Redox Reactions — Page 6 | NEET Chemistry

Q51

Which of the following options are correct for the reaction

2\left[\mathrm{Au}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \rightarrow 2 \mathrm{Au}(\mathrm{s})+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq})

A. Redox reaction B. Displacement reaction C. Decomposition reaction D. Combination reaction Choose the correct answer from the options given below:

A A and B only

B C and D only

C A only

D A and D only

Correct Answer

Option A

Solution

$2\left[\stackrel{+1}{\mathrm{Au}}(\mathrm{CN})_2\right]^{-}+\stackrel{0}{\mathrm{Z}} \mathrm{n}(\mathrm{s}) \longrightarrow 2 \stackrel{0}{\mathrm{Au}}+\left[\stackrel{+2}{\mathrm{Zn}}(\mathrm{CN})_4\right]^{-2}$ $\mathrm{Zn}$ displaced $\mathrm{Au}^{+}$ Reduction and Oxidation both are taking place.

Q52

The oxidation states of nitrogen in NO, NO2, N2O and NO

_3^ -

are in the order of :

A N2O > NO2 > NO > NO

_3^ -

B NO > NO2 > N2O > NO

_3^ -

C NO2 > NO

_3^ -

> NO > N2O

D NO

_3^ -

> NO2 > NO > N2O

Correct Answer

Option D

Solution

To determine the oxidation states of nitrogen in each compound, we consider the overall charge of the compound and the known oxidation states of other elements involved, typically oxygen.

NO (Nitric Oxide): Oxygen generally has an oxidation state of $-2$ .

Let the oxidation state of nitrogen in NO be $x$ .

The equation: $x + (-2) = 0$ .

Solving for $x$ , we find $x = +2$ .

NO $_2$ (Nitrogen Dioxide): Oxygen has an oxidation state of $-2$ .

Let the oxidation state of nitrogen be $x$ .

The equation: $x + 2(-2) = 0$ .

Solving for $x$ , we find $x = +4$ .

N $_2$ O (Dinitrogen Monoxide or Nitrous Oxide): Oxygen has an oxidation state of $-2$ .

Let each nitrogen atom have an oxidation state of $x$ .

The equation: $2x + (-2) = 0$ .

Solving for $x$ , we find $x = +1$ .

NO $_3^-$ (Nitrate Ion): Oxygen has an oxidation state of $-2$ .

The overall charge of the ion is $-1$ .

Let the oxidation state of nitrogen be $x$ .

The equation: $x + 3(-2) = -1$ .

Solving for $x$ , we find $x = +5$ .

Therefore, the oxidation states of nitrogen are: NO: $+2$ NO $_2$ : $+4$ N $_2$ O: $+1$ NO $_3^-$ : $+5$ The order of oxidation states from highest to lowest is: NO $_3^-$ ( $+5$ ) > NO $_2$ ( $+4$ ) > NO ( $+2$ ) > N $_2$ O ( $+1$ ) Thus, the correct option is: Option D NO

_3^-

> NO $_2$ > NO > N $_2$ O

Q53

The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is :

A NO2 < NO < N2O3 < N2O

B NO2 < N2O3 < NO < N2O

C N2O < NO < N2O3 < NO2

D N2O < N2O3 < NO < NO2

Correct Answer

Option C

Solution

In N2O oxidation states of nitrogen = +1 In NO oxidation states of nitrogen = +2 In N2O3 oxidation states of nitrogen = +3 In NO2 oxidation states of nitrogen = +4

Q54

Given below are two statements: Statement I : In redox titration, the indicators used are sensitive to change in

\mathrm{pH}

of the solution. Statement II : In acid-base titration, the indicators used are sensitive to change in oxidation potential. In the light of the above statements, choose the most appropriate answer from the options given below

A Both Statement I and Statement II are incorrect

B Statement I is correct but Statement II is incorrect

C Statement I is incorrect but Statement II is correct

D Both Statement I and Statement II are correct

Correct Answer

Option A

Solution

The correct answer is Option A: Both Statement I and Statement II are incorrect.

