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NEET • CHEMISTRY • REDOX REACTIONS

Redox reactions practice · Mixed difficulty

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91 Medium

91 of 184

Application +4 | -1

The number of mole of oxalate ions oxidized by one mole of

\mathrm{MnO}_4^ -

ion is:

View answer and explanation

Correct option: C

\frac{5}{2}

Option C (

\frac{5}{2}

) is correct because each permanganate ion gains 5 electrons, while each oxalate ion loses 2 electrons, requiring a ratio of 2 permanganate ions to 5 oxalate ions.

92 Easy

92 of 184

Analyzing +4 | -1

Which of the following reactions does not involve either oxidation or reduction

View answer and explanation

Correct option: C

CrO_4^{ - 2} \to C{r_2}O_7^{ - 2}

Option C (

CrO_4^{ - 2} \to C{r_2}O_7^{ - 2}

) is correct because the oxidation state of Cr remains the same (+6) in both species.

93 Medium

93 of 184

Remembering +4 | -1

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

acts as ……. indicator in its redox titrations.

View answer and explanation

Correct option: A

Self

Option A (Self) is correct because KMnO₄ acts as a self-indicator in redox titrations due to its intense purple color.

94 Medium

94 of 184

Remembering +4 | -1

The algebraic sum of the oxidation number of all the atoms in a compound must be

View answer and explanation

Correct option: C

zero

Option C (zero) is correct because the sum of oxidation numbers in a neutral compound must equal zero.

95 Hard

95 of 184

Remembering +4 | -1

Solution of potassium chloride or ammonium nitrate in salt-bridge usually solidified by boiling with

View answer and explanation

Correct option: A

agar-agar

Option A (agar-agar) is correct because agar-agar is commonly used to solidify the electrolyte solution in a salt bridge, allowing for ion transport while preventing mixing of the solutions.

96 Medium

96 of 184

Remembering +4 | -1

The stable oxidation states of Mn are∶

View answer and explanation

Correct option: C

+ 2,\mathrm{ } + 7

Option C (

+2, +7

) is correct because these are the most stable and commonly observed oxidation states for Manganese.

97 Easy

97 of 184

Application +4 | -1

Oxidation state of oxygen is −1 in the compound :

View answer and explanation

Correct option: D

\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}

Option D (

\mathrm{N}{\mathrm{a}_2}{\mathrm{O}_2}

) is correct because in Na2O2 (sodium peroxide), the oxidation state of oxygen is -1.

98 Easy

98 of 184

Understanding +4 | -1

Which quantities are conserved in all oxidation-reduction reactions?

View answer and explanation

Correct option: A

Charge only

Option A (Charge only) is correct because in all chemical reactions, including redox reactions, charge is always conserved, but mass may not be conserved in nuclear reactions.

99 Medium

99 of 184

Remembering +4 | -1

The maximum oxidation number of transition metals may be:

View answer and explanation

Correct option: C

+ 8

Option C (+8) is correct because some transition metals, like Osmium (Os) and Ruthenium (Ru), can exhibit a maximum oxidation number of +8.

100 Hard

100 of 184

Analyzing +4 | -1

In basic medium,

{H_2}{O_2}

exhibits which of the following reactions?
A)

M{n^{2 + }} \to M{n^{4 + }}

\,\,\,\,\,\,\,\,\,\,

{I_2} \to {I^ - }

\,\,\,\,\,\,\,\,\,\,\,

PbS \to PbS{O_4}

Choose the most appropriate answer from the options given below.

View answer and explanation

Correct option: D

A and B

Option D (A and B) is correct because H2O2 can act as both an oxidizing and reducing agent in basic medium, oxidizing Mn2+ to Mn4+ and reducing I2 to I-.

101 Easy

101 of 184

Understanding +4 | -1

In nitric oxide (NO), the oxidation state of nitrogen is :

View answer and explanation

Correct option: D

+ 2

Option D (

+2

) is correct because in nitric oxide (NO), oxygen has an oxidation state of -2, and the overall molecule is neutral, so nitrogen must have an oxidation state of +2.

