Redox Reactions — Page 5 | NEET Chemistry

Q41

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.

Q42

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.

Q43

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.

Q44

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.}}

Q45

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.

Q46

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⁺

Q47

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.

Q48

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.

Q49

Which of the following cannot function as an oxidising agent?

A

\mathrm{SO}_4^{2-}

B

\mathrm{MnO}_4^{-}

C

\mathrm{N}^{3-}

D

\mathrm{BrO}_3^{-}

Correct Answer

Option C

Solution

The ability of a species to function as an oxidizing agent depends on its ability to gain electrons (be reduced).

For a species to be an oxidizing agent, it must contain an element that has a positive oxidation state and is capable of being reduced (gain electrons), thereby oxidizing another species.

Let's look at each of the options : Option A:

\mathrm{SO}_4^{2-}

(sulfate ion) has sulfur in a +6 oxidation state.

Sulfate is the fully oxidized form of sulfur in aqueous solution, and while it can undergo certain reductions under specific conditions (to form, for example, $\mathrm{SO}_3^{2-}$ or sulfite), it is generally stable and not commonly considered a strong oxidizing agent.

Option B:

\mathrm{MnO}_4^{-}

(permanganate ion) has manganese in a +7 oxidation state.

This ion is a very strong oxidizing agent and is well known for its ability to oxidize a wide range of organic and inorganic substances, with manganese being reduced to a lower oxidation state (often to +2 as $\mathrm{Mn}^{2+}$ in acidic solution).

Option C:

\mathrm{N}^{3-}

is the nitride ion, with nitrogen in a -3 oxidation state.

Since nitrogen is in its lowest oxidation state, it cannot be further reduced (gain more electrons), and hence cannot act as an oxidizing agent.

In fact, it would more likely act as a reducing agent because it has room to be oxidized (lose electrons).

Option D:

\mathrm{BrO}_3^{-}

(bromate ion) has bromine in a +5 oxidation state.

This ion can act as an oxidizing agent because the bromine can be reduced to a lower oxidation state (often to -1 as $\mathrm{Br}^{-}$ ).

Therefore, the species that cannot function as an oxidizing agent due to its inability to undergo reduction is Option C, the

\mathrm{N}^{3-}

ion.

Q50

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 Answer

Option C

Solution

2H{I^{ - 1}} + {H_2}\mathop {S{O_4}}\limits^{ + 6} \to I_2^0 + \mathop {S{O_2}}\limits^{ + 4} + 2{H_2}O

in this reaction oxidation number of

S

is decreasing from

+6

to

+4

hence undergoing reduction and for

HI

oxidation Number of

I

is increasing from

-1

to

0

hence undergoing oxidation therefore

{H_2}S{O_4}

is acting as oxidising agent.