Q4. Given E⊖: K+/K −2.93 V, Ag+/Ag +0.80 V, Hg2+/Hg +0.79 V, Mg2+/Mg −2.37 V and Cr3+/Cr −0.74 V, arrange the metals in increasing order of reducing power (NCERT Exercise 7.29).
The more negative E⊖, the stronger the reducing agent: Ag < Hg < Cr < Mg < K. One caution on the data: +0.79 V is in fact the value for Hg22+/Hg (+0.797 V). For Hg2+/Hg it is +0.85 V, which would make mercury a weaker reducing agent than silver: Hg < Ag < Cr < Mg < K.
Q5. A cell is made from the couples Ni2+/Ni (−0.25 V) and Ag+/Ag (+0.80 V). Which metal is the anode, and what is E⊖cell?
Nickel, with the lower E⊖, is oxidised, so it is the anode (negative): Ni + 2Ag+ → Ni2+ + 2Ag. E⊖cell = 0.80 − (−0.25) = 1.05 V.
Q6. Fluorine is made from fluorides by electrolysis, not by oxidising F− with an ordinary chemical oxidising agent. Use NCERT Table 7.1 to explain why.
F2/F− has the highest standard potential in the table, +2.87 V, so F2 is the strongest oxidant among the listed aqueous couples, and ordinary aqueous chemical oxidants do not liberate F2 from fluoride. Specialised nonaqueous chemical syntheses exist (Christe,1986); the table does not prohibit them.
Q7. In permanganate titrations the solution is acidified with dilute sulphuric acid, not hydrochloric acid. Explain with electrode potentials.
MnO4−/Mn2+ (+1.51 V) is above Cl2/Cl− (+1.36 V), so acidified permanganate oxidises chloride to chlorine. Some permanganate would be used up on the HCl instead of the sample, and the titre would be too high (see redox titrations).
Q8. Using electrode potentials, justify that fluorine is the best oxidant among the halogens and that hydroiodic acid is the best reductant among the hydrohalic acids (NCERT Exercise 7.15).
E⊖: F2/F− +2.87, Cl2/Cl− +1.36, Br2/Br− +1.09, I2/I− +0.54 V. The highest value makes F2 the strongest oxidising agent; it oxidises even water, 2F2 + 2H2O → 4HF + O2. The lowest value makes I− the strongest reducing agent: HI is oxidised by Cl2 and Br2, for example Cl2 + 2I− → 2Cl− + I2.
Q9. Predict the products of electrolysis of (i) aqueous AgNO3 with silver electrodes; (ii) aqueous AgNO3 with platinum electrodes; (iii) dilute H2SO4 with platinum electrodes; (iv) aqueous CuCl2 with platinum electrodes (NCERT Exercise 7.27).
(i) Silver deposits at the cathode and the silver anode dissolves as Ag+. (ii) Silver at the cathode; oxygen at the anode, as water is oxidised, and the solution becomes acidic. (iii) Hydrogen at the cathode, oxygen at the anode. (iv) Copper at the cathode and, under the usual textbook conditions with sufficient chloride, chlorine at the anode; dilute chloride and electrode conditions can favour oxygen evolution instead. Electrolysis is treated fully in Class 12.