NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

1. Assign oxidation numbers to the underlined elements in each of the following species:

(a) ${\mathrm{NaH}}_2{\mathrm{PO}}_4$

Ans: P's oxidation number will be $x$.

We are aware of this.

Oxidation number of $\mathrm{Na}=+1$

Oxidation number of $\mathrm{H}=+1$

Oxidation number of $\mathrm{O}=-2$

$\begin{array}{rl}& (+1)(+1)(\mathrm{x})(-2)={\mathrm{NaH}}_2{\mathrm{PO}}_4\end{array}$

Then there's

$1(+1)+2(+1)+1(x)+4(-2)=0$

$1+2+x-8=0$

$x=+5$

As a result, P's oxidation number is $+5$

(b) ${\mathrm{NaHSO}}_4$

$(+1)(+1)(x)(-2)$

Ans: ${\mathrm{NaHSO}}_4$

Then there's

$1(+1)+1(+1)+1(x)+4(-2)=0$

$1+1+x-8=0$

$x=+6$

As a result, S's oxidation number is $+6$

(c) ${\mathrm{H}}_4{\mathrm{P}}_2{\mathrm{O}}_7$

Ans: ${\mathrm{H}}_4{\mathrm{P}}_2{\mathrm{O}}_7$

Then, there's

$4(+1)+2(x)+7(-2)=0$

$4+2x-14=0$

$2x=+10$

$x=+5$

As a result, P's oxidation number is $+5$

(d) ${\mathrm{K}}_2{\mathrm{MnO}}_4$

Ans: Then, there's

$2(+1)+x+4(-2)=0$

$2+x-8=0$

$x=+6$

As a result, Mn 's oxidation number is $+6$

(e) ${\mathrm{CaO}}_2$

Ans:

Then, there's

$(+2)+2(x)=0$

$2+2x=0$

x=-1

As a result, O's oxidation number is $-1$

(f) ${\mathrm{NaBH}}_4$

Ans: Then, there's

$1(+1)+1(x)+4(-1)=0$

$1+x-4=0$

$x=+3$

As a result, B's oxidation number is $+3$

(g) ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_7$

Ans: ${\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_7$

Then, there's

$2(+1)+2(x)+7(-2)=0$

$2+2x-14=0$

$2x=12$

$x=+6$

As a result, S's oxidation number is $+6$

(h) $\mathrm{KAl}{\left({\mathrm{SO}}_4\right)}_2.12\mathrm{H}2\mathrm{O}$

Ans: $\mathrm{KAl}{\left({\mathrm{SO}}_4\right)}_2.12{\mathrm{H}}_2\mathrm{O}$

Then, there's

$1(+1)+1(+3)+2(x)+8(-2)+24(+1)+12(-2)=0$

$1+3+2x-16+24-24=0$

$2x=12$

$x=+6$

As a result, S's oxidation number is $+6$

Because water is a neutral molecule, we can disregard it. The sum of all atoms in the water molecule's oxidation numbers can then be considered as zero. As a result of disregarding the water molecule, we now have

2. What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your results?

(a) ${\mathrm{KI}}_3$

Ans: In $K{I}_3$ K has an oxidation number (O.N.) of one. As a result, I's average oxidation number is ${\displaystyle \frac{1}{3}}.0.N$, on the other hand, cannot be fractional. To determine the oxidation states, we must first study the structure of ${\mathrm{KI}}_3$.

An iodine atom makes a coordinate covalent link with an iodine molecule in a ${\mathrm{KI}}_3$ molecule.

$\stackrel{+1}{{\text{K}}^{+}}[\stackrel{0}{\text{I}}-\stackrel{0}{\text{I}}\leftarrow \stackrel{-1}{\text{I}}]$

As a result, the O.N. of the two atoms that make up the $I_2$ molecule in a ${\mathrm{KI}}_3$ molecule is 0, whereas the $0.N$. of the I atom that makes up the coordinate bond is $-1.$

(b) ${\mathrm{H}}_2{\mathrm{S}}_4{\mathrm{O}}_6$

Ans: $\text{ Now, }2(+1)+4(x)+6(+2)=0$

$\Rightarrow 2+4x-12=0$

$\Rightarrow 4x=10$

$\Rightarrow x=+2\frac{1}{2}$

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

The O.N. of two of the four $\mathrm{S}$ atoms is $+5$ and the O.N. of the other two $S$ atoms is $0.$

(c) ${\mathrm{Fe}}_3{\mathrm{O}}_4$

Ans: When the $O.N$. of $O$ is set to $-2$, the O.N. of $\mathrm{Fe}$ is found to be $+2{\displaystyle \frac{2}{3}}$. O.N., on the other hand, cannot be fractional.

One of the three Fe atoms in this example has an O.N. of $+2$, whereas the other two $\mathrm{Fe}$ atoms have an O.N. of $+3$.

