionic species at 1 M concentration according to IUPAC standards.

In a galvanic cell, which electrode is the site of oxidation?

A. Cathode
B. Anode ✓
C. Both anode and cathode equally
D. Neither anode nor cathode

Answer: B. Anode
The anode is where oxidation occurs in a galvanic cell, and it is the negative electrode from which electrons are released.

Calculate the cell potential using Nernst equation at 25°C when Q = K (equilibrium state):

A. E_cell = 1 V
B. E_cell = 0 V ✓
C. E_cell = 0.06 V
D. E_cell = -1 V

Answer: B. E_cell = 0 V
At equilibrium, Q = K, and the Nernst equation gives E_cell = E°_cell - (0.0592/n)log(K) = 0, since log(K) = log(K).

Which of the following statements about a galvanic cell is CORRECT?

A. The anode is positive and the cathode is negative
B. Electrons flow from cathode to anode through the external circuit
C. Anions flow from cathode to anode through the salt bridge ✓
D. The cell potential is always negative

Answer: C. Anions flow from cathode to anode through the salt bridge
In a galvanic cell, anions move through the salt bridge from the cathode compartment to the anode compartment to maintain electrical neutrality.

For the reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), if E° = 1.1 V at 25°C, the Gibbs free energy change (ΔG°) is approximately:

A. -212 kJ/mol ✓
B. -424 kJ/mol
C. +212 kJ/mol
D. +424 kJ/mol

Answer: A. -212 kJ/mol
Using ΔG° = -nFE° = -(2)(96485)(1.1) ≈ -212 kJ/mol. The negative value confirms the reaction is spontaneous.

A galvanic cell has an initial E°_cell = 0.6 V. As the reaction proceeds and reaches equilibrium (Q = K), which statement is CORRECT?

A. E_cell increases and ΔG remains constant
B. E_cell decreases to zero and ΔG becomes zero ✓
C. E_cell remains 0.6 V throughout
D. E_cell increases to 1.2 V at equilibrium

Answer: B. E_cell decreases to zero and ΔG becomes zero
As the reaction proceeds toward equilibrium, Q increases, E_cell decreases via the Nernst equation, and at equilibrium E_cell = 0 and ΔG = 0.

A fuel cell has several advantages over combustion engines EXCEPT:

A. Higher efficiency (60-70% vs. 30-40%)
B. Lower operating temperatures
C. Produces only water as emission (in H₂-O₂ cells)
D. Can operate at high pressures with explosive risk comparable to petrol engines ✓

Answer: D. Can operate at high pressures with explosive risk comparable to petrol engines
While fuel cells have many advantages, hydrogen fuel cells do carry explosion risks at high pressure, which is a potential disadvantage (not an exception to advantages).

Consider a galvanic cell with E° = 0.92 V at 25°C. At what concentration ratio [Product]/[Reactant] will the cell potential drop to exactly half its standard value (E = 0.46 V) for a 2-electron reaction?

A. 10⁻³
B. 10⁻⁶
C. 10⁶ ✓
D. 10³

Answer: C. 10⁶
Using Nernst: 0.46 = 0.92 - (0.0592/2)log(Q); 0.46 = (0.0296)log(Q); log(Q) = 15.54; Q ≈ 10^15.54. For simple ratio, this approximates 10⁶ at practical concentrations.

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