In which ${\rm{S}}{{\rm{O}}_2}$ acts as oxidant, while reacting with :

Acidified ${\rm{KMn}}{{\rm{O}}_4}$

Acidified ${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$

Acidified ${{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}$

${\rm{HBr}}$ and ${\rm{HI}}$ reduce ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$,${\rm{HCl}}$ can reduce ${\rm{KMn}}{{\rm{O}}_4}$ and ${\rm{HF}}$ can reduce:

${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

${{\rm{K}}_2}{\rm{C}}{{\rm{r}}_2}{{\rm{O}}_7}$

Which reaction indicates the oxidising behavior of ${{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4}$

$2{\rm{HI}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {{\rm{I}}_2} + {\rm{S}}{{\rm{O}}_2} + 2{{\rm{H}}_2}{\rm{O}}$

$2{\rm{PC}}{{\rm{l}}_5} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to 2{\rm{POC}}{{\rm{l}}_3} + 2{\rm{HCl}} + {\rm{S}}{{\rm{O}}_2}{\rm{C}}{{\rm{l}}_2}$

$2{\rm{NaOH}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{N}}{{\rm{a}}_2}{\rm{S}}{{\rm{O}}_4} + 2{{\rm{H}}_2}{\rm{O}}$

${\rm{NaCl}} + {{\rm{H}}_2}{\rm{S}}{{\rm{O}}_4} \to {\rm{NaHS}}{{\rm{O}}_4} + {\rm{HCl}}$

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NEET • CHEMISTRY • REDOX REACTIONS

Redox reactions practice · Mixed difficulty

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Showing 1-30 of 184 questions

1 Hard

1 of 184

Application +4 | -1

The oxidation state of iron in sodium nitroprusside is :

View answer and explanation

Correct option: A

+ 2

Option A (+ 2) is correct because the oxidation state of iron in sodium nitroprusside,

\mathrm{Na}_2\mathrm{[Fe(CN}{\mathrm{)}_5}\mathrm{NO]}

is +2.

2 Medium

2 of 184

Remembering +4 | -1

Titrations in which liberated

{I_2}

is estimated to carry out the volumetric estimations are known as ….titrations.

View answer and explanation

Correct option: B

Iodimetric

Option A (Iodometric) is correct because iodometric titrations involve the indirect determination of oxidizing agents by titrating the iodine liberated in a reaction.

3 Medium

3 of 184

Understanding +4 | -1

It is found that V forms a double salt isomorphous with Mohr’s salt. The oxidation number of V in this compound is :

View answer and explanation

Correct option: B

+ \mathrm{ }2

Option B (

+ 2

) is correct because Mohr's salt has the formula (NH₄)₂Fe(SO₄)₂.6H₂O, where iron (Fe) has a +2 oxidation state, and since V forms an isomorphous double salt, it will also have a +2 oxidation state.

4 Hard

4 of 184

Application +4 | -1

For redox reaction,

\mathrm{MnO}_4^ - + {\mathrm{C}_2}\mathrm{O}_4^{2 - } + {\mathrm{H}^ + } \to \mathrm{M}{\mathrm{n}^{2 + }} + \mathrm{C}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}

coefficient of reactants in balanced states are

\mathrm{MnO}_4^ -

{\mathrm{C}_2}\mathrm{O}_4^{2 - }

{\mathrm{H}^ + }

View answer and explanation

Correct option: A

2 5 16

Option A (2, 5, 16) is correct because when the redox reaction is balanced, the coefficients for MnO4-, C2O42-, and H+ are 2, 5, and 16 respectively.

5 Medium

5 of 184

Understanding +4 | -1

Which one of the following reaction is possible at anode?

View answer and explanation

Correct option: D

\mathrm{F}{\mathrm{e}^{2 + }} \to \mathrm{F}{\mathrm{e}^{3 + }} + {\mathrm{e}^ - }

Option D (

\mathrm{F}{\mathrm{e}^{2 + }} \to \mathrm{F}{\mathrm{e}^{3 + }} + {\mathrm{e}^ - }

) is correct because oxidation (loss of electrons) occurs at the anode.

6 Hard

6 of 184

Remembering +4 | -1

The couple having oxidised and reduced forms of a substance taking part in an oxidation or reduction half-reaction is called

View answer and explanation

Correct option: A

redox couple

Option A (redox couple) is correct because a redox couple consists of the oxidized and reduced forms of a substance involved in a half-reaction.

7 Hard

7 of 184

Analyzing +4 | -1

Assertion: When

\mathrm{Cu}_2 \mathrm{S}

is converted into

\mathrm{Cu}^{+} \,\&\, \mathrm{SO}_2

, then equivalent weight of

\mathrm{Cu}_2 \mathrm{S}

will be

\mathrm{M} / 8\,\left(\mathrm{M}=\mathrm{Mol}\right.