${\mathrm{FeO}}^{+2}{\mathrm{Fe}}_2{\mathrm{O}}_3$

(d) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$

Ans: $x+1-2$

$C_2{H}_{6}O$

$2(x)+6(+1)+1(-2)=0$

$2x+6-2=0$

$x=-2$

This molecule's two carbon atoms are found in two separate settings. As a result, their oxidation numbers cannot be the same.

As a result, C has the oxidation states of -3 and -1.

(e) ${\mathrm{CH}}_3\mathrm{COOH}$

Ans:

$x+1-2$

$C_2{H}_4{O}_2$

$2(x)+4(+1)+2(-2)=0$

$2x+4-4=0$

$x=0$

The average O.N. of $C$, on the other hand, is 0 . This molecule's two carbon atoms are found in two separate settings. As a result, their oxidation numbers cannot be the same.

In ${\mathrm{CH}}_3\mathrm{COOH},\mathrm{C}$ has the oxidation states of $+3$ and $-3$.

3. Justify that the following reactions are redox reactions:

(a) $\mathrm{CuO}(\mathrm{s})+{\mathrm{H}}_2(\mathrm{g})\to \mathrm{Cu}(\mathrm{s})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$

Ans: Let's write the oxidation number of each element in the reaction as follows:

$\mathrm{CuO}(\mathrm{s})+{\mathrm{H}}_2(\mathrm{g})\to \mathrm{Cu}(\mathrm{s})+1-2$

Cu 's oxidation number falls from $+2$ in $\mathrm{CuO}$ to 0 in $\mathrm{Cu}$, implying that $\mathrm{CuO}$ is reduced to $\mathrm{Cu}.$ In addition, the oxidation number of $\mathrm{H}$ in ${\mathrm{H}}_2$ increases from 0 to $+1$ in ${\mathrm{H}}_2\mathrm{O}$, indicating that ${\mathrm{H}}_2$ is oxidized to ${\mathrm{H}}_2\mathrm{O}$. As a result, this is a redox reaction.

(b) ${\mathrm{Fe}}_2{\mathrm{O}}_3(\mathrm{s})+3\mathrm{CO}(\mathrm{g})\to 2\mathrm{Fe}(\mathrm{s})+3{\mathrm{CO}}_2(\mathrm{g})$

Ans: Let's write the oxidation number of each element in the reaction as follows:

${\mathrm{Fe}}_2{\mathrm{O}}_3(\mathrm{s})+3\mathrm{CO}(\mathrm{g})\to 2\mathrm{Fe}(\mathrm{s})+3{\mathrm{CO}}_2(\mathrm{g})$

Fe 's oxidation number falls from $+3$ in ${\mathrm{Fe}}_2{\mathrm{O}}_3$ to 0 in Fe, implying that ${\mathrm{Fe}}_2{\mathrm{O}}_3$ is reduced to Fe. The oxidation number of $\mathrm{C}$, on the other hand, increases from $+2$ in $\mathrm{CO}$ to $+4$ in ${\mathrm{CO}}_2$, indicating that $\mathrm{CO}$ is oxidized to ${\mathrm{CO}}_2$. As a result, the reaction in question is a redox reaction.

(c) $4{\mathrm{BCl}}_3(\mathrm{g})+3{\mathrm{LiAlH}}_4(\mathrm{s})\to 2{\mathrm{B}}_2{\mathrm{H}}_6(\mathrm{g})+3\mathrm{LiCl}(\mathrm{s})+3{\mathrm{AlCl}}_3(\mathrm{s})$

Ans: Let's write the oxidation number of each element in the reaction as follows:

$4{\mathrm{BCl}}_3(\mathrm{g})+3{\mathrm{LiAlH}}_4(\mathrm{s})\to 2{\mathrm{B}}_2{\mathrm{H}}_6(\mathrm{g})+3\mathrm{LiCl}(\mathrm{s})+3{\mathrm{AlCl}}_3(\mathrm{s})$

The oxidation number of $\mathrm{B}$ drops from $+3$ in ${\mathrm{BCl}}_3$ to $-3$ in ${\mathrm{B}}_2{\mathrm{H}}_6$ in this reaction. ${\mathrm{BCl}}_3$ is reduced to ${\mathrm{B}}_2{\mathrm{H}}_6$ in this way. In addition, the oxidation number of $\mathrm{H}$ in ${\mathrm{LiAlH}}_4$ increases to $-1$ in ${\mathrm{B}}_2{\mathrm{H}}_6$, indicating that LiAlH 4 is oxidized to ${\mathrm{B}}_2{\mathrm{H}}_6.$ As a result, the reaction in question is a redox reaction.

(d) $2\mathrm{K}(\mathrm{s})+{\mathrm{F}}_2(\mathrm{g})\to 2\mathrm{K}+\mathrm{F}-(\mathrm{s})$

Ans: Let's write the oxidation number of each element in the reaction as follows:

$2\mathrm{K}(\mathrm{s})+{\mathrm{F}}_2(\mathrm{g})\to 2\mathrm{K}+\mathrm{F}-(\mathrm{s})$

The oxidation number of $K$ increases from 0 in to $+1$ in $KF$ in this reaction, indicating that $\mathrm{K}$ is oxidized to $\mathrm{KF}$. The oxidation number of $\mathrm{F}$, on the other hand, decreases from 0 in ${\mathrm{F}}_2$ to $-1$ in $\mathrm{KF}$, indicating that ${\mathrm{F}}_2$ is reduced to $\mathrm{KF}$.

