Redox Reactions

(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.

(a) In Kl₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K⁺ [I —I <— I]^–. Here, a coordinate bond is formed between I₂ molecule and I^– ion. The oxidation number of two iodine atoms forming the I₂ molecule is zero, while that of iodine forming the coordinate bond is -1. Thus, the oxidation number of the three I atoms, atoms in Kl₃ is 0, 0 and -1, respectively.

(b) By conventional method O.N. of S in H₂S₄O₆is calculated as:

2 (+1) +4x + 6) (-2) = 0

Or x = +2.5

But all the four S atoms cannot be in the same oxidation state.

So by chemical bonding method:

The O.N. of each of the S-atoms linked with each other in the middle is zero while that of each of the remaining two S-atoms is +5.

(c) By conventional method O.N. of Fe in Fe₃O₄ is calculated as:

3x + 4 (-2) = 0 or x = 8/3

By stoichiometry O.N. of Fe in Fe₃O₄ is +2 and +3.

(d) By conventional method, O.N. of C in CH3CH2OH is calculated as:

2x + 6 (+1) +1 (-2) = 0 or x = -2

(e) By conventional method, O.N. of C in CH₃COOH is calculated as:

2x + 4 - 4 = 0 or x = 0

By chemical bonding method, C2 is attached to three H-atoms (less electronegative than carbon) and one -COOH group (more electronegative than carbon).

Therefore, O.N. of C₂ = 3 (+1) + x + 1 (-1) = 0 or x = -2

But C₁ is attached to one oxygen atom by a double bond, one OH (-1) and one CH3 (+1) group, therefore, O.N. of C₁ = +1 + x+ 1 (-2) + 1 (-1) = 0 or x = +2