NCERT Solutions for Class 11 Chemistry Chapter 8 Redox Reactions

Question 7.1 Assign oxidation number to the underlined elements in each of the following species

$(a)Na{H}_2\overline{P}{O}_4$

$(b)NaH\overline{S}{O}_4$

$(c)H_4\overline{P_2}{O}_7$

$(d)K_2\overline{Mn}{O}_4$

$(e)Ca\overline{O_2}$

$(f)Na\overline{B}{H}_4$

$(g)H_2\overline{S_2}{O}_7$

$(h)KAl(\overline{S}{O}_4{)}_2.12H_{2}O$

Answer : O.N is the oxidation number

O.N of Oxygen( $O$ ) = -2 ( In the case of peroxide and superoxide it wil be different ON)

O.N of hydrogen( $H$ )= +1 (In case of metalic hydride, -1)

O.N of sodium ( $Na$ ) = +1

O.N of aluminium ( $Al$ ) = +3

O.N of potassium ( $K$ )= +1

O.N of calcium ( $Ca$ ) = +2

In neutral compounds the sum of O.N of all the atoms is zero.

(a) Let the O.N of P be x

$\therefore 1\ast 1+2\ast 1+x+4\ast (-2)=0\Rightarrow x=+5$

(b) Let the O.N of S be x

$\therefore 1\ast 1+1\ast 1+x+4\ast (-2)=0\Rightarrow x=+6$

(c) Let the O.N of P be x

$\therefore 4\ast 1+2\ast x+7\ast (-2)=0\Rightarrow x=+5$

(d) Let the O.N of Mn be x

$\therefore 2\ast 1+x+4\ast (-2)=0\Rightarrow x=+6$

(e) Let the O.N of O be x

Ca is an alkaline earth metal so its O.N. is +2

$\therefore 1\ast 2+2\ast x=0\Rightarrow x=-1$

(f) Let the O.N of B be x

Note that in this H exists as hydride ion $H^{-}$ so its O.N. is -1

$\therefore 1\ast 1+x+4\ast (-1)=0\Rightarrow x=+3$

(g) Let the O.N of S be x

$\therefore 2\ast 1+2\ast x+7\ast (-2)=0\Rightarrow x=+6$

(h) Let the O.N of S be x

$\therefore 1\ast 1+1\ast 3+[x+(-2)\ast 4]\ast 2+12\ast [1\ast 2+(-2)]=0\Rightarrow x=+6$

Question 7.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalize your results

$(a)K{\underset{―}{I}}_3$

$(b)H_2\underset{―}{S_4}{O}_6$

$(c){\underset{―}{Fe}}_3{O}_4$

$(d)\underset{―}{C}{H}_3\underset{―}{C}{H}_{2}OH$

$(e)\underset{―}{C}{H}_3\underset{―}{C}OOH$

Answer :

Solution-

O.N of potassium ( $K$ )= +1

O.N of hydrogen( $H$ )= +1 (In case of metalic hydride, -1)

O.N of Oxygen( $O$ ) = -2 ( In the case of peroxide and superoxide it wil be different ON)

$(a)K{\underset{―}{I}}_3$

1*1 + 3*x = 0

x = (-1/3)

average O. N. Of $I$ is $-\frac{1}{3}$ . But it is wrong because O.N cannot be fractional. So lets try with structure of $K{I}_3$

$K^{+}(I-I\leftarrow I{)}^{-1}$

O.N of $I$ = -1 (because a coordinate bond is formed between $I_2$ and $I^{-}$ ion. Hence O. N of three $I$ atoms are 0,0 and -1, O.N 0 in $I_2$ moleculeand -1 in $I^{-}$ ion.)

$(b)H_2\underset{―}{S_4}{O}_6$

Assume O.N of S is x

$2\ast 1+4\ast x+6\ast (-2)=0\Rightarrow x=2.5$

Fractional O.N is not possible so try with structure -

The two S atom present in the middle has zero (0) O.N and other two have (+5) O.N [ -2 from two O atom and -1 from OH]

$(c){\underset{―}{Fe}}_3{O}_4$

If you calculate the oxidation number of Fe in $F{e}_3{O}_4$ it would be 8/3 and however, O.N cannot be in fractional.

$F{e}_3{O}_{4}isanequimolarmixtureof(FeO)and(F{e}_2{O}_3)$ Here one iron atom has +2 O.N and the other two are of +3 O.N.

