Q.Answer the following question.
Justify that the following reactions are redox reaction; identify the species oxidized/reduced, which acts as an oxidant and which act as a reductant.
a. 2Cu2O (s) + Cu2S (s) → 6Cu (s) + SO2

a. 2Cu2O(s) + Cu2S(s) → 6Cu(s) + SO2(g). In Cu2O, O=-2 gives Cu=+1 (per atom). In Cu2S, matching Cu=+1, S works out to -2 (2(+1)+ON(S)=0). In the products, Cu(s)=0 and in SO2, O=-2 x2=-4 gives S=+4. So Cu falls from +1 to 0 (reduced, 6 atoms total across both reactants) while S rises from -2 to +4 (oxidised, losing 6 electrons from the one S atom) -- the electron counts match (6 lost = 6 gained), confirming redox. Cu(I), supplied by both Cu2O and Cu2S, is reduced (so both compounds act as the oxidant for copper), while the sulfide ion within Cu2S is oxidised (so Cu2S is the reductant).

b. HF(aq) + OH⁻(aq) → H2O(l) + F⁻(aq). H stays +1 throughout, F stays -1 throughout, O stays -2 throughout. No oxidation number changes anywhere -- this is a simple acid-base neutralisation (H⁺ from HF combining with OH⁻), NOT a redox reaction. …

Assigning oxidation numbers to every atom before and after each reaction and checking for a change; identifying oxidant/reduc …

• Assuming every acid + base or ion-exchange-looking reaction (like b) must be redox just because ions are involved -- oxidation-number change, not ion involvement, is the actual test. …