EMF of a Cell - UNSOLVED PRACTICE SET
Chapter: Electrochemistry | Topic: EMF of a Cell
EMF OF A CELL - UNSOLVED PRACTICE SET
Topic: EMF of a Cell
Multiple Choice Questions
Q1. The standard EMF of a cell is calculated as:
- E°cell = E°cathode + E°anode
- E°cell = E°cathode - E°anode
- E°cell = E°anode - E°cathode
- E°cell = E°cathode × E°anode
Q2. For a spontaneous cell reaction, E°cell must be:
- Negative
- Positive
- Zero
- Infinite
Q3. The EMF of a concentration cell depends on:
- The nature of the electrodes
- The difference in concentrations of the electrolyte
- The temperature only
- The volume of the solution
Q4. A Daniel cell has E°cell = +1.10 V. This means:
- The cell reaction is non-spontaneous
- The cell reaction is spontaneous under standard conditions
- The cell requires external energy
- The cell is at equilibrium
Q5. The EMF of a cell at equilibrium is:
- Equal to E°cell
- Equal to zero
- Maximum
- Negative
Q6. For the cell: Mg | Mg²⁺ || Ag⁺ | Ag, the standard EMF is:
- E°(Ag⁺/Ag) - E°(Mg²⁺/Mg)
- E°(Mg²⁺/Mg) - E°(Ag⁺/Ag)
- E°(Ag⁺/Ag) + E°(Mg²⁺/Mg)
- Zero
Short Answer Questions
Q7. Calculate the standard EMF of the cell: Ni | Ni²⁺ (1 M) || Cu²⁺ (1 M) | Cu. [E°(Ni²⁺/Ni) = -0.25 V, E°(Cu²⁺/Cu) = +0.34 V]
Q8. What is a concentration cell? Give one example and explain why the EMF depends only on concentration differences.
Q9. For the cell: Zn | Zn²⁺ (0.1 M) || Ag⁺ (0.01 M) | Ag, E°cell = +1.56 V. Predict whether the cell EMF will increase or decrease from the standard value and explain.
Q10. Why does the EMF of a cell decrease as the cell operates? What happens when Ecell becomes zero?
Q11. Your wristwatch battery is marked 1.5 V. Explain what this voltage represents and how it relates to the chemical reactions occurring inside.
Q12. Calculate the equilibrium constant for the reaction: Cu + 2Ag⁺ → Cu²⁺ + 2Ag at 25°C. [E°cell = +0.46 V]
Long Answer Questions
Q13. Discuss the EMF of a cell in detail:
(a) Definition of EMF and its significance
(b) Calculation of standard EMF: E°cell = E°cathode - E°anode
(c) Relationship between EMF and spontaneity: ΔG = -nFE
(d) Effect of concentration on EMF — Nernst equation for complete cells
(e) Concentration cells — definition, examples, and EMF calculation
(f) EMF at equilibrium and its relationship to equilibrium constant
Q14. Explain the factors affecting the EMF of a cell:
(a) Nature of the electrodes — standard electrode potentials
(b) Concentration of electrolytes — Nernst equation
(c) Temperature
(d) Number of electrons transferred
(e) pH (for cells involving H⁺ or OH⁻)
Illustrate with numerical examples showing how each factor influences EMF.
Q15. Cell EMF measurements drive innovation in India's energy sector. Discuss:
(a) How EMF calculations guide the design of lithium-ion batteries for India's electric vehicle market (Ola, Ather, Tata)
(b) The role of high EMF cells in India's solar energy storage systems
(c) How fuel cell EMF determines the efficiency of hydrogen-powered buses in Delhi and other cities
(d) The development of solid-state batteries with higher EMF and safety for Indian consumers
Numerical / Application-Based Problems
Q16. For the cell: Fe | Fe²⁺ (0.01 M) || Cu²⁺ (0.1 M) | Cu
Given: E°(Fe²⁺/Fe) = -0.44 V, E°(Cu²⁺/Cu) = +0.34 V
(a) Calculate the standard EMF of the cell.
(b) Calculate the cell EMF at 25°C using the Nernst equation.
(c) Calculate the equilibrium constant for the cell reaction.
Q17. A concentration cell is set up as: Ag | Ag⁺ (0.001 M) || Ag⁺ (1 M) | Ag
(a) Calculate the EMF of this cell at 25°C.
(b) Write the cell reaction and identify the direction of the spontaneous reaction.
(c) If the volume of each half-cell is 100 mL, calculate the mass of silver deposited at the cathode when the cell reaches equilibrium.
Q18. A lead-acid battery in an Indian truck has six cells connected in series, each with E°cell = +2.04 V.
(a) Calculate the total voltage of the fully charged battery.
(a) When the battery is 50% discharged, the acid concentration drops from 6 M to 3 M. Calculate the new cell voltage per cell using the Nernst equation for: Pb + PbO₂ + 4H⁺ + 2SO₄²⁻ → 2PbSO₄ + 2H₂O
(c) Calculate the total energy stored in a fully charged 100 Ah battery and compare it with the energy needed to run a 1-ton AC for 8 hours. (Assume 1 V × 1 A × 1 h = 1 Wh)