102 Easy

102 of 184

Understanding +4 | -1

The halogen that most easily reduced is

View answer and explanation

Correct option: A

{F_2}

Option A (

{F_2}

) is correct because fluorine has the highest reduction potential among the halogens, making it the easiest to reduce.

103 Easy

103 of 184

Application +4 | -1

Carbon is in the lowest oxidation state in

View answer and explanation

Correct option: A

C{H_4}

Option A (

CH_4

) is correct because carbon is bonded to hydrogen, which is less electronegative, giving carbon a -4 oxidation state, the lowest among the options.

104 Hard

104 of 184

Application +4 | -1

On balancing the given redox reaction,

\mathrm{aCr}_2 \mathrm{O}_7^{2-}+\mathrm{bSO}_3^{2-}(\mathrm{aq})+\mathrm{cH}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{aCr}^{3+}(\mathrm{aq})+\mathrm{bSO}_4^{2-}(\mathrm{aq})+\frac{\mathrm{c}}{2} \mathrm{H}_2 \mathrm{O}(\mathrm{l})

the coefficients

\text{a, b}

and

\text{c}

are found to be, respectively

View answer and explanation

Correct option: C

1,3,8

Option C (

1,3,8

) is correct because these are the coefficients that balance the given redox reaction by ensuring both mass and charge are conserved.

105 Medium

105 of 184

Analyzing +4 | -1

Identify the correct statement.

View answer and explanation

Correct option: D

The oxidation states of chromium in

Cr\,O_4^{2 - }\,\,\mathrm{and}\,\,C{r_2}\,O_7^{2 - }

are not the same.

Option A is correct because Cr2+ (d4) is a stronger reducing agent than Fe2+ (d6) in water due to its tendency to achieve a stable d5 configuration.

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106 Easy

106 of 184

Analyzing +4 | -1

Which reaction involves neither oxidation nor reduction?

View answer and explanation

Correct option: A

\mathrm{CrO}_4^{2 - } \to \mathrm{C}{\mathrm{r}_2}\mathrm{O}_7^{2 - }

Option A (CrO4^2- → Cr2O7^2-) is correct because the oxidation state of Cr remains +6 in both ions.

107 Hard

107 of 184

Application +4 | -1

For the reaction,

\mathrm{N}{\mathrm{H}_3} + \mathrm{OC}{\mathrm{l}^ - } \to {\mathrm{N}_2}{\mathrm{H}_4} + \mathrm{C}{\mathrm{l}^ - }

occurring in basic medium, the coefficient of

{\mathrm{N}_2}{\mathrm{H}_4}

in the balanced equation will be

View answer and explanation

Correct option: A

Option A (1) is correct because the balanced equation has a coefficient of 1 for

{\mathrm{N}_2}{\mathrm{H}_4}

108 Medium

108 of 184

Analyzing +4 | -1

Which change requires a reducing agent?

View answer and explanation

Correct option: B

\mathrm{BrO}_3^ - \to \mathrm{Br}{\mathrm{O}^ - }

Option B (

\mathrm{BrO}_3^ - \to \mathrm{Br}{\mathrm{O}^ - }

) is correct because the oxidation state of Br decreases, indicating reduction, which requires a reducing agent.

109 Medium

109 of 184

Understanding +4 | -1

Volumetric estimation of

\mathrm{CuS}{\mathrm{O}_4}

using hypo as intermediate solution along with KI solution and starch as indicator is an example of :

View answer and explanation

Correct option: A

Redox titration

Option A (Redox titration) is correct because the reaction involves both oxidation and reduction processes during the volumetric estimation of CuSO4 using hypo.

110 Medium

110 of 184

Application +4 | -1

Moles of

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

required for decolorizing 1 mole of acidified

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

are :

View answer and explanation

Correct option: C

\frac{5}{2}

Option C (

\frac{5}{2}

) is correct because the balanced redox reaction requires 5 moles of H2O2 for every 2 moles of KMnO4, making the ratio 5/2.

