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)

<p><strong> (a)</strong> Here, O is removed from CuO, therefore, it is reduced to Cu, while O is added to H₂&nbsp;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&nbsp;to +1 in&nbsp;H₂0. Therefore, CuO is reduced to Cu but H₂&nbsp;is oxidised to H₂O. Thus, this is a redox reaction.<br><br><strong> (b) </strong>Here O.N. of Fe decreases from +3 in&nbsp;Fe₂O₃&nbsp;to 0 in Fe while that of C increases from +2 in CO to +4 in CO₂. Further, oxygen is removed from Fe₂O₃&nbsp;and added to CO, therefore, Fe₂O₃&nbsp;is reduced while CO is oxidised. Thus, this is a redox reaction.<br><br><strong> (c)</strong> Here, O.N. of B decreases from +3 in BrCl₃to -3 in B₂H₆&nbsp;while that of H increases from -1 in LiAlH₄to +1 in B₂H₆. Therefore, BCl₃&nbsp;is reduced while LiAlH₄&nbsp;is oxidised. Further, H is added to BCl₃&nbsp;but is removed from LiAlH₄, therefore, BC1₃ is reduced while LiAlH₄&nbsp;is oxidised. Thus, it is a redox reaction.<br><br></p><p><strong> (d)</strong> Here, each K atom as lost one electron to form K⁺ while F₂&nbsp;has gained two electrons to form two F^–&nbsp;ions. Therefore, K is oxidised while F₂&nbsp;is reduced. Thus, it is a redox reaction.</p><p><strong>&nbsp;</strong></p><p><strong> (e) </strong>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.</p>

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