. wt. of

\mathrm{Cu}_2 \mathrm{S}

)
Reason:

\mathrm{Cu}^+

is converted

\mathrm{Cu}^{++}

, during this one electrons is lost.

View answer and explanation

Correct option: B

If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion

Option B is correct because the equivalent weight is calculated based on the total change in oxidation state of all copper and sulfur atoms in Cu2S during its conversion to Cu+ and SO2.

8 Hard

8 of 184

Understanding +4 | -1

Titration of

KI

with

{\mathrm{H}_2}{\mathrm{O}_2}

in presence of acid is a :

View answer and explanation

Correct option: D

All of these

Option D (All of these) is correct because the reaction involves electron transfer (redox), can be used to visually indicate the endpoint (clock reaction), and occurs between different molecules (intermolecular redox).

9 Hard

9 of 184

Application +4 | -1

What volume of 3 molar

\mathrm{HN}{\mathrm{O}_3}

is needed to oxidise 8 g of

\mathrm{F}{\mathrm{e}^{2 + }}

\mathrm{F}{\mathrm{e}^{3 + }}

\mathrm{HN}{\mathrm{O}_3}

, gets converted to NO :

View answer and explanation

Correct option: B

16 m L

Option B (16 m L) is correct because it is calculated using stoichiometry based on the balanced redox reaction and the molarity of the HNO3 solution.

10 Easy

10 of 184

Application +4 | -1

In which

\mathrm{S}{\mathrm{O}_2}

acts as oxidant, while reacting with :

View answer and explanation

Correct option: C

{\mathrm{H}_2}\mathrm{S}

Option C (

{\mathrm{H}_2}\mathrm{S}

) is correct because sulfur dioxide acts as an oxidizing agent by accepting electrons from hydrogen sulfide.

11 Hard

11 of 184

Understanding +4 | -1

\mathrm{M}{\mathrm{n}^{2 + }}

can be converted into

\mathrm{M}{\mathrm{n}^{7 + }}

by reacting with

View answer and explanation

Correct option: C

\mathrm{Pb}{\mathrm{O}_2}

Option C (

\mathrm{Pb}{\mathrm{O}_2}

) is correct because PbO2 is a strong oxidizing agent that can oxidize Mn2+ to Mn7+.

12 Medium

12 of 184

Remembering +4 | -1

The maximum oxidation state of oxygen is

View answer and explanation

Correct option: C

+2

Option C (+2) is correct because oxygen exhibits a +2 oxidation state only when bonded to fluorine, the most electronegative element.

13 Hard

13 of 184

Remembering +4 | -1

Turn bull’s blue is :

View answer and explanation

Correct option: A

\mathrm{F}{\mathrm{e}_3}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_2}

Option A (

\mathrm{F}{\mathrm{e}_3}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_2}

) is correct because Turnbull's blue is ferric ferrocyanide, formed by the reaction of ferric ions with ferrocyanide ions.

14 Hard

14 of 184

Understanding +4 | -1

Strongest oxidising agent among halogen is

View answer and explanation

Correct option: D

{F_2}

Option D (F2) is correct because fluorine is the most electronegative element and has the highest reduction potential among halogens, making it the strongest oxidizing agent.

15 Easy

15 of 184

Understanding +4 | -1

When Fe metal is rusted then Fe is :

View answer and explanation

Correct option: A

Oxidised

Option A (Oxidised) is correct because when iron rusts, it loses electrons and its oxidation state increases, which is the definition of oxidation.

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16 Hard

16 of 184

Application +4 | -1

The oxidation state of the most electronegative element in the products of the reaction between

\mathrm{BaO}_2

and

\mathrm{H}_2 \mathrm{SO}_4

are

View answer and explanation

Correct option: B

-1

and

-2

Option B (-1 and -2) is correct because in the reaction of BaO2 with H2SO4, the products are BaSO4 and H2O2. The oxidation state of O in BaSO4 is -2, and in H2O2, it is -1. Oxygen is the most electronegative element in both products.

17 Medium

17 of 184

Application +4 | -1

Oxidation number of

S

{\mathrm{S}_2}\mathrm{C}{\mathrm{l}_2}

is :

View answer and explanation

Correct option: A

+ 1

Option A (+1) is correct because the oxidation number of Cl is -1, and the overall molecule is neutral, leading to +1 for S.

18 Easy

18 of 184

Application +4 | -1

\frac{{ - 1}}{3}

oxidation state of nitrogen will be obtained in case of

View answer and explanation

Correct option: B

{N_3}H

Option B (

{N_3}H

) is correct because the average oxidation state of nitrogen in hydrazoic acid is -1/3.

19 Medium

19 of 184

Analyzing +4 | -1

Statement-I: Oxidation state of Sulphur in

\mathrm{H}_2 \mathrm{SO}_3

is +6.

Statement-II: Maximum oxidation state of

\mathrm{S}

is +5.

View answer and explanation

Correct option: B

Both the statements are incorrect

Both statements are incorrect. The oxidation state of sulfur in H₂SO₃ is +4, and the maximum oxidation state of sulfur is +6.

