Gibbs Energy and Cell Potential - UNSOLVED PRACTICE SET
Chapter: Electrochemistry | Topic: Gibbs Energy and Cell Potential
GIBBS ENERGY AND CELL POTENTIAL - UNSOLVED PRACTICE SET
Topic: Gibbs Energy and Cell Potential
Multiple Choice Questions
Q1. The relationship between Gibbs free energy change and cell EMF is:
- ΔG = nFEcell
- ΔG = -nFEcell
- ΔG = nF/Ecell
- ΔG = Ecell/nF
Q2. For a spontaneous reaction, the value of ΔG is:
- Positive
- Negative
- Zero
- Infinite
Q3. The standard Gibbs free energy change is related to the equilibrium constant by:
- ΔG° = -RT ln K
- ΔG° = RT ln K
- ΔG° = -RT/K
- ΔG° = RT × K
Q4. The relationship between E°cell and K is:
- E°cell = (RT/nF) ln K
- E°cell = (nF/RT) ln K
- E°cell = -(RT/nF) ln K
- E°cell = RT × nF × K
Q5. If E°cell = +0.50 V for a two-electron process at 25°C, the equilibrium constant is approximately:
- 10⁸
- 10¹⁷
- 10²⁵
- 10³⁴
Q6. The maximum electrical work obtainable from a cell is equal to:
- ΔH
- ΔG
- TΔS
- -TΔS
Short Answer Questions
Q7. Derive the relationship between ΔG° and E°cell. What does the negative sign signify?
Q8. For the reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, E°cell = +1.10 V. Calculate ΔG° and predict whether the reaction is spontaneous.
Q9. Calculate the equilibrium constant for the reaction: 2Fe³⁺ + Sn²⁺ → 2Fe²⁺ + Sn⁴⁺ at 25°C. [E°cell = +0.62 V]
Q10. Why is ΔG more negative than ΔH for many electrochemical reactions? What role does entropy play?
Q11. Your teacher explains that a dead battery has Ecell = 0 but ΔG° remains negative. Explain this apparent contradiction.
Q12. Calculate the maximum work that can be obtained from a Daniel cell producing 1 mole of Cu. [E°cell = +1.10 V]
Long Answer Questions
Q13. Discuss the thermodynamics of galvanic cells:
(a) Relationship between ΔG and Ecell: ΔG = -nFEcell
(b) Relationship between ΔG° and E°cell: ΔG° = -nFE°cell
(c) Relationship between ΔG° and equilibrium constant K: ΔG° = -RT ln K
(d) Combined relationship: E°cell = (RT/nF) ln K = (0.0591/n) log K at 25°C
(e) Maximum electrical work from a cell: Wmax = -ΔG
(f) Efficiency of a cell and comparison with heat engines
Q14. Explain the relationship between cell potential, Gibbs energy, and equilibrium:
(a) How E°cell determines the spontaneity and extent of a reaction
(b) How K relates to the position of equilibrium
(c) The effect of concentration on ΔG and Ecell
(d) Why a reaction with positive E°cell may not proceed if kinetic barriers exist
(e) Numerical examples connecting all three quantities
Q15. Thermodynamic principles guide India's energy transition. Discuss:
(a) Why fuel cells (ΔG → electrical work) are more efficient than combustion engines (ΔH → heat → work)
(b) How ΔG calculations help design better catalysts for hydrogen production in India's green hydrogen mission
(c) The theoretical maximum efficiency of lithium-ion batteries and why real efficiency is lower
(d) How understanding ΔG vs ΔH explains why some reactions are spontaneous despite being endothermic
Numerical / Application-Based Problems
Q16. For the cell reaction: 2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu
Given: E°(Al³⁺/Al) = -1.66 V, E°(Cu²⁺/Cu) = +0.34 V
(a) Calculate E°cell.
(b) Calculate ΔG° for the reaction.
(c) Calculate the equilibrium constant K at 25°C.
[Given: F = 96500 C/mol, R = 8.314 J/K·mol]
Q17. A hydrogen-oxygen fuel cell operates at 25°C with E°cell = +1.23 V.
(a) Calculate ΔG° for the reaction: 2H₂ + O₂ → 2H₂O
(b) Calculate the maximum electrical work obtainable from 1 kg of H₂.
(c) If the fuel cell operates at 80% efficiency, calculate the actual electrical energy produced. Compare this with the energy from burning the same amount of H₂ in a heat engine with 30% efficiency.
[Given: ΔH°f(H₂O) = -286 kJ/mol]
Q18. India's National Hydrogen Mission aims to produce 5 million tonnes of green hydrogen annually by 2030.
(a) Calculate the total electrical energy required if water electrolysis operates at 70% voltage efficiency (theoretical voltage = 1.23 V, actual = 1.75 V).
(b) If solar electricity costs ₹ 3 per kWh, calculate the cost of producing 1 kg of green hydrogen.
(c) Calculate the ΔG° savings if a catalyst reduces the overpotential by 0.3 V, and determine the annual cost savings for 5 million tonnes production.
[Given: Molar mass of H₂ = 2 g/mol, F = 96500 C/mol]