Explanation : In redox titrations, the indicators used are sensitive to a change in oxidation potential, not pH.

A redox titration (also called an oxidation-reduction titration) can accurately determine the concentration of an unknown analyte by measuring the amount of an oxidizing or reducing agent that it consumes.

These indicators work by undergoing a definite color change at a particular electrode potential.

In contrast, in acid-base titrations, the indicators used are sensitive to a change in pH, not oxidation potential.

Acid-base titrations are based on the neutralization reaction between the acid and the base.

The point at which all the acid or base has reacted with the other component is called the equivalence point, and it often results in a sudden change in pH.

This sudden change in pH can be detected using a pH-sensitive indicator.

Q55

Match List I with List II. .tg .tg LIST I Reaction LIST II Type of redox reaction A.

\mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \rightarrow 2 \mathrm{NO}_{(\mathrm{g})}

I. Decomposition B.

2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \rightarrow 2 \mathrm{PbO}_{(\mathrm{s})}+4 \mathrm{NO}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}

II. Displacement C.

2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}_{(\mathrm{l})} \rightarrow 2 \mathrm{NaOH}_{(\mathrm{aq} .)}+\mathrm{H}_{2(\mathrm{~g})}

III. Disproportionation D.

2 \mathrm{NO}_{2(\mathrm{~g})}+2^{-} \mathrm{OH}(\text{ aq. }) \rightarrow \mathrm{NO}_{2(\mathrm{aq} .)}^{-}+\mathrm{NO}_{3(\text{ aq. })}^{-}+\mathrm{H}_2 \mathrm{O}_{(\mathrm{l})}

IV. Combination Choose the correct answer from the options given below :

A (A)-(IV), (B)-(I), (C)-(II), (D)-(III)

B (A)-(I), (B)-(II), (C)-(III), (D)-(IV)

C (A)-(II), (B)-(III), (C)-(IV), (D)-(I)

D (A)-(III), (B)-(II), (C)-(I), (D)-(IV)

Correct Answer

Option A

Solution

(A)

\stackrel{\circ}{\mathrm{N}}_2+\stackrel{\circ}{\mathrm{O}}_2 \longrightarrow 2 \mathrm{NO}_{(\mathrm{g})}

Combination reaction (B)

2 \mathrm{~Pb}\left(\mathrm{NO}_3\right)_{2(\mathrm{~s})} \longrightarrow 2 \mathrm{PbO}+4 \mathrm{NO}_2+\mathrm{O}_2

Decomposition reaction (C)

2 \mathrm{Na}_{(\mathrm{s})}+2 \mathrm{H}_2 \mathrm{O}(\mathrm{I}) \longrightarrow 2 \mathrm{NaOH}_{(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{~g})}

Displacement reaction (D)

2 \mathrm{NO}_{2(\mathrm{~g})}+2 \mathrm{OH}_{(\mathrm{g})}^{-} \longrightarrow \mathrm{NO}_{2 \text{ (aq) }}^{-}+\mathrm{NO}_{3(\text{ (qq) }}^{-}+\mathrm{H}_2 \mathrm{O}

Nitrogen oxidises and reduces both. So it is a disproportionation reaction.

\mathrm{A} \rightarrow(\mathrm{IV}), \mathrm{B} \rightarrow(\mathrm{I}), \mathrm{C} \rightarrow(\mathrm{II}), \mathrm{D} \rightarrow(\mathrm{III})

Q56

Which one of the following is an example of disproportionation reaction ?

A

3MnO_4^{2 - } + 4{H^ + } \to 2MnO_4^ - + Mn{O_2} + 2{H_2}O

B

MnO_4^ - + 4{H^ + } + 4{e^ - } \to Mn{O_2} + 2{H_2}O

C

10{I^ - } + 2MnO_4^ - + 16{H^ + } \to 2M{n^{2 + }} + 8{H_2}O + 5{I_2}

D

8MnO_4^ - + 3{S_2}O_3^{2 - } + {H_2}O \to 8Mn{O_2} + 6SO_4^{2 - } + 2O{H^ - }

Correct Answer

Option A

Solution

MnO_4^{2 - }

is an intermediate oxidation state and is converted into compounds having higher and lower oxidation states.