As a result, the preceding reaction is a redox reaction.

(e) $4{\mathrm{NH}}_3(\mathrm{g})+5{\mathrm{O}}_2(\mathrm{g})\to 4\mathrm{NO}(\mathrm{g})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$

Ans:

Let's write the oxidation number of each element in the reaction as follows:

$-3+10$

$4{\mathrm{NH}}_3(\mathrm{g})+5{\mathrm{O}}_2(\mathrm{g})\to 4$ $\mathrm{NO}(\mathrm{g})+6{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$

The oxidation number of $\mathrm{N}$ rises from $-3$ in ${\mathrm{NH}}_3$ to $+2$ in $\mathrm{NO}$ in this case. The oxidation number of ${\mathrm{O}}_2$ drops from 0 in ${\mathrm{O}}_2$ to $-2$ in $\mathrm{NO}$ and ${\mathrm{H}}_2\mathrm{O}$, indicating that ${\mathrm{O}}_2$ is reduced.

As a result, the reaction in question is a redox reaction.

4. Fluorine reacts with ice and results in the change:

${\mathrm{H}}_2\mathrm{O}(\mathrm{s})+{\mathrm{F}}_2(\mathrm{g})\to \mathrm{HF}(\mathrm{g})+\mathrm{HOF}(\mathrm{g})$

Justify that this reaction is a redox reaction.

Ans:

Let's write the oxidation number of each atom in the reaction above its symbol as follows:

$+1-20+1\cdot 1+1\cdot 2+1$

${\mathrm{H}}_2\mathrm{O}(\mathrm{s})+{\mathrm{F}}_2(\mathrm{g})\to$$\mathrm{HF}(\mathrm{g})+\mathrm{HOF}(\mathrm{g})$

Here, we have observed that the oxidation number of $\mathrm{F}$ increases from 0 in ${\mathrm{F}}_2$ to $+1$ in HQF . Also, the oxidation number decreases from 0 in ${\mathrm{F}}_2$ to $-1$ in $\mathrm{HF}$. Thus, in the above reaction, $F$ is both oxidized and reduced. Hence, the given reaction is a redox reaction.

5. Calculate the oxidation number of Sulphur, chromium, and nitrogen in ${\mathrm{H}}_2{\mathrm{SO}}_5$, ${\mathrm{Cr}}_2{\mathrm{O}}_7{}^{2-}$ and ${\mathrm{NO}}_3^{-}$. Suggest structure of these compounds. Count for the fallacy.

Ans: 

$\begin{array}{cc}+1& \times -2\end{array}$

(i) ${\mathrm{H}}_2{\mathrm{SO}}_5$

$2(+1)+1(x)+5(-2)=0$

$2+x-10=0$

$x=+8$

S's O.N., on the other hand, cannot be $+8.$ S has six electrons in its valence shell. As a result, S's O.N. cannot be greater than $+6$.

The structure of ${\mathrm{H}}_2{\mathrm{SO}}_5$ is depicted in the diagram below.

$2(\mathrm{H})+1(\mathrm{S})+\beta (0)+2(0$ in peroxy linkage $2(+1)+1(x)+3(-2)+2(-1)=0$

$2+x-6-2=0$

$x=+6$

As a result, S's O.N. is $+6$.

(ii)

$x-2$

${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$

$2(x)+7(-2)=-2$

$2x-14=-2$

$x=+6$

The O.N. of $\mathrm{Cr}$ in ${\mathrm{Cr}}_2{\mathrm{O}}_7{}^{2-}$ is not a fallacy in this case.

The structure of ${\mathrm{Cr}}_2{\mathrm{O}}_7{}^{2-}$ is depicted in the diagram below

Each of the two $\mathrm{Cr}$ atoms here have an O.N. of $+6$.

(iii) ${\mathrm{NO}}_3$

$1(x)+3(-2)=-1$

$x-6=-1$

$x=+5$

The O.N. of $\mathrm{N}$ in ${\mathrm{NO}}_3^{\circ }$ is not a fallacy in this case.

The structure of ${\mathrm{NO}}_3^{\circ }$ is depicted in the diagram below

The O.N. value of the N atom is +5.

6. Write the formula for the following compounds:

(a) Mercury (II) chloride

Ans: ${\mathrm{H}}_{\mathrm{g}}{\mathrm{Cl}}_2$

(b) Nickel (II) sulphate

Ans: ${\mathrm{NiSO}}_4$

(c) Tin (IV) oxide

Ans: ${\mathrm{SnO}}_2$

(d) Thallium(I) sulphate

Ans: ${\mathrm{Tl}}_2{\mathrm{SO}}_4$

(e) Iron (III) sulphate

Ans: ${\mathrm{Fe}}_2{\left({\mathrm{SO}}_4\right)}_3$

(f) Chromium (III) oxide

Ans: ${\mathrm{Cr}}_2{\mathrm{O}}_3$

7. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.