$(d)\overline{C}{H}_3\overline{C}{H}_{2}OH$

let assume carbon has x oxidation Number

So,[ x + 1(3) +x +1(2) +(-2)+1 = 0]

2x = -4

x=-2

In this molecule two carbon atoms present in different environments. Hence, they cannot have the same O.N.Thus, C exhibits the O. S of -3 and -1.

$(e)\overline{C}{H}_3\overline{C}OOH$

suppose the oxidation number of Carbon is x.

If we calculate the O.N of x we get x=0

However, 0 is average O.N. of C atoms. In this molecule, two carbon atoms present in different enviroments. Hence, they cannot have the same O.N.Thus, C exhibits the O.S of +3 and –3 in CH3COOH This can be more understood by structure-

Here we can see that at right C, +3 O.N (-1 from OH and -2from O atom) and in left C, -3 O.N(contribution from H atom only)

Question 7.3(a) Justify that the following reactions are redox reaction

$(\mathrm{a})\mathrm{CuO}{\mathrm{O}}_{(\mathrm{s})}+{\mathrm{H}}_{2(\mathrm{g})}\to {\mathrm{Cu}}_{(s)}+{\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{g})}$

Answer : Let us write the O.N of each element

$\stackrel{2+-2}{CuO}+\stackrel{0}{H_2}\to \stackrel{0}{Cu}+\stackrel{+1-2}{H_{2}O}$

Here, the O.N of Cu decreases from +2 to 0 i.e., CuO is reduced to Cu. Also, the O.N of H increases from 0 to +1 i.e., H2 is oxidized to H2O. Hence it is a redox reaction.

Question 7.3(b) Justify that the following reactions are redox reaction

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

Answer : Let us write the O.N of each element

$\stackrel{+3-2}{F{e}_2{O}_3}+3\stackrel{+2-2}{CO}\to 2\stackrel{0}{Fe}+$ $3\stackrel{+4-2}{C{O}_2}$

Here, the O.N of Fe decreases from +3 to 0. Also, the O.N of C increases from +2 to +4 . Hence it is a redox reaction.

Question 7.3(c) Justify that the following reactions are redox reaction

$4{\mathrm{BCl}}_{3(\mathrm{g})}+3{\mathrm{LiAlH}}_{4(\mathrm{s})}\to 2{\mathrm{B}}_2{\mathrm{H}}_{6(\mathrm{g})}+3{\mathrm{LiCl}}_{(\mathrm{s})}+3{\mathrm{AlCl}}_{3(\mathrm{s})}$

Answer : Let us write the O.N of each element

$4\stackrel{+3-1}{BC{l}_3}+3\stackrel{+1+3-1}{LiAl{H}_4}\to 2\stackrel{-3+1}{B_2{H}_6}+3\stackrel{+1-1}{LiCl}+3\stackrel{+3-1}{AlC{l}_3}$

Here, the O.N of Fe decreases from +3 in BCl3 to –3 in B2H6. And, the O.N of H increases from –1 in $LiAl{H}_4$ to +1 in $B_2{H}_6$ . Hence it is a redox reaction.

Question 7.3(d) Justify that the following reactions are redox reaction

$2K_{(s)}+F_{2(g)}\to 2K^{+}{F}_{(s)}^{-}$

Answer : We know that oxidation = loosing of $e^{-}$ by atom

and reduction = gaining of $e^{-}$ by another atom

here $K$ lose its electron and $F$ accept it, Hence it is a redox reaction

Question 7.3(e) Justify that the following reactions are redox reaction

$4{\mathrm{NH}}_{3(g)}+5{\mathrm{O}}_{2(\mathrm{g})}\to 4{\mathrm{NO}}_{(g)}+6{\mathrm{H}}_2{\mathrm{O}}_{(g)}$

Answer : Here $N(-3)\to N(+2)$ oxidation reaction

and $O(0)\to O(-2)$ reduction reaction (oxidation state of oxygen is zero at molecular state )

hence it's a redox reaction

Question 7.4 Fluorine reacts with ice and results in the change

${\mathrm{H}}_2{\mathrm{O}}_{(s)}+{\mathrm{F}}_{2(g)}\to {\mathrm{HF}}_{(g)}+{\mathrm{HOF}}_{(g)}$

Justify that this reaction is a redox reaction

Answer : $F_2$ $HF$ $HOF$

Oxidation state of $F$ $\to$ $0$ $-1$ $+1$

Here $F$ is oxidized and reduced as well. So, it is a redox reaction.

