Redox Reactions

(a) Toluene can be oxidised to benzoic acid in acidic, basic and neutral media according to the following redox equations:

In the laboratory, benzoic acid is usually prepared by alcoholic KMnO₄ oxidation of toluene. However, in industry alcoholic KMnO₄ is preferred over acidic or alkaline KMnO₄ because of the following reasons:
(i) The cost of adding an acid or the base is avoided because in the neutral medium, the base (OH^- ions) are produced in the reaction itself.
(ii) Since reactions occur faster in homogeneous medium than in heterogeneous medium, therefore, alcohol helps in mixing the two reactants, i.e., KMnO₄  (due to its polar nature) and toluene (because of its being an organic compound).

(b) KBr + H₂? SO₄? → KHSO₄? + HBr
HBr is a strong reducing agent which further reduces H₂? SO₄?  as follows
2HBr + H₂? SO₄? → Br₂? + SO₂? + H₂? O

But in case of mixture containing Chloride the reaction ends up on forming HCl as HCl is a weak reducing agent which can't further reducethe sulphuric acid further.
KCl+H₂? SO₄? →KHSO₄? +HCl? (colourless pungent smell gas)

The three examples are:

(i) When excess P_4?   (reducing agent) reacts with F_2? (oxidizing agent),  PF_3?  is produced in which P has +3 oxidation number.
                        P_4? (excess) + F₂? → PF₃?
But if fluorine is in excess,  PF₅?  is formed in which P has oxidation number of +5.
                        P_4? ? + F₂? (excess) → PF₅?

(ii) Oxidizing agent is oxygen and reducing agent is K. When excess K reacts with oxygen,  K_2? O is formed in which oxygen has oxidation number of -2.
                        4K (excess) + O₂? → 2K₂? O

But if oxygen is in excess, then K_2? O₂?  is formed in which O has oxidation number of -1.
                        2K + O₂? (excess) → K₂? O₂?

(iii) The oxidizing agent is oxygen and the reducing agent is C. When an excess of C reacts with oxygen, CO is formed in which C has +2 oxidation number.
                        C (excess) + O_2? → CO

When excess of oxygen is used,  CO₂?  is formed in which C has +4 oxidation number.
                        C + O₂? (excess) → CO₂

(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.