111 Easy

111 of 184

Understanding +4 | -1

Reduction is a process which involves :

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because reduction involves gaining electrons, adding hydrogen/removing oxygen, and adding metal/removing non-metal.

112 Easy

112 of 184

Application +4 | -1

Which compound shows highest oxidation number for chlorine?

View answer and explanation

Correct option: D

\mathrm{KCl}{\mathrm{O}_4}

Option D (

\mathrm{KCl}{\mathrm{O}_4}

) is correct because in KClO4, the oxidation state of chlorine is +7, which is the highest among the given options.

113 Medium

113 of 184

Understanding +4 | -1

The oxidation number of Ba in barium peroxide is :

View answer and explanation

Correct option: A

+ 2

Option A (+2) is correct because barium is an alkaline earth metal and always exhibits a +2 oxidation state in its compounds.

114 Easy

114 of 184

Understanding +4 | -1

Fluorine does not undergo disproportionation because.

View answer and explanation

Correct option: B

F exhibit -1 and +1 oxidation numbers

Option A is correct because fluorine is the most electronegative element and only exhibits a -1 oxidation state.

115 Easy

115 of 184

Understanding +4 | -1

Stannous chloride gives a white precipitate with a solution of mercuric chloride. In this process mercuric chloride is :

View answer and explanation

Correct option: B

Reduced

Option B (Reduced) is correct because stannous chloride (SnCl2) reduces mercuric chloride (HgCl2) to mercurous chloride (Hg2Cl2) or mercury (Hg).

116 Easy

116 of 184

Analyzing +4 | -1

The conversion of

\mathbf{Pb}{\mathbf{O}_\mathbf{2}}\,\,to\,\mathbf{Pb}{\left( {\mathbf{N}{\mathbf{O}_\mathbf{3}}} \right)_\mathbf{2}}

View answer and explanation

Correct option: B

Reduction

Reduction. Option B is correct because the oxidation number of Pb decreases from +4 in

\mathbf{Pb}{\mathbf{O}_\mathbf{2}}

to +2 in

\mathbf{Pb}{\left( {\mathbf{N}{\mathbf{O}_\mathbf{3}}} \right)_\mathbf{2}}

117 Easy

117 of 184

Application +4 | -1

Total number of

\mathrm{Al}{\mathrm{F}_3}

molecules in a sample of

\mathrm{Al}{\mathrm{F}_3}

containing

3.01 \times {10^{23}}

ions of

{\mathrm{F}^ - }

is :

View answer and explanation

Correct option: D

{10^{23}}

Option D (

{10^{23}}

) is correct because each AlF3 molecule contains 3 fluoride ions, so the number of AlF3 molecules is one-third the number of fluoride ions.

118 Hard

118 of 184

Analyzing +4 | -1

All the following can undergo disproportion except

View answer and explanation

Correct option: A

MnO_4^ -

Option A (

MnO_4^ -

) is correct because Mn in

MnO_4^ -

is in its highest oxidation state (+7), and therefore cannot be further oxidized, preventing it from undergoing disproportionation.

119 Easy

119 of 184

Understanding +4 | -1

Oxidation number of sodium in sodium amalgam is :

View answer and explanation

Correct option: D

Zero

Option D (Zero) is correct because sodium in sodium amalgam (NaHg) exists in its metallic state, hence its oxidation number is zero.

120 Easy

120 of 184

Analyzing +4 | -1

In which reaction there is no change in valency and the oxidation state?

View answer and explanation

Correct option: C

N{a_2}{O_2} + {H_2}S{O_4} \to N{a_2}S{O_4} + {H_2}{O_2}

Option C (Na2O2 + H2SO4 → Na2SO4 + H2O2) is correct because the oxidation states of all elements remain unchanged in this reaction.

Revision notes and key takeaways

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