Redox Reactions

(i) In SO₂, O.N. of S is +4. In principle, S can have a minimum O.N. of -2 and maximum of +6. Therefore, S in SO₂ can either decrease or increase its O.N. and hence can act both as an oxidising as well as a reducing agent.

(ii) In H₂O₂, the O.N. of O is -1. In principle, O can have a minimum O.N. of -2 and maximum of zero (+1 is possible in O₂F₂ and +2 in OF₂). Therefore, O in H₂O₂ can either decrease its O.N. from -1 to -2 or can increase its O.N. from -1 to zero. Therefore, H₂O₂ acts both as an oxidising as well as a reducing agent.

(iii) In O₃, the O.N. of O is zero. It can only decrease its O.N. from zero to -1 or -2, but cannot increase to +2. Therefore, O₃ acts only as an oxidant.

(iv) In HNO₃, O.N. of N is +5 which is maximum. Therefore, it can only decrease its O.N. and hence it acts as an oxidant only.

(a) H_2? SO₅?  by conventional method.
Let x be the oxidation number of S
2 (+1) + x + 5 (−2) = 0
x = +8
+8 oxidation state of S is not possible as S cannot have an oxidation number more than 6. The fallacy is overcome if we calculate the oxidation number from its structure HO−S (O₂)−O−O−H.
−1+X+2 (−2)+2 (−1)+1=0
x=+6

(b) Dichromate ion
Let x be the oxidation number of Cr in dichromate ion
2x+7 (−2)=−2
x=+6
Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy.

(c) Nitrate ion, by conventional method
Let x be the oxidation number of N in nitrate ion.
x+3 (−2)=−1
From the structure O−−N+ (O)−O−
x+1 (−1)+1 (−2)+1 (−2)=0
x=+5
Thus there is no fallacy.

(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