Question 7.5 Calculate the oxidation number of sulphur, chromium, and nitrogen in $H_{2}S{O}_5$ , $C{r}_2{O}_7^{2-}$ and $N{O}_3^{-}$ . Suggest the structure of these compounds. Count for the fallacy.

Answer :

(i) $H_{2}S{O}_5$ let the oxidation number of sulphur be x

So,

$$\begin{gathered} 2\ast 1+x+5(-2)=0 \\ x=+8 \end{gathered}$$

There is a fallacy Sulphur cannot have +8 oxidation state because it has maximum +6 oxidation number, not more than that.The structure of $H_{2}S{O}_5$ is shown as follows:

$2(H)+1(S)+3(O)+2(Oinperoxylinkage)$

$\Rightarrow 2(+1)+1(x)+3(-2)+2(-1)=0$

$\Rightarrow x=+6(Answer))$

(ii) $C{r}_2{O}_7^{2-}$

let the oxidation number of chromium be x

now

$$\begin{gathered} 2x+7\ast (-2)=-2 \\ x=(-2+14)/2 \\ x=+6 \end{gathered}$$

There is no fallacy here

(iii) $N{O}_3^{-}$

let assume oxidation number of N is x

Now, $$\begin{gathered} x+(-2)\ast 3=-1 \\ (x-6)=-1 \\ x=+5 \end{gathered}$$

here is no fallacy about the O.N of N in $N{O}_3^{-}$

Question 7.6(a) Write formulas for the following compounds:

Mercury(II) chloride

Answer : $HgC{l}_2$

in this formula, we can see that mercury has $+2$ oxidation state

Question 7.6(b) Write formulas for the following compounds:

Nickel(II) sulphate

Answer : $NiS{O}_4$

sulphate has $-2$ oxidation state

Question 7.6(c) Write formulas for the following compounds:

Tin(IV) oxide

Answer : $Sn{O}_2$

Oxygen has $-2$ oxidation state

Question 7.6(d) Write formulas for the following compounds:

Thallium(I) sulphate

Answer : $T{l}_{2}S{O}_4$

Question 7.6(e) Write formulas for the following compounds:

Iron(III) sulphate

Answer : Formula of the compounds: Iron(III) sulphate is

$F{e}_2(S{O}_4{)}_3$

Question 7.6(f) Write formulas for the following compounds:

Chromium(III) oxide

Answer :

Answer- Formula of the Chromium(III) oxide compounds is

$C{r}_2{O}_3$

Question 7.7 Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5

Answer :

the substance of Carbon -

substance for Nitrogen-

Question 7.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

Answer :

• sulphur dioxide $(S{O}_2)$ here oxidation state of sulphur is +4 and the range of oxidation number of S is from -2 to +6 . It means it can accept an electron and lose as well, therefore, it can behave as oxidant and reductant both.
• In case of hydrogen peroxide $(H_2{O}_2)$ oxidation state is -1 and the oxidation state of O can vary from 0 to -2 . So it shows both oxidizing and reducing properties.
• For $HN{O}_3$ , Nitrogen has +5 oxidation state and it varies from +5 to -3. So it only accepts electrons. The oxidation state of N only decreases. Hence it acts as only oxidants.
• And in case of $O_3$ , the oxidation state of O is zero(0) and the range of oxidtion number of O is 0 to -2. It only decreases in this case also so therefore it acts as only oxidants.

Question 7.9 Consider the reactions:

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

$$(b){\mathrm{O}}_{3(g)}+{\mathrm{H}}_2{\mathrm{O}}_{2(l)}\to {\mathrm{H}}_2\mathrm{O}(l)+2{\mathrm{O}}_2(\mathrm{g})$$

Why it is more appropriate to write these reactions as :

$(\mathrm{a})6{\mathrm{CO}}_{2(g)}+12{\mathrm{H}}_2{\mathrm{O}}_{(l)}\to {\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_{6(aq)}+6{\mathrm{H}}_2{\mathrm{O}}_{(\mathrm{l})}+6{\mathrm{CO}}_{2(g)}$
(b) ${\mathrm{O}}_{3(g)}+{\mathrm{H}}_2{\mathrm{O}}_{2(l)}\to {\mathrm{H}}_2\mathrm{O}(\mathrm{l})+{\mathrm{O}}_2(\mathrm{g})+{\mathrm{O}}_{2(g)}$

Also, suggest a technique to investigate the path of the above (a) and (b) redox reactions.

