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Redox Reactions MCQs for NEET — Chemistry Questions with Answers

Practice free Redox Reactions (Chemistry) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

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Sulphurous acid can be used as:

oxidising agent

reducing agent

bleaching agent

all of these

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Explanation

(d) S in H₂SO₃ is in +4 oxidation state. It lies in between its maximum and minimum oxidation state, i.e., +6 and -2 and thus, S can increase or decrease its ox. no. as the case may be.

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When SO₂ is passed through acidified solution of potassium dichromate, then chromium sulphate is formed. The change in oxidation number of chromium is:

+4 to+2

+5 to+3

+6 to+3

+7 to+2

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Explanation

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In which transfer of five electrons takes place?

Mn $O_{4}^{-}$ $\to$ Mn²⁺

Cr $O_{4}^{2 -}$ $\to$ Cr³⁺

Mn $O_{4}^{-}$ $\to$ MnO₂

Cr₂ $O_{7}^{2 -}$ $\to$ 2Cr³⁺

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Explanation

(a) Mn⁷⁺ + 5e $\to$ Mn²⁺

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In the reaction;

2Ag + 2H₂SO₄ $\to$ Ag₂SO₄ + 2H₂O + SO₂,H₂SO₄ acts as:

oxidising agent

reducing agent

dehydrating agent

none of these

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Explanation

(a) 2e + S⁶⁺ $\to$ S⁴⁺

S of H₂SO₄ is reduced.

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Which of the following shows highest ox. no. in combined state?

All of these

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Explanation

(d) All of these show +8 ox. no.

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In the preparation of chlorine from HCl; MnO₂ acts as:

reducing agent

oxidising agent

catalytic agent

dehydrating agent

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Explanation

(b) Mn⁴⁺ + 2e $\to$ Mn²⁺;

MnO₂ is itself reduced.

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Which combination is odd with respect to oxidation numbers of S, Cr, N and H respectively:\

H₂SO₅, H₂S₂O₈, H₂SO₄, SF₆

K₂Cr₂O₇, K₂CrO₄, CrO₅, CrO₂Cl₂

NH₃ , NH₄⁺, N₃H, N $O_{2}^{-}$

CaH₂ , NaH , LiH, MgH₂

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Explanation

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In which reaction H₂O₂ acts as a reducing agent?

Ag₂O + H₂O₂ $\to$ 2Ag + H₂O + O₂

2KI + H₂O_{2 $\to$}2KOH + I₂

PbS + 4H₂O₂_$\to$PbSO₄ + 4H₂O₂

H₂O₂+ SO₂ $\to$ H₂SO₄

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Explanation

(a) Ag⁺ is reduced to Ag.

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Excess of KI reacts with CuSO₄ solution and then Na₂S₂O₃ solution is added to it. Which of the statement is incorrect in this reaction?

Evolved I₂ is reduced

CuI₂ is formed

Na₂S₂O₃ is oxidised

Cu₂I₂ is formed

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Explanation

(b) This liberated I₂ is reduced and Na₂S₂O₃ is oxidise.

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In the equation,

N $O_{2}^{-}$ + H₂O $\to$ N $O_{3}^{-}$ + 2H⁺ + ne^-

n stands for:

H⁺

e^-

2e^-

3e^-

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Explanation

-1 = -1+2+(-n) $∴$ n=2

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