Ans:

The substances where nitrogen can exhibit oxidation states from –3 to +5 are listed in the following table.

The oxidation number (O.N.) of $\mathrm{S}$ in sulphur dioxide $\left({\mathrm{SO}}_2\right)$ is $+4$, while the O.N. of $\mathrm{S}$ can range from $+6$ to $-2$.

As a result, ${\mathrm{SO}}_2$ can function as both an oxidising and a reducing agent.

The O.N. of $\mathrm{O}$ in hydrogen peroxide $\left({\mathrm{H}}_2{\mathrm{O}}_2\right)$ is $-1$, and the range of $\mathrm{O}.\mathrm{N}$. that $\mathrm{O}$ can have is 0 to $-2$. The oxidation values $+1$ and $+2$ are also possible for $\mathrm{O}$.

As a result, ${\mathrm{H}}_2{\mathrm{O}}_2$ can function as both an oxidizing and a reducing agent.

As a result, in this scenario, the O.N. of $\mathrm{O}$ can only drop. As a result, ${\mathrm{O}}_3$ serves solely as an oxidant.

The O.N. of $\mathrm{N}$ in nitric acid $\left({\mathrm{HNO}}_3\right)$ is $+5$, and the range of $\mathrm{O}.\mathrm{N}$ that $\mathrm{N}$ can have from $+5$ to $-3$. As a result, in this scenario, the O.N. of $\mathrm{N}$ can only drop. As a result, HNO, serves solely as an oxidant.

9. Consider the reactions:

(a) $6{\mathrm{CO}}_2(\mathrm{g})+6\mathrm{H}2\mathrm{O}(\mathrm{l})\to {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6(\mathrm{aq})+6{\mathrm{O}}_2(\mathrm{g})$

(b) ${\mathrm{O}}_3(\mathrm{g})+{\mathrm{H}}_2{\mathrm{O}}_2(1)\to {\mathrm{H}}_2\mathrm{O}(1)+2{\mathrm{O}}_2(\mathrm{g})$

Why it is more appropriate to write these reactions as:

(a) $6{\mathrm{CO}}_2(\mathrm{g})+12{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6(\mathrm{aq})+6{\mathrm{H}}_2\mathrm{O}(1)+6\mathrm{O}2(\mathrm{g})$

Ans: The ${\mathrm{H}}_2$ produced in step 1 reduces ${\mathrm{CO}}_2$, thereby producing glucose $\left({\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6\right)$ and ${\mathrm{H}}_2\mathrm{O}$.

$6{\mathrm{CO}}_{2(\mathrm{g})}+12{\mathrm{H}}_{2(\mathrm{g})}\to {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_{6(s)}+6{\mathrm{H}}_2{\mathrm{O}}_{(i)}$

Now, the net reaction of the process is given as: $2{\mathrm{H}}_2{\mathrm{O}}_{(t)}\to 2{\mathrm{H}}_{2(\mathrm{g})}+{\mathrm{O}}_{2(\mathrm{g})}]\times 6$

${\displaystyle \frac{6C{O}_{2(g)}+12H_{2(g)}\to C_6{H}_{12}{O}_{6(s)}+6H_2{O}_{(l)}}{6C{O}_{2(g)}+12H_2{O}_{(l)}\to C_6{H}_{12}{O}_{6(g)}+6H_2{O}_{(l)}+6O_{2(g)}}}$

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis. The path of this reaction can be investigated by using radioactive ${\mathrm{H}}_2{\mathrm{O}}_{18}$ in place of ${\mathrm{H}}_2\mathrm{O}$

(b) ${\mathrm{O}}_3(\mathrm{g})+{\mathrm{H}}_2{\mathrm{O}}_2(\mathrm{l})\to {\mathrm{H}}_2\mathrm{O}(1)+{\mathrm{O}}_2(\mathrm{g})$

Ans: ${\mathrm{O}}_2$ is produced from each of the two reactants ${\mathrm{O}}_3$ and ${\mathrm{H}}_2{\mathrm{O}}_2.$ For this reason, ${\mathrm{O}}_2$ is written twice.