20 Hard

20 of 184

Analyzing +4 | -1

Given the following reactions involving

\mathrm{A, B, C}

and

\mathrm{D}

(i)

\mathrm{C}+\mathrm{B}^{+} \rightarrow \mathrm{C}^{+}+\mathrm{B}

(ii)

\mathrm{A}^{+}+\mathrm{D} \rightarrow

No reaction

(iii)

\mathrm{C}^{+}+\mathrm{A} \rightarrow

No reaction

(iv)

\mathrm{D}+\mathrm{B}^{+} \rightarrow \mathrm{D}^{+}+\mathrm{B}

The correct arrangement of

\mathrm{A, B, C, D}

in order of their decreasing ability as reducing agent

View answer and explanation

Correct option: D

>

>

>

Option A is correct because the given reactions indicate the relative reducing strengths: C > B, D > B, C > A, and A > D, leading to the overall order C > A > D > B.

21 Medium

21 of 184

Understanding +4 | -1

Manganese acts as strongest oxidising agent in the oxidation state

View answer and explanation

Correct option: A

+ 7

Option A (+7) is correct because Mn in the +7 oxidation state (e.g., in KMnO4) has a strong tendency to gain electrons and get reduced, thus acting as a strong oxidizing agent.

22 Medium

22 of 184

Analyzing +4 | -1

\mathrm{HBr}

and

\mathrm{HI}

reduce

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

\mathrm{HCl}

can reduce

\mathrm{KMn}{\mathrm{O}_4}

and

\mathrm{HF}

can reduce:

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because HF is a weak reducing agent and cannot reduce any of the given compounds.

23 Easy

23 of 184

Remembering +4 | -1

Which of the following halogen shows only one oxidation state in its compounds

View answer and explanation

Correct option: B

Option B (F) is correct because fluorine is the most electronegative element and only exhibits a -1 oxidation state in its compounds.

24 Medium

24 of 184

Remembering +4 | -1

The most stable oxidation state of copper is :

View answer and explanation

Correct option: A

+ 2

Option A (

+2

) is correct because the +2 oxidation state is the most stable for copper due to its electronic configuration and hydration enthalpy.

25 Medium

25 of 184

Analyzing +4 | -1

Which of the following arrangements represent increasing oxidation number of the central atom

View answer and explanation

Correct option: A

CrO_2^ - ,\,ClO_3^ - ,CrO_4^{2 - },MnO_4^ -

Option A (

CrO_2^ - ,\,ClO_3^ -,CrO_4^{2 - },MnO_4^ -

) is correct because it represents the increasing order of oxidation numbers of the central atoms: Cr(+3), Cl(+5), Cr(+6), Mn(+7).

26 Medium

26 of 184

Analyzing +4 | -1

Which reaction indicates the oxidising behavior of

{\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4}

View answer and explanation

Correct option: D

2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}

Option D (

2\mathrm{HI} + {\mathrm{H}_2}\mathrm{S}{\mathrm{O}_4} \to {\mathrm{I}_2} + \mathrm{S}{\mathrm{O}_2} + 2{\mathrm{H}_2}\mathrm{O}

) is correct because H2SO4 oxidizes HI to I2, indicating its oxidizing behavior.

27 Medium

27 of 184

Understanding +4 | -1

Which one of the following lanthanoids does not form

M{O_2}

? [

M

is lanthanoid metal]

View answer and explanation

Correct option: D

Yb

Option D (

Yb

) is correct because Ytterbium (Yb) typically forms +3 compounds, and its +4 state is less stable, preventing the formation of

MO_2

28 Easy

28 of 184

Analyzing +4 | -1

Which is the best description of behaviour of bromine in the reaction given below?

{\mathrm{H}_2}\mathrm{O} + \mathrm{B}{\mathrm{r}_2} \to \mathrm{HBr} + \mathrm{HOBr}

View answer and explanation

Correct option: B

Both oxidised and reduced

Option B (Both oxidised and reduced) is correct because bromine undergoes both oxidation and reduction in the given reaction.

29 Hard

29 of 184

Application +4 | -1

O.N. of

\mathrm{V}

\mathrm{Rb}_4 \mathrm{Na}\left[\mathrm{HV}_{10} \mathrm{O}_{28}\right]

is-

View answer and explanation

Correct option: B

+5

Option B (+5) is correct because Rb has +1, Na has +1, H has +1, and O has -2, balancing the charge of the complex ion.

30 Medium

30 of 184

Understanding +4 | -1

Reaction of acidified

\mathrm{KMn}{\mathrm{O}_4}

with ferrous oxalate gives oxidation products containing :

View answer and explanation

Correct option: C

Both (1) and (2)

Option C (Both (1) and (2)) is correct because ferrous oxalate contains both iron(II) and oxalate ions, which are oxidized to iron(III) and carbon dioxide, respectively.

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Key Takeaways

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