More practice questions

4. Q4. Find the potential of a zinc electrode in 0.001 M Zn²⁺ at 298 K (E^⊖ = −0.76 V, NCERT Table 2.1, page 37).

   Show answerHide answer

   E = −0.76 − (0.059/2) log(1/0.001) = −0.76 − 0.0295 × 3 = −0.85 V (NCERT Eq. (2.8), page 36).

5. Q5. Find the emf of a Daniell cell at 298 K with [Zn²⁺] = 0.1 M and [Cu²⁺] = 0.001 M (E^⊖_cell = 1.10 V).

   Show answerHide answer

   Q = 0.1/0.001 = 100, so E_cell = 1.10 − 0.0295 × 2 = 1.041 V (NCERT Eq. (2.12), page 38).

6. Q6. Find the equilibrium constant at 298 K of Zn(s) + Fe²⁺(aq) → Zn²⁺(aq) + Fe(s). (E^⊖: Fe²⁺/Fe −0.44 V, Zn²⁺/Zn −0.76 V, NCERT Table 2.1, page 37.)

   Show answerHide answer

   E^⊖_cell = −0.44 − (−0.76) = 0.32 V and n = 2, so log K_c = (2 × 0.32)/0.059 = 10.85 and K_c ≈ 7 × 10¹⁰ (NCERT Eq. (2.14), page 39).

7. Q7. Find Δ_rG^⊖ for Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s), E^⊖_cell = 0.46 V, and check it against K_c = 3.92 × 10¹⁵ (NCERT Example 2.2).

   Show answerHide answer

   Δ_rG^⊖ = −2 × 96 487 × 0.46 = −88 768 J mol⁻¹ ≈ −88.8 kJ mol⁻¹. From K: −RT ln K = −8.314 × 298 × ln(3.92 × 10¹⁵) ≈ −89.0 kJ mol⁻¹. The small difference comes from rounding 0.0591 to 0.059 (NCERT Eqs. (2.14) and (2.16)).

8. Q8. Assertion: when the cell reaction is written with all coefficients doubled, Δ_rG doubles but E_cell does not. Reason: E_cell is an intensive parameter, but Δ_rG is an extensive thermodynamic property whose value depends on n. Are both true, and does the reason explain the assertion?

   Show answerHide answer

   Both are true, and the reason explains the assertion: n doubles in Δ_rG = −nFE_cell, while E_cell stays the same (NCERT page 40). Strictly, Δ_rG is per mole of reaction as written (see the body; reasoned).

9. Q9. A concentration cell has two copper electrodes: Cu|Cu²⁺(0.01 M)||Cu²⁺(0.1 M)|Cu. Find its emf at 298 K.

   Show answerHide answer

   Beyond the NCERT text, but it follows from NCERT Eq. (2.13): both half-cells are the same couple, so E^⊖_cell = 0. The net cell reaction consumes Cu²⁺ in the 0.1 M half-cell and generates it in the 0.01 M half-cell, so Q = 0.01/0.1 and E_cell = 0 − (0.059/2) log 0.1 = 0.0295 V. As the cell runs, the stronger solution gets weaker and the weaker one stronger, until they are equal and the emf is zero.

10. Q10. At what pH does a hydrogen electrode (hydrogen at 1 bar, 298 K) have a potential of −0.236 V?

    Show answerHide answer

    E = −0.059 pH, so pH = 0.236/0.059 = 4 (NCERT Eq. (2.13), page 38).

11. Q11. Given E^⊖ = 0.77 V for Fe³⁺/Fe²⁺ and −0.44 V for Fe²⁺/Fe (NCERT Table 2.1, page 37), find E^⊖ for Fe³⁺/Fe.

    Show answerHide answer

    Potentials do not add, but Gibbs energies do (reasoned from NCERT Eq. (2.16), page 40). So the nE^⊖ values add: 1 × 0.77 + 2 × (−0.44) = 0.77 − 0.88 = −0.11 V. Divide by n = 3: −0.037 V.