The given reaction involves two steps. First, ${\mathrm{O}}_3$ decomposes to form $\mathrm{O}2$ and $\mathrm{O}$. In the second step, ${\mathrm{H}}_2{\mathrm{O}}_2$ reacts with the O produced in the first step, thereby producing ${\mathrm{H}}_2\mathrm{O}$ and ${\mathrm{O}}_2$

$O_{3(g)}\to O_{2(g)}+O_{(g)}$

$\frac{{\mathrm{H}}_2{\mathrm{O}}_{2(t)}+{\mathrm{O}}_{(g)}\to {\mathrm{H}}_2{\mathrm{O}}_{(t)}+{\mathrm{O}}_{2(g)}}{2{\mathrm{H}}_2{\mathrm{O}}_{2(i)}+{\mathrm{O}}_{3(g)}\to {\mathrm{H}}_2{\mathrm{O}}_{(t)}+{\mathrm{O}}_{2(g)}+{\mathrm{O}}_{2(\mathrm{g})}}$

The path of this reaction can be investigated by using, ${\mathrm{H}}_2{\mathrm{O}}_2^{18}$ or ${\mathrm{O}}_3^{18}$

10. The compound ${\mathrm{AgF}}_2$ is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?

Ans:

$\mathrm{Ag}$ in ${\mathrm{AgF}}_2$ has an oxidation state of $+2$. However, $\mathrm{Ag}$ 's oxidation state of $+2$ is unstable.

As a result, silver quickly takes an electron to create $\mathrm{Ag}+$ whenever ${\mathrm{AgF}}_2$ is formed. This helps to reduce $\mathrm{Ag}$ 's oxidation state from $+2$ to $+1$, which is a more stable condition. As a result, ${\mathrm{AgF}}_2$ is an extremely powerful oxidizing agent.

11. Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidizing agent is in excess. Justify this statement giving three illustrations.

Ans:

When an oxidizing agent and a reducing agent react, a lower oxidation state compound is formed if the reducing agent is in excess, and a higher oxidation state compound is formed if the oxidizing agent is in excess. As an example, consider the following:

(i) Reducing and oxidising agents, respectively, are $P_4$ and $F_2$.

When an excess of ${\mathrm{P}}_4$ is treated with ${\mathrm{F}}_2,{\mathrm{PF}}_3$ is formed, with a positive oxidation number (O.N.) for $\mathrm{P}$.

However, if $P_4$ is treated with an excessive amount of $F_2,P{F}_5$ is formed, with a P.N. of $+5.$

(ii) ${\mathrm{O}}_2$ is an oxidising agent, whereas $\mathrm{K}$ is a reducing agent.

${\mathrm{K}}_2\mathrm{O}$ is generated when an excess of $\mathrm{K}$ reacts with ${\mathrm{O}}_2$, with the O.N. of $\mathrm{O}$ being $-2.$

$4\mathrm{K}(\text{ excess })+{\mathrm{O}}_2\to 2{\mathrm{K}}_2\mathrm{O}$

When $K$ reacts with an excess of ${\mathrm{O}}_2$, however, $2{\mathrm{K}}_2{\mathrm{O}}_2$ is produced, with the O.N. of $\mathrm{O}$ being -

$4\mathrm{K}+{\mathrm{O}}_2(\text{ excess })\to 2{\mathrm{K}}_2{\mathrm{O}}_2$

While $C$ is a reducing agent, ${\mathrm{O}}_2$ is an oxidizing agent.

$\mathrm{CO}$ is created when an excess of $\mathrm{C}$ is burned in the presence of inadequate ${\mathrm{O}}_2$, with the O.N. of $\mathrm{C}$ being $+2$

$\mathrm{C}(\text{ excess })+{\mathrm{O}}_2\to \mathrm{CO}$

If there is an excess of ${\mathrm{O}}_2$ in the combustion of $\mathrm{C},{\mathrm{CO}}_2$ is generated, with the O.N. of C being $+4$

$\mathrm{C}+{\mathrm{O}}_2(\text{ excess })\to {\mathrm{CO}}_2$

12. How do you count for the following observations? 

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colorless pungent smelling gas $\mathrm{HCl}$, but if the mixture contains bromide then we get red vapour of bromine. Why?

Ans:  (a) Alcoholic potassium permanganate is utilized as an oxidant in the production of benzoic acid from toluene for the following reasons.

(i) In a neutral medium, OH- ions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced.

(ii) Because both ${\mathrm{KMnO}}_4$ and alcohol are polar, they are homogenous. Because they a both organic molecules, toluene and alcohol are also homogenous.

In a homogeneous medium, reactions can proceed more quickly than in a heterogeneous one. As a result, ${\mathrm{KMnO}}_4$ and toluene might react more quickly in alcohol.

For the reaction in a neutral medium, the balanced redox equation is as follows:

${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_3(l)+2{\mathrm{MnO}}_4^{-}(\text{alcoholic })\to {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}\text{(alcoholic) }$

$+2{\mathrm{MnO}}_2(s)+{\mathrm{H}}_2\mathrm{O}(l)+{\mathrm{OH}}^{-}(aq)$

When concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ is introduced to an inorganic bromide mixture, $\mathrm{HBr}$ is generated at first. With the formation of red bromine vapour, $\mathrm{HBr}$, as a powerful reducing agent, lowers ${\mathrm{H}}_2{\mathrm{SO}}_4$ to ${\mathrm{SO}}_2$.

