Redox Reactions — Page 3 | JEE Chemistry

Q21

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 Answer

Option D

Solution

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

\left( 1 \right)\,\,\,\,\,\,{H_2}\mathop {{O_2}}\limits^{ - 1} + 2{H^ + } + 2{e^ - }\,\,\overset{\,}\longrightarrow \,\,2{H_2}\mathop O\limits^{ - 2} \,\,({\mathop{\rm Re}\nolimits} d.)

\left( 2 \right)\,\,\,\,\,\,{H_2}\mathop {{O_2}}\limits^{ - 1} \,\,\overset{\,}\longrightarrow \mathop {{O_2}}\limits^0 + 2{H^ + } + 2{e^ - }\,\,(Ox.)

\left( 3 \right)\,\,\,\,\,\,{H_2}\mathop {{O_2}}\limits^{ - 1} + 2{e^ - }\,\,\overset{\,}\longrightarrow \,\,2\mathop O\limits^{ - 2} {H^ - }\,\,(Red.)

\left( 4 \right)\,\,\,\,\,\,{H_2}O_2^{ - 1} + 2O{H^ - }\,\overset{\,}\longrightarrow \mathop {{O_2}}\limits^0 + {H_2}O + 2{e^ - }\,\,\left( {Ox.} \right)

Q22

Which of the given reactions is not an example of disproportionation reaction?

A

2 \mathrm{H}_{2} \mathrm{O}_{2} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}

B

2 \mathrm{NO}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{HNO}_{3}+\mathrm{HNO}_{2}

C

\mathrm{MnO}_{4}^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}

D

3 \mathrm{MnO}_{4}^{2-}+4 \mathrm{H}^{+} \rightarrow 2 \mathrm{MnO}_{4}^{-}+\mathrm{MnO}_{2}+2 \mathrm{H}_{2} \mathrm{O}

Correct Answer

Option C

Solution

\begin{aligned} & 2 \mathrm{H}_2 \overset{-1}{\mathrm{O}_2} \longrightarrow 2 \mathrm{H}_2 \overset{2-}{\mathrm{O}}+\overset{0}{\mathrm{O}_2}: \text{ Disproportionation } \\\\ & 2 \overset{+4}{\mathrm{NO}_2}+\mathrm{H}_2 \mathrm{O} \rightarrow \overset{+5}{\mathrm{HNO}_3}+\overset{+3}{\mathrm{HNO}_2} \text{ : Disproportionation } \\\\ & \mathrm{MnO}_4^{-}+4 \mathrm{H}^{+}+3 \mathrm{e}^{-} \rightarrow \mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}: \text{ reduction } \\\\ & 3 \overset{+6}{\mathrm{MnO}_4^{2-}}+4 \mathrm{H}^{+} \rightarrow 2 \overset{+7}{\mathrm{MnO}_4^{-}}+\overset{+4}{\mathrm{MnO}_2}+2 \mathrm{H}_2 \mathrm{O}: \text{ Disproportionation } \end{aligned}

Q23

The redox reaction among the following is :

A reaction of H2SO4 with NaOH.

B formation of ozone form atmosphere oxygen in the presence of sunlight.

C combination of dinitrogen with dioxygen at 2000 K

D reaction of [Co(H2O)6]Cl3 With AgNO3

Correct Answer

Option C

Solution

N2 + O2 $\to$ 2NO during the reaction, oxidation of nitrogen take place from 0 to 2 and reduction of oxygen take place from 0 to –2.

It means this reaction is redox reaction.

3O2 $\to$ 2O3 (Non - redox reaction) 2NaOH + H2SO4 $\to$ Na2SO4 + 2H2O (neutralization reaction) [Co(H2O)6]Cl3 + 3AgNO3 $\to$ 3AgCl $\downarrow$ + [Co(H2O)6](NO3)3 Reaction of [CO(H2O)6]Cl3 with AgNO3 is not redox reaction.

It is a precipitation reaction.

Q24

0.1 M solution of KI reacts with excess of

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

and

\mathrm{KIO}_3

solutions. According to equation

5 \mathrm{I}^{-}+\mathrm{IO}_3^{-}+6 \mathrm{H}^{+} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}

Identify the correct statements : (A) 200 mL of KI solution reacts with 0.004 mol of

\mathrm{KIO}_3

(B) 200 mL of KI solution reacts with 0.006 mol of

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

(C) 0.5 L of KI solution produced 0.005 mol of

\mathrm{I}_2

(D) Equivalent weight of

\mathrm{KIO}_3

is equal to (

\dfrac{\text{ Molecular weight }}{5}

) Choose the correct answer from the options given below :

A (C) and (D) only

B (A) and (B) only

C (B) and (C) only

D (A) and (D) only

Correct Answer

Option D

Solution

Molarity of KI = 0.1 M

5{I^ - } + IO_3^ - + 6{H^ + } \to 3{I_2} + 3{H_2}O

(balanced chemical equation)

5KI + KI{O_3} + 3{H_2}S{O_4} \to 3{I_2} + 3{K_2}S{O_4} + 3{H_2}O

(A) Volume given (KI) = 200 mL = 0.2 L Molarity (KI) = 0.1 M or mol L $^{-1}$ The molarity formula is used here to calculate the number of moles of K.I.

