1. ⇒ (JEE Main 2020 (Online) 9th January Morning Slot)

The compound that cannot act both as oxidising and reducing agent is :

A. H₃PO₄

B. H₂SO₃

C. H₂O₂

D. HNO₂

Correct option is (A)

In H₃PO₄, P is present in +5 oxidation state and it can act as reducing agent only.

2. ⇒ ( JEE Main 2019 (Online) 10th January Evening Slot)

In the reaction of oxalate with permanganate in acidic medium, the number of electrons involved in producing one molecule of CO₂ is :

A. 10

B. 2

C. 1

D. 5

Correct option is (C)

The balanced reaction of oxalate with permanganate in acidic medium is :

$5 C_{2} O_{4}^{2 -} + 2 {MnO}_{4}^{-} + 16 H^{+} \to 10 {CO}_{2} + 2 {Mn}^{2 +} + 8 H_{2} O$

In this redox reaction, the oxalate ion $C_{2} O_{4}^{2 -}$ is oxidized to carbon dioxide ${CO}_{2}$ , and the permanganate ion ${MnO}_{4}^{-}$ is reduced to manganese(II) ion ${Mn}^{2 +}$ .

The oxidation half-reaction, representing the change for the oxalate ion, is :

$C_{2} O_{4}^{2 -} \to 2 {CO}_{2} + 2 e^{-}$

From this half-reaction, it's clear that each oxalate ion produces two CO₂ molecules and releases two electrons in the process.

So, the number of electrons involved in producing one molecule of CO₂ is :

$\frac{2 e^{-}}{2 {CO}_{2}} = 1 e^{-} / {CO}_{2}$

Therefore, the correct answer is Option C: One electron is involved in the production of one molecule of CO₂.

3. ⇒ ( JEE Main 2017 (Offline))

Which of the following reactions is an example of a redox reaction?

A. $X e F_{6}$ + H₂O $\to$ $X e O F_{4}$ + 2HF

B. $X e F_{6}$ + 2H₂O $\to$ $X e O_{2} F_{2}$ + 4HF

C. $X e F_{4}$ + O₂F₂ $\to$ $X e F_{6}$ + O₂

D. $X e F_{2}$ + PF₅ $\to$ ${[X e F]}^{+} P F_{6}^{-}$

Correct option is (C)

JEE Main 2017 (Offline) Chemistry - Redox Reactions Question 41 English Explanation

4. ⇒ (JEE Main 2014 (Offline))

In which of the following reactions H₂O₂ acts as a reducing agent?
1. H₂O₂ + 2H⁺ + 2e^- $\to$ 2H₂O
2. H₂O₂ - 2e^- $\to$ O₂ + 2H⁺
3. H₂O₂ + 2e^- $\to$ 2OH^-
4. H₂O₂ + 2OH^- - 2e^- $\to$ O₂ + 2H₂O

A. 1,2

B. 3,4

C. 1,3

D. 2,4

Correct option is (D)

The reducing agent loses electron during redox reaction i.e. oxidises itself.

$(1) H_{2} \overset{- 1}{O_{2}} + 2 H^{+} + 2 e^{-} \overset{}{⟶} 2 H_{2} \overset{- 2}{O} (Re d .)$

$(2) H_{2} \overset{- 1}{O_{2}} \overset{}{⟶} \overset{0}{O_{2}} + 2 H^{+} + 2 e^{-} (O x .)$

$(3) H_{2} \overset{- 1}{O_{2}} + 2 e^{-} \overset{}{⟶} 2 \overset{- 2}{O} H^{-} (R e d .)$

$(4) H_{2} O_{2}^{- 1} + 2 O H^{-} \overset{}{⟶} \overset{0}{O_{2}} + H_{2} O + 2 e^{-} (O x .)$

5. ⇒ (AIEEE 2006)

Which of the following chemical reactions depicts the oxidizing behaviour of H₂SO₄?

A. NaCl + H₂SO₄ $\to$ NaHSO₄ + HCl

B. 2PCl₅ + H₂SO₄ $\to$ 2POCl₃ + 2HCl + SO₂Cl₂

C. 2HI + H₂SO₄ $\to$ I₂ + SO₂ + 2H₂O

D. Ca(OH)₂ + H₂SO₄ $\to$ CaSO₄ + 2H₂O

Correct option is (C)

$2 H I^{- 1} + H_{2} \overset{+ 6}{S O_{4}} \to I_{2}^{0} + \overset{+ 4}{S O_{2}} + 2 H_{2} O$

in this reaction oxidation number of $S$ is decreasing from $+ 6$ to $+ 4$ hence undergoing reduction and for $H I$ oxidation Number of $I$ is increasing from $- 1$ to $0$ hence undergoing oxidation therefore $H_{2} S O_{4}$ is acting as oxidising agent.

⇒Redox

Topic 5 : Oxidizing and Reducing Agent in Reaction