$\begin{array}{rl}2\mathrm{NaBr}+2{\mathrm{H}}_2{\mathrm{SO}}_4& \to 2{\mathrm{NaHSO}}_4+2\mathrm{HBr}\\ 2\mathrm{HBr}+{\mathrm{H}}_2{\mathrm{SO}}_4\to & {\mathrm{Br}}_2+{\mathrm{SO}}_2+2{\mathrm{H}}_2\mathrm{O}\\ & \text{ (red vapour) }\end{array}$

When concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ is added to an inorganic combination with chloride, a pungent-smelling gas ( ${\mathrm{HCl}}^{\prime }$ ) is produced. Because $\mathrm{HCl}$ is a poor reducing agent, it cannot convert ${\mathrm{H}}_2{\mathrm{SO}}_4$ to ${\mathrm{SO}}_2$

$2\mathrm{NaCl}+2{\mathrm{H}}_2{\mathrm{SO}}_4\to 2{\mathrm{NaHSO}}_4+2\mathrm{HCl}$

13. Identify the substance oxidized, reduced, oxidizing agent, and reducing agent for each of the following reactions:

(a) $2\mathrm{AgBr}(\mathrm{s})+{\mathrm{C}}_6{\mathrm{H}}_6{\mathrm{O}}_2\text{ (aq) }\to 2\mathrm{Ag}(\mathrm{s})+2\mathrm{HBr}(\mathrm{aq})+{\mathrm{C}}_6{\mathrm{H}}_4{\mathrm{O}}_2(\mathrm{aq})$

Ans: 

$\text{ Oxidized substance }\to {\mathrm{C}}_6{\mathrm{H}}_6{\mathrm{O}}_2$

Reduced substance $\to \mathrm{AgBr}$

Oxidizing agent $\to \mathrm{AgBr}$

Reducing agent $\to {\mathrm{C}}_6{\mathrm{H}}_6{\mathrm{O}}_2$

(b) $\mathrm{HCHO}(1)+2{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}(\mathrm{aq})+3{\mathrm{OH}}^{\circ }(\mathrm{aq})\to 2\mathrm{Ag}(\mathrm{s})+{\mathrm{HCOO}}^{-}(\mathrm{aq})+4{\mathrm{NH}}_3(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{I})$

Ans: Oxidized substance $\to$ HCHO 

Reduced substance $\to {\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}$ Oxidising agent $\to {\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}$ Reducing agent $\to \mathrm{HCH}$ 

(c) $\mathrm{HCHO}(\mathrm{I})+2{\mathrm{Cu}}^{2+}(\mathrm{aq})+5{\mathrm{OH}}^{\circ }(\mathrm{aq})\to {\mathrm{Cu}}_2\mathrm{O}(\mathrm{s})+{\mathrm{HCOO}}^{\circ }$ (aq) $+3{\mathrm{H}}_2\mathrm{O}(\mathrm{I})$

Ans: Oxidised substance $\to$ HCHO

Reduced substance $\to {\mathrm{Cu}}^{2+}$

Oxidising agent $\to {\mathrm{Cu}}^{2+}$

Reducing agent $\to$ HGHO

(d) ${\mathrm{N}}_2{\mathrm{H}}_4(\mathrm{I})+2{\mathrm{H}}_2{\mathrm{O}}_2$ (I) $\to {\mathrm{N}}_2(\mathrm{g})+4{\mathrm{H}}_2\mathrm{O}(\mathrm{I})$

Ans: Oxidised substance $\to {\mathrm{N}}_2{\mathrm{H}}_4$

Reduced substance $\to {\mathrm{H}}_2{\mathrm{O}}_2$

Oxidising agent $\to {\mathrm{H}}_2{\mathrm{O}}_2$

Reducing agent $\to {\mathrm{N}}_2{\mathrm{H}}_4$

(e) $\mathrm{Pb}(\mathrm{s})+\mathrm{Pb}{\mathrm{O}}_2(\mathrm{s})+2{\mathrm{H}}_2{\mathrm{SO}}_4(\mathrm{aq})\to 2{\mathrm{PbSO}}_4(\mathrm{s})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{I})$

Ans: Oxidised substance $\to$ Reduced substance $\to {\mathrm{PbO}}_2$

Oxidising agent $\to \mathrm{PbO}$

14. Consider the reactions:

$2{\mathrm{S}}_2{\mathrm{O}}_3^{2\cdot }(\mathrm{aq})+{\mathrm{I}}_2(\mathrm{s})\to {\mathrm{S}}_4{\mathrm{O}}_6{}^{2-}(\mathrm{aq})-2{\mathrm{I}}^{-}(\mathrm{aq})$

${\mathrm{S}}_2{\mathrm{O}}_3{}^{2-}(\mathrm{aq})+2{\mathrm{Br}}_2(1)+5$ ${\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to 2$ ${\mathrm{SO}}_4^{2-}(\mathrm{aq})+4{\mathrm{Br}}^{-}(\mathrm{aq})+10$ ${\mathrm{H}}^{+}(\mathrm{aq})$

Why does the same reductant, thiosulphate react differently with iodine and bromine?