Q57

Which of the following oxidation reactions are carried out by both

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

and

\mathrm{KMnO}_4

in acidic medium? A.

\Gamma^{-} \rightarrow \mathrm{I}_2

B.

\mathrm{S}^{2-} \rightarrow \mathrm{S}

C.

\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}

D.

\mathrm{I}^{-} \rightarrow \mathrm{IO}_3^{-}

E.

\mathrm{S}_2 \mathrm{O}_3{ }^{2-} \rightarrow \mathrm{SO}_4^{2-}

Choose the correct answer from the options given below :

A A, B and C Only

B C, D and E Only

C B, C and D Only

D A, D and E Only

Correct Answer

Option A

Solution

$\mathrm{I}^{-} \xrightarrow{\mathrm{H}^{+}} \mathrm{I}_2\qquad \mathrm{I}^{-} \xrightarrow{\mathrm{OH}^{-}} \mathrm{IO}_3^{-}$ $\mathrm{S}^{-2} \xrightarrow{\mathrm{H}^{+}} \mathrm{S} \quad \mathrm{S}_2 \mathrm{O}_3^{2-} \xrightarrow{\mathrm{OH}^{-}} \mathrm{SO}_4^{2-}$

\begin{aligned} & \mathrm{Fe}^{+2} \longrightarrow \mathrm{Fe}^{+3} \\ & \mathrm{~S}_2 \mathrm{O}_3^{2-} \xrightarrow{\mathrm{H}^{+}} \mathrm{S} \downarrow+\mathrm{SO}_4^{2-} \end{aligned}

Q58

2 \mathrm{IO}_{3}^{-}+x \mathrm{I}^{-}+12 \mathrm{H}^{+} \rightarrow 6 \mathrm{I}_{2}+6 \mathrm{H}_{2} \mathrm{O}

What is the value of

x

?

A 10

B 2

C 12

D 6

Correct Answer

Option A

Solution

The reaction given is a disproportionation reaction where iodine ( $\mathrm{I}_2$ ) undergoes both oxidation and reduction.

Disproportionation reactions are a type of redox reaction where an element is simultaneously oxidized and reduced.

In the process, iodine gets oxidized from 0 oxidation state (in $\mathrm{I}_2$ ) to +5 oxidation state (in $\mathrm{IO}_3^{-}$ ), and reduced from 0 oxidation state (in $\mathrm{I}_2$ ) to -1 oxidation state (in $\mathrm{I}^{-}$ ).

The n-factor is the total change in oxidation state per molecule that undergoes the redox reaction.

In this case, the n-factor for $\mathrm{IO}_3^{-}$ is 5 (as iodine goes from 0 to +5) and for $\mathrm{I}^{-}$ , it's 1 (as iodine goes from 0 to -1).

Now, to balance the redox reaction, the total increase in oxidation state (total oxidation) must equal the total decrease in oxidation state (total reduction).

Hence, the molar ratio of $\mathrm{IO}_3^{-}$ to $\mathrm{I}^{-}$ must be 1:5.

So, the balanced reaction would be: $\mathrm{IO}_3^{-}+6 \mathrm{H}^{+}+5 \mathrm{I}^{-} \rightarrow 3 \mathrm{I}_2+3 \mathrm{H}_2 \mathrm{O}$ This equation yields 3 moles of $\mathrm{I}_2$ , but the original equation needs to produce 6 moles of $\mathrm{I}_2$ , so the entire equation is multiplied by 2: $2 \mathrm{IO}_3^{-}+12 \mathrm{H}^{+}+10 \mathrm{I}^{-} \rightarrow 6 \mathrm{I}_2+6 \mathrm{H}_2 \mathrm{O}$ This tells us that to get 6 moles of $\mathrm{I}_2$ , you need 10 moles of $\mathrm{I}^{-}$ .

So, $x = 10$ .