Q4. 0.654 g of zinc dissolves from a zinc rod placed in silver nitrate solution. Write the half reactions and find the mass of silver deposited. (Zn = 65.4, Ag = 107.9 g mol−1)
Zn → Zn2+ + 2e− and 2Ag+ + 2e− → 2Ag. Moles of Zn = 0.654/65.4 = 0.0100 mol, so 0.0200 mol Ag forms: 0.0200 × 107.9 ≈ 2.16 g. If all silver stays attached and there are no other losses, the rod gains 2.16 − 0.654 ≈ 1.50 g.
Q5. Zinc is left in blue copper nitrate solution. Why does the blue colour fade, and how can you show that Zn2+ ions have formed?
The blue colour is due to Cu2+, which is reduced to copper metal on the zinc, so the colour fades as Cu2+ is used up. Passing H2S through the colourless solution and making it alkaline with ammonia gives white ZnS, showing Zn2+.
Q6. When cobalt is placed in nickel sulphate solution, both Co2+ and Ni2+ are present at moderate concentrations at equilibrium. What does this tell you about cobalt and nickel?
That their tendencies to release electrons are similar, so neither the reactants nor the products are greatly favoured. With zinc and copper, by contrast, the difference is large and the reaction goes almost completely.
Q7. Justify that 2K(s) + F2(g) → 2K+F−(s) is a redox reaction (NCERT Exercise 7.3 (d)).
The product is ionic. Half reactions: 2K → 2K+ + 2e− (oxidation) and F2 + 2e− → 2F− (reduction). Potassium loses electrons and fluorine gains them, so it is a redox reaction; F2 is the oxidising agent.
Q8. Classify each half reaction as oxidation or reduction: (a) Fe3+ + e− → Fe2+; (b) 2Br− → Br2 + 2e−; (c) Sn2+ → Sn4+ + 2e−; (d) O2 + 4e− → 2O2−.
(a) Reduction (gain of an electron). (b) Oxidation (loss). (c) Oxidation (loss). (d) Reduction (gain).