Ans: $2{\mathrm{S}}_2{\mathrm{O}}_3^{2-}(aq)+{\mathrm{I}}_2(s)⟶{\mathrm{S}}_4{\mathrm{O}}_6^{2-}(aq)+2{\mathrm{I}}^{-}(aq)$

$\stackrel{+2-2}{{\mathrm{S}}_2{\mathrm{O}}_3^{2-}(aq)+2{\mathrm{Br}}_2(l)+.5{\mathrm{H}}_2\mathrm{O}(l)⟶}$

$2{\mathrm{SO}}_4^{+6-2}(aq)+4{\mathrm{Br}}^{-}(aq)+10{\mathrm{H}}^{+}(aq)$

When compared to ${\mathrm{I}}_2$, bromine is a more powerful oxidizer. In ${\mathrm{SO}}_4{}^{2-}$, it oxidises the $\mathrm{S}$ of ${\mathrm{S}}_2{\mathrm{O}}_3^2$ to a higher oxidation state $+6$.

In ${\mathrm{S}}_4{\mathrm{O}}_6{}^{2-},{\mathrm{I}}_2$ oxidises S from ${\mathrm{S}}_2{\mathrm{O}}_3{}^2$ to a lower oxidatión state of $2.5.$ As a result, the same reductant, thiosulphate, reacts with bromine and iodine in distinct ways.

15. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Ans:  ${\mathrm{F}}_2$ can also oxidize $\mathrm{Cl}$ to ${\mathrm{Cl}}_2,{\mathrm{Br}}^{-}$to ${\mathrm{Br}}_2$ and ${\mathrm{I}}^{-}$to ${\mathrm{I}}_2$

${\mathrm{F}}_{2(\mathrm{aq})}+2{\mathrm{Cl}}_{(s)}\to 2{\mathrm{F}}_{(\mathrm{aq})}+{\mathrm{Cl}}_{(\mathrm{g})}$

${\mathrm{P}}_{2(\mathrm{aq})}+2{\mathrm{I}}_{(\mathrm{aq})}\to 2{\mathrm{F}}_{(\mathrm{aq})}+{\mathrm{I}}_{2(\mathrm{s})}$

${\mathrm{Cl}}_2,\mathrm{B}{\mathrm{r}}_2$, and ${\mathrm{I}}_2$, on the other hand, are unable to convert ${\mathrm{F}}^{-}$to ${\mathrm{F}}_2$. Halogens have an oxidizing power of ${\mathrm{I}}_2<{\mathrm{Br}}_2<{\mathrm{Cl}}_2<{\mathrm{F}}_2$. Fluorine, as a result, is the best halogen oxidant.

${\mathrm{H}}_2{\mathrm{SO}}_4$ can be converted to ${\mathrm{SO}}_2$ using $\mathrm{HI}$ and $\mathrm{HBr}$, but not with $\mathrm{HCl}$ or $\mathrm{HF}$. HI and $\mathrm{HBr}$ are thus more effective reductants than $\mathrm{HCl}$ and $\mathrm{HF}$

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

$2\mathrm{HBr}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{Br}}_2+{\mathrm{SO}}_2+2{\mathrm{H}}_2\mathrm{O}$

I can reduce ${\mathrm{Cu}}^{2+}$ to ${\mathrm{Cu}}^{+}$once more, whereas ${\mathrm{Br}}^{-}$cannot.

$4{\mathrm{I}}_{(\mathrm{aq})}+2{\mathrm{Cu}}^{2+}(aq)\to {\mathrm{Cu}}_2{\mathrm{I}}_{2(\mathrm{f})}+{\mathrm{I}}_{2(\mathrm{aq})}$

As a result, among hydrohalic compounds, hydroiodic acid is the best reductant.

Hydrohalic acids' reducing power thus grows in the order of $\mathrm{HF}<\mathrm{HCl}<\mathrm{HBr}<\mathrm{HI}$

16. Why does the following reaction occur?

${\mathrm{XeO}}_6^{4-}(\mathrm{aq})+2{\mathrm{F}}^{-}(\mathrm{aq})\to {\mathrm{XeO}}_3(\mathrm{g})+{\mathrm{F}}_2(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

What conclusion about the compound ${\mathrm{NaXeO}}_6$ (of which ${\mathrm{XeO}}_6^4$ is a part) can be drawn from the reaction?

Ans: Because ${\mathrm{XeO}}_6^{4-}$ oxidizes ${\mathrm{F}}^{-}$and ${\mathrm{F}}^{-}$decreases ${\mathrm{XeO}}_6{}^{4-}$, the stated reaction occurs.

