Nernst Equation - UNSOLVED PRACTICE SET
Chapter: Electrochemistry | Topic: Nernst Equation
NERNST EQUATION - UNSOLVED PRACTICE SET
Topic: Nernst Equation
Multiple Choice Questions
Q1. The Nernst equation for a half-cell reaction at 25°C is:
- E = E° + (RT/nF) ln Q
- E = E° - (0.0591/n) log Q
- E = E° + (0.0591/n) log [oxidised]/[reduced]
- Both (b) and (c) are correct forms
Q2. For the electrode reaction Mⁿ⁺ + ne⁻ → M, the Nernst equation is:
- E = E° + (0.0591/n) log [Mⁿ⁺]
- E = E° - (0.0591/n) log [Mⁿ⁺]
- E = E° + (0.0591/n) log 1/[Mⁿ⁺]
- E = E° - (0.0591/n) [Mⁿ⁺]
Q3. When the concentration of the metal ion in a half-cell is decreased to 1/10th, the electrode potential:
- Increases by 0.0591/n V
- Decreases by 0.0591/n V
- Remains unchanged
- Becomes zero
Q4. The Nernst equation is applicable to:
- Only galvanic cells
- Only electrolytic cells
- Both galvanic and electrolytic cells
- Neither galvanic nor electrolytic cells
Q5. For a cell reaction at equilibrium, the Nernst equation gives:
- E = E°
- E = 0
- E = ∞
- E = -E°
Q6. The term (2.303 RT/F) at 298 K is approximately:
- 0.0257 V
- 0.0591 V
- 0.118 V
- 1.0 V
Short Answer Questions
Q7. Write the Nernst equation for the half-cell reaction: Fe³⁺ + e⁻ → Fe²⁺ at 25°C.
Q8. Calculate the electrode potential of a zinc electrode dipped in 0.01 M ZnSO₄ solution at 25°C. [E°(Zn²⁺/Zn) = -0.76 V]
Q9. For the cell reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, write the Nernst equation and explain what happens to cell EMF as the cell discharges.
Q10. Why does the voltage of a battery decrease as it is used? Explain using the Nernst equation.
Q11. Your mobile phone battery shows 4.2 V when fully charged and 3.5 V when nearly discharged. Explain this voltage drop in terms of concentration changes using the Nernst equation concept.
Q12. At what concentration of Ag⁺ ions will the electrode potential of Ag⁺/Ag be equal to that of Cu²⁺/Cu (0.1 M)? [E°(Ag⁺/Ag) = +0.80 V, E°(Cu²⁺/Cu) = +0.34 V]
Long Answer Questions
Q13. Discuss the Nernst equation in detail:
(a) Derivation of the Nernst equation from thermodynamic principles
(b) Nernst equation for half-cells: E = E° - (RT/nF) ln Q
(c) Nernst equation at 25°C (298 K): E = E° - (0.0591/n) log Q
(d) Nernst equation for complete cells
(e) Relationship between Nernst equation and equilibrium constant
(f) Solved examples for metal ion electrodes, gas electrodes, and concentration cells
Q14. Explain the applications of the Nernst equation:
(a) Calculating electrode potential under non-standard conditions
(b) Calculating cell EMF under non-standard conditions
(c) Determining equilibrium constants from standard cell potentials
(d) Calculating pH using a hydrogen electrode or a quinhydrone electrode
(e) Understanding concentration cells and their working
Illustrate each application with a numerical example.
Q15. The Nernst equation has transformed how we understand batteries and sensors in India. Discuss:
(a) How the Nernst equation explains why lithium-ion batteries lose voltage as they discharge
(b) The design of pH meters and ion-selective electrodes used in Indian laboratories and industries
(c) How potentiometric titrations using Nernst equation principles provide precise endpoint detection
(d) The role of the Nernst equation in designing glucose biosensors for diabetic patients in India
Numerical / Application-Based Problems
Q16. For the half-cell reaction: Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V.
(a) Calculate the electrode potential when [Cu²⁺] = 0.001 M at 25°C.
(b) Calculate the electrode potential when [Cu²⁺] = 10 M at 25°C.
(c) At what concentration of Cu²⁺ will the electrode potential be exactly zero?
Q17. For the cell: Zn | Zn²⁺ (0.1 M) || Cu²⁺ (0.01 M) | Cu, E°cell = +1.10 V.
(a) Calculate the cell EMF at 25°C using the Nernst equation.
(b) Calculate the equilibrium constant for the cell reaction.
(c) If the cell is allowed to operate until equilibrium is reached, what will be the final concentrations of Zn²⁺ and Cu²⁺ if the initial volume of each half-cell is 100 mL?
Q18. A hydrogen electrode is immersed in a solution of unknown pH. The measured electrode potential is -0.18 V at 25°C. [E°(H⁺/H₂) = 0 V, PH₂ = 1 atm]
(a) Calculate the pH of the solution.
(b) If this solution is mixed with an equal volume of 0.1 M HCl, calculate the new pH and the new electrode potential.
(c) A pH meter in a dairy plant in Gujarat uses this principle. If milk sours (pH decreases from 6.6 to 4.5), calculate the change in electrode potential and explain how this detects spoilage.