Test: Oxidation Reduction & Disproportionation Reactions - NEET MCQ

Oxidation number increases, oxygen gained and loss of electrons is oxidation.

• Reduction involves gaining electrons or hydrogen.
• It is characterized by a decrease in oxidation number.
• Therefore, options I (electronation) and IV (decrease in oxidation number) are correct.

Options II (deelectronation and gain of oxygen) and III (increase in oxidation number) describe oxidation, not reduction.

CN^- → CNO^-
4+ x = -1      4+ x-2 = -1
∴ x = -5    ∴ x = -3   
Oxidation number increases. This is oxidation.

CN^- → CNO^- + 2e^-

Com pounds having oxidising and reducing nature in given reaction are said to have oxidation-reduction duality. Such compounds are said to undergo disproportionation reaction.

(This is called Cannizzaro reaction.)

Note Such compounds have O.N. of the affected atoms intermediate of oxidation part and reduction part

If balanced, then to balance charge
x = - 1

To balance oxidation number, cross-multiply by change in oxidation number

H is reduced to H₂ and O is oxidised to O₂.

Ag⁺ is reduced to Ag by oxygen



Thus, except in (b), oxygen behaves as reducing agent.

To balance electron  (II)  is to be multiplied by (3). Thus,

Cross -multiply by change in oxidation number and balance H by H⁺ ions.

Thus, x = 2, y = 5 and z = 6 

Maximum oxidation number of Cl = + 7

Thus, species with + 7 oxidation number can be reduced but not oxidised.

Oxidising power of F₂ > CI₂ > Br₂ > l₂


Thus, (II) is true.

In experiment (III), orange colour of Br₂ does not appear, as it also oxidises Kl to l₂. Hence , only violet colour appears in CHCI₃ layer. Thus, (III) is also true.

Since, ΔG ° < 0, hence it is spontaneous in forward direction . Oxidation number of Xe decreases hence, it is an oxidising agent and oxidation number of F increases, hence it is a reducing agent.

To balance a redox reaction , balance effected atoms

II. Take oxidation number of two Cr-atom

Take total (oxidation number) of each atom (oxidised, re duced ) cross - multiply by change in oxidation number and add
I x 6 + II x 1

Thus, x = 6
y = 1
z = 14

Fe 2⁺ and Cr³⁺ are oxidised (increase in oxidation number)
O₂, is reduced .

A species which is oxidised (deelectronation) is a reducing agent.

Thus, (ii) and (iv) represent reducing behaviour of H₂O₂.

Both elements are present in the same substance . Such reaction is called intramolecular redox reaction.

(a) Zn is oxidised to Zn²⁺ in complex

(b) Au⁺ is reduced to Au — True
(c) Au (s) is displaced
(d) Since, it is a spontaneous reaction ΔG° < 0

Carbon is reducing agent (R) and oxygen is oxidising agent (O). CO with oxidation number of C (= + 2) is formed.
(R) is inexcess hence, product of lower (oxidation number) is formed.
Thus, (a) is correct.

(O) is in excess hence product of higher oxidation number is formed. Thus, (b) is also correct.

In all of the above reactions, oxidised and reduced part are in same species, hence, intramolecular redox reactions.