${\mathrm{XeO}}_6^{4-}(\mathrm{aq})+2{\mathrm{F}}^{-}(\mathrm{aq})\to {\mathrm{XeO}}_3(\mathrm{g})+{\mathrm{F}}_2(\mathrm{g})+3{\mathrm{H}}_2\mathrm{O}(\mathrm{l})$

Xe 's oxidation number (O.N.) falls from $+8$ in ${\mathrm{XeO}}_6^4$ to $+6$ in ${\mathrm{XeO}}_3$, while $\mathrm{F}$ 's $\mathrm{O}.\mathrm{N}$. rises from $-1$ in ${\mathrm{F}}^{-}$to $\mathrm{O}$ in ${\mathrm{F}}_2$

As a result, we can deduce that ${\mathrm{NaXeO}}_6{}^4$ is a more powerful oxidizer than $\mathrm{F}$

17. Consider the reactions:

(a) ${\mathrm{H}}_3{\mathrm{PO}}_2(\mathrm{aq})+4{\mathrm{AgNO}}_3(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to {\mathrm{H}}_3{\mathrm{PO}}_4(\mathrm{aq})+4\mathrm{Ag}(\mathrm{s})+4\mathrm{H}{\mathrm{NO}}_3(\mathrm{aq})$

(b) ${\mathrm{H}}_3{\mathrm{PO}}_2(\mathrm{aq})+2{\mathrm{CuSO}}_4(\mathrm{aq})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{l})\to {\mathrm{H}}_3{\mathrm{PO}}_4(\mathrm{aq})+2\mathrm{Cu}(\mathrm{s})+{\mathrm{H}}_2{\mathrm{SO}}_4(\mathrm{aq})$

(c) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}(\mathrm{l})+2{\left[\mathrm{Ag}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}(\mathrm{aq})+3{\mathrm{OH}}^{-}(\mathrm{aq})\to {\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}(\mathrm{aq})+2\mathrm{Ag}(\mathrm{s})+4{\mathrm{NH}}_3(\mathrm{aq})$

(d) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}(\mathrm{l})+2{\mathrm{Cu}}^{2+}(\mathrm{aq})+5{\mathrm{OH}}^{-}(\mathrm{aq})\to$ No change observed.

What inference do you draw about the behaviour of ${\mathrm{Ag}}^{+}$ and ${\mathrm{Cu}}^{2+}$ from these reactions?

Ans: In reactions (a) and (b), ${\mathrm{Ag}}^{+}$and ${\mathrm{Cu}}^{2+}$, respectively, act as oxidizing agents.

Ag+ oxidizes ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$ to ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{COO}}^{-}$in reaction $(\mathrm{c})$, but ${\mathrm{Cu}}^{2+}$ cannot oxidize ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CHO}$ in reaction $(\mathrm{d})$

As a result, $\mathrm{Ag}+$ is a more powerful oxidizing agent than ${\mathrm{Cu}}^{2+}$

18. Balance the following redox reactions by ion-electron method:

(a) ${\mathrm{MnO}}^{-}(\mathrm{aq})+{\mathrm{I}}^{-}(\mathrm{aq})\to {\mathrm{MnO}}_2(\mathrm{s})+{\mathrm{I}}_2(\mathrm{s})$

Ans: Step 1 : The following are the two half reactions involved in the given reaction:

Half-reaction of oxidation ${\mathrm{I}}_{(\mathrm{a}\varphi })\to {\mathrm{I}}_{2(\mathrm{s})}$

Half-reaction of reduction ${\mathrm{MnO}}_4$ (aq) $\to {\mathrm{MnO}}_{2(\mathrm{aq})}$

Step 2: We have the following equation for balancing $\mid$ in the oxidation half reaction:

$2{\mathrm{F}}_{(\mathrm{aq})}\to {\mathrm{I}}_{2(\mathrm{s})}$

To balance the charge, we add $2{\mathrm{e}}^{-}$to the reaction's RHS.

$2{\mathrm{I}}_{(\mathrm{aq})}^{-}\to {\mathrm{I}}_{2(\mathrm{s})}+2{\mathrm{e}}^{-}$

Step 3: Mn 's oxidation state has decreased from+7 to+4 throughout the reduction half reaction.

${\mathrm{MnO}}_4^{-}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_{2(\mathrm{aq})}$

Step 4: Six O atoms are on the RHS and four $\mathrm{O}$ atoms are on the LHS in this equation. As a result, the LHS is given two water molecules.

${\mathrm{MnO}}_{4(\mathrm{aq})}^{-}+2{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\to {\mathrm{MnO}}_{2(\mathrm{aq})}+4{\mathrm{OH}}^{-}$

Step 5: By multiplying the oxidation half reaction by 3 and the reduction half reaction by 2 , we may equalize the quantity of electrons.

$6{\mathrm{I}}_{(\mathrm{aq})}^{-}\to 3{\mathrm{I}}_{2(\mathrm{s})}+6{\mathrm{e}}^{-}$

$2{\mathrm{MnO}}_4{}^{-}(\mathrm{aq})+4{\mathrm{H}}_2\mathrm{O}+6{\mathrm{e}}^{-}\to 2{\mathrm{MnO}}_{2(\mathrm{aq})}+8{\mathrm{OH}}_{(aq)}^{-}$