Molarity,

M = {{number\,of\,moles} \over {volume\,in\,L}}

So, number of moles = Molarity $\times$ Volume in L For the given volume of KI, number of moles = 0.1 mol L $^{-1}$ $\times$ 0.2 L = 0.02 mol The stoichiometric ratio between KI and KIO $_3$ is

KI:KI{O_3}

{I^ - }:IO_3^ -

5:1

1:{1 \over 5}

For one mol

{I^ - },{1 \over 5}

mol

IO_3^ -

is used. So, for 0.2 mol I $^-$ , Moles of

IO_3^ - = 0.02 \times {1 \over 5} = 0.004

mol The statement (A) is correct.

(B) Volume given (KI) = 200 mL = 0.2 L Molarity (KI) = 0.1 M number of moles of KI (I $^-$ ) = Molarity $\times$ Volume (L)

= 0.1\,mol\,{L^{ - 1}} \times 0.2\,L = 0.02\,mol

The stoichiometric ratio between KE and

{H_2}H{O_4}

is

KI:{H_2}H{O_4}

5:3

1:{3 \over 5}

{3 moles

{H_2}S{O_4}

can give 6H $^+$ }

{I^ - },{3 \over 5}

For one mol

{I^ - },{3 \over 5}

mol

{H_2}S{O_4}

is used. So, for 0.02 mol I $^-$ , moles of

{H_2}S{O_4} = 0.02 \times {3 \over 5} = 0.012

mol This statement is not correct.

(C) Volume gien (KI) = 0.5 L Molarity (KI) = 0.1 M Number of moles of KI = Molarity $\times$ Volume

= 0.1\,mol\,{L^{ - 1}} \times 0.5\,L = 0.05\,mol

The stoichiometric ratio between I $^-$ and I $_2$ is

KI:{I_2}

{I^ - }:{I_2}

5:3

1:{3 \over 5}

For 1 mol

{I^ - },{3 \over 5}

mol I $_2$ is produced. So, for 0.05 mol, moles of

{I_2} = 0.05 \times {3 \over 5} = 0.03

mol This statement is not correct. (D) Equivalent weight of

KI{O_3} = {{Molecular\,weight} \over 5}

This statement is correct. Equivalent weight

= {{Molecular\,weight} \over {Valency\,factor\, \to number\,of\,electrons}}

Here, valency factor = 5 Each KIO $_3$ molecule gains 5 electrons in this vacation.

KIO $_3$ acts as the ordinating agent, gains electrons from the iodide ion in KI.

When KIO $_3$ is redued to I $_2$ each molecule gains 5 electrons.

So, the valency factor is 5 and equivalent weight is lower than molecular weight.

This statement is correct.

Correct statements are A and D.

Q25

Given below are two statements: Statement I: In the oxalic acid vs

\mathrm{KMnO}_4

(in the presence of dil

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

) titration the solution needs to be heated initially to

60^{\circ} \mathrm{C}

, but no heating is required in Ferrous ammonium sulphate (FAS) vs

\mathrm{KMnO}_4

titration (in the presence of dil

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

) Statement II: In oxalic acid vs

\mathrm{KMnO}_4

titration, the initial formation of

\mathrm{MnSO}_4

takes place at high temperature, which then acts as catalyst for further reaction. In the case of FAS vs

\mathrm{KMnO}_4

, heating oxidizes

\mathrm{Fe}^{2+}

into

\mathrm{Fe}^{3+}

by oxygen of air and error may be introduced in the experiment. In the light of the above statements, choose the correct answer from the options given below

A Statement I is false but Statement II is true

B Statement I is true but Statement II is false

C Both Statement I and Statement II are true

D Both Statement I and Statement II are false

Correct Answer

Option C

Solution

For the titration: Oxalic acid v/s $\mathrm{KMnO}_4$

\begin{aligned} & 2 \mathrm{MnO}_4^{-}+5(\mathrm{COO})_2^{2-}+16 \mathrm{H}^{+} \rightarrow \\ & 10 \mathrm{CO}_2+2 \mathrm{Mn}^{2+}+8 \mathrm{H}_2 \mathrm{O} \end{aligned}

This reaction is slow at room temperature, but becomes fast at $60^{\circ} \mathrm{C}$ .

Manganese(II) ions catalyse the reaction; thus, the reaction is autocatalytic; once manganese(II) ions are formed, it becomes faster and faster.

The titration of FAS v/s $\mathrm{KMnO}_4$ do not require heating because at higher temperature the oxidation of $\mathrm{Fe}^{+2}$ to $\mathrm{Fe}^{+3}$ by atmospheric $\mathrm{O}_2$ will be prominent.

