8.3. Justify that the following reactions are redox reactions: (a) CuO (s) + H2(g) → Cu(s) + H2O(g) (b) Fe2O3(s) +3CO(g) → 2Fe(s) + 3CO2(g) (c) 4BCl3(g) +3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3AlCl3(s) (d) 2K(s) +F2(g) → 2K+F–(s) (e) 4 NH3 (g) + 5 O2 → 4NO (g) + 6H2O (g)

(a) Here, O is removed from CuO, therefore, it is reduced to Cu, while O is added to H2 to form H2O, therefore, it is oxidised. Further, O.N. of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H2 to +1 in H20. Therefore, CuO is reduced to Cu but H2 is oxidised to H2O. Thus, this is a redox reaction. (b) Here O.N. of Fe decreases from +3 in Fe2O3 to 0 in Fe while that of C increases from +2 in CO to +4 in CO2. Further, oxygen is removed from Fe2O3 and added to CO, therefore, Fe2O3 is reduced while CO is oxidised. Thus, this is a redox reaction. (c) Here, O.N. of B decreases from +3 in BrCl3to -3 in B2H6 while that of H increases from -1 in LiAlH4to +1 in B2H6. Therefore, BCl3 is reduced while LiAlH4 is oxidised. Further, H is added to BCl3 but is removed from LiAlH4, therefore, BC13 is reduced while LiAlH4 is oxidised. Thus, it is a redox reaction. (d) Here, each K atom as lost one electron to form K+ while F2 has gained two electrons to form two F– ions. Therefore, K is oxidised while F2 is reduced. Thus, it is a redox reaction. (e) Here, the oxidation number of N increases from –3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2 decreases from 0 in O2 to –2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.

8.3. Justify that the following reactions are redox reactions:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O₃(s) +3CO(g) → 2Fe(s) + 3CO₂(g)
(c) 4BCl₃(g) +3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s)
(d) 2K(s) +F₂(g) → 2K⁺F^–(s)

(e) 4 NH₃(g) + 5 O₂ → 4NO (g) + 6H₂O (g)

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Vishal Baghel | Contributor-Level 10

10 months ago

(a) Here, O is removed from CuO, therefore, it is reduced to Cu, while O is added to H₂ to form H₂O, therefore, it is oxidised. Further, O.N. of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H₂ to +1 in H₂0. Therefore, CuO is reduced to Cu but H₂ is oxidised to H₂O. Thus,