Q26

The species which does not undergo disproportionation reaction is :

A

\mathrm{ClO}_4^{-}

B

\mathrm{ClO}_3^{-}

C

\mathrm{ClO}^{-}

D

\mathrm{ClO}_2^{-}

Correct Answer

Option A

Solution

\textbf{Oxidation States:}

In

\mathrm{ClO_4^-}

(perchlorate), chlorine is in the +7 oxidation state. In

\mathrm{ClO_3^-}

(chlorate), chlorine is in the +5 oxidation state. In

\mathrm{ClO_2^-}

(chlorite), chlorine is in the +3 oxidation state. In

\mathrm{ClO^-}

(hypochlorite), chlorine is in the +1 oxidation state.

\textbf{Disproportionation Reaction:}

A disproportionation reaction is one in which a species simultaneously undergoes oxidation and reduction.

For this to occur, the element must be in an intermediate oxidation state such that it can be oxidized to a higher state and reduced to a lower one.

In

\mathrm{ClO^-}

and

\mathrm{ClO_2^-}

, chlorine is in lower oxidation states (+1 and +3, respectively), making them susceptible to disproportionation.

For example, hypochlorite can disproportionate in basic solution as follows:

3\,\mathrm{ClO^-} \rightarrow 2\,\mathrm{Cl^-} + \mathrm{ClO_3^-}

In

\mathrm{ClO_3^-}

, chlorine is in an intermediate oxidation state (+5) that, under certain conditions, can undergo disproportionation.

However, in

\mathrm{ClO_4^-}

, chlorine is in its highest possible oxidation state (+7) and cannot be oxidized further.

Since disproportionation requires one part of the species to be oxidized and the other reduced,

\mathrm{ClO_4^-}

is thermodynamically stable and does not undergo disproportionation.

\boxed{\mathrm{ClO_4^-} \text{ (perchlorate ion) does not undergo disproportionation.}}

Q27

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 Answer

Option C

Solution

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

5 \, \text{C}_2\text{O}_4^{2-} + 2 \, \text{MnO}_4^- + 16 \, \text{H}^+ \rightarrow 10 \, \text{CO}_2 + 2 \, \text{Mn}^{2+} + 8 \, \text{H}_2\text{O}

In this redox reaction, the oxalate ion $\text{C}_2\text{O}_4^{2-}$ is oxidized to carbon dioxide $\text{CO}_2$ , and the permanganate ion $\text{MnO}_4^-$ is reduced to manganese(II) ion $\text{Mn}^{2+}$ .

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

\text{C}_2\text{O}_4^{2-} \rightarrow 2 \, \text{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 \, \text{CO}_2} = 1 \, e^-/\text{CO}_2

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

Q28

In acidic medium,

\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7

shows oxidising action as represented in the half reaction:

\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+\mathrm{XH}^{+}+\mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A}+\mathrm{ZH}_2 \mathrm{O}

\mathrm{X}, \mathrm{Y}, \mathrm{Z}

and

\mathrm{A}

are respectively are :

A

14,7,6

and

\mathrm{Cr}^{3+}

B

14,6,7

and

\mathrm{Cr}^{3+}

C

8,4,6

and

\mathrm{Cr}_2 \mathrm{O}_3

D

8,6,4

and

\mathrm{Cr}_2 \mathrm{O}_3

Correct Answer

Option B

Solution

Balancing the Half-Reaction Chromium (Cr) : Already balanced with 2 Cr on each side. Oxygen (O): Add 7 H₂O to the right:

\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + \mathrm{XH}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}

Hydrogen (H) : Add 14 H⁺ to the left:

\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + \mathrm{Ye}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}

Charge : Add 6e⁻ to the left:

\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{~A} + 7\mathrm{H}_2 \mathrm{O}

Identifying A : The reduction of Cr₂O₇²⁻ in acidic medium forms Cr³⁺ ions. Final Balanced Equation :

\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-} + 14\mathrm{H}^{+} + 6\mathrm{e}^{\ominus} \rightarrow 2 \mathrm{Cr}^{3+} + 7\mathrm{H}_2 \mathrm{O}

Answer X = 14 Y = 6 Z = 7 A = Cr³⁺

Q29

The number of ions from the following that are expected to behave as oxidising agent is :

\mathrm{Sn}^{4+}, \mathrm{Sn}^{2+}, \mathrm{Pb}^{2+}, \mathrm{Tl}^{3+}, \mathrm{Pb}^{4+}, \mathrm{Tl}^{+}

A 3

B 4

C 2

D 1

Correct Answer

Option C

Solution

Due to inert pair effect

\mathrm{Pb}^{2+}

is more stable than

\mathrm{Pb}^{4+}

and

\mathrm{Tl}^{+}

is more stable than

\mathrm{Tl}^{3+}

. Therefore,

\mathrm{Pb}^{4+}

and

\mathrm{T}^{3+}

will function as oxidising agents and easily get reduced to

\mathrm{Pb}^{2+}

and

\mathrm{TI}^{+}

respectively.

Q30

While titrating dilute HCl solution with aqueous NaOH, which of the following will not be required?

A Pipette and distilled water

B Clamp and phenolphthalein

C Burette and porcelain tile

D Bunsen burner and measuring cylinder

Correct Answer

Option D

Solution

Bunsen Burner and measuring cylinder is not required for titration.