Answer :

(a) In the photosynthesis process-

step 1- the liberation of $O_2$ and $H_2$ --> $2H_{2}O\to O_2+2H_2$

step-2 The $H_2$ produced in above reduces the $C{O}_2$ into glucose $(C_6{H}_{12}{O}_6)$ and water $(H_{2}O)$

$6C{O}_2+12H_2\to C_6{H}_{12}{O}_6+6H_{2}O$

So, the final net reaction is

$$\begin{gathered} 2H_{2}O\to C{O}_2+2H_2]\ast 6 \\ +6C{O}_2+12H_2\to C_6{H}_{12}{O}_6+6H_{2}O \\ ------------------ \\ 6C{O}_2+12H_{2}O\to C_6{H}_{12}{O}_6+6H_{2}O \end{gathered}$$

It is more appropriate to write the reaction as above because water $(H_{2}O)$ molecule also produced in photosynthesis reaction.

The path of reaction can be investigated by using the radioactive $H_2{O}^{18}$ instead of $(H_{2}O)$

(b)

$$\begin{gathered} {O}_3(g)\to O_2(g)+O(g) \\ {H}_2{O}_2(l)+O(g)\to H_{2}O(l)+O_2(g) \\ --------------- \\ {H}_2{O}_2(l)+O_3(g)\to H_{2}O(l)+O_2(g)+O_2(g) \end{gathered}$$ (the final net reaction)

Dioxygen is produced from both steps, one from the decomposition of ozone ( $O_3$ ) and other is from the reaction of hydrogen peroxide with(O)

• The path of the reaction can be investigated by using $O_3^{18}/H_2{O}^{18}$ .

Question 7.11 Whenever a reaction between an oxidising 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 oxidising agent is in excess. Justify this statement giving three illustrations?

Answer :

These can be understood by the following examples-

• $P_4$ is reducing agent and $C{l}_2$ is an oxidizing agent

$P_4+10C{l}_2(Excess)\to PC{l}_5$ [O.N of phosphorus +5] $\Rightarrow$ Higher O.S of P

$P_4(Excess)+6C{l}_2\to 4PC{l}_3$ [O.N of phosphorus +3] $\Rightarrow$ Lower O.S of P

• is an $O_2$ is a reducing agent and $C$ oxidising agent

$C+O_2\to C{O}_2$ (O is in excess) [O.N of C +4]

$C+O_2\to CO$ (C is in excess) [O.N of C +2]

• $K$ is a reducing agent and $O_2$ is an oxidizing agent

$K+O_2\to K_{2}O$ (K is in excess) [O.N of O -2] (lower O.S.)

$K+O_2\to K_2{O}_2$ (O is in excess) [O.N of O -1] (lower O.S.)

Question 7.12(a) How do you count for the following observations?

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.

Answer :

Alcohol and $KMn{O}_4$ both are polar in nature and alcohol is homogenous to toluene because both are organic compounds. So the reaction is faster in the homogenous medium rather than heterogeneous medium. And hence all the compounds react at a faster rate.

Chemical equation-

$C_6{H}_{6}C{H}_3+Mn{O}_4^{-}(alc.)\to C_6{H}_{6}CO{O}^{-}+Mn{O}_2+H_{2}O(l)+O{H}^{-}(aq)$

Question 7.12(b) How do you count for the following observations?

When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why ?

Answer :

• Concentrated sulphuric acid is added to an inorganic mixture containing chloride-

$2NaCl+2H_{2}S{O}_4\to 2NaHS{O}_4+2HCl$

HCl is a weak reducing agent and it cannot reduce $H_{2}S{O}_4$ to $S{O}_2$ thats why we get colourless pungent smelling gas HCl.

• Concentrated sulphuric acid is added to an inorganic mixture containing bromide-

$2NaBr+2H_{2}S{O}_4\to 2NaHS{O}_4+2HBr$

$2HBr+H_{2}S{O}_4\to B{r}_2(RedVapour)+S{O}_2+2H_{2}O$

When conc. sulphuric acid is added to an inorganic mixture containing bromide initially it produces $HBr$ and it is a strong reducing agent so it reduces $H_{2}S{O}_4$ to $S{O}_2$ with evolution of is a red vapor of bromine.

Question 7.13(a) Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions

$(a)2AgBr(s)+C_6{H}_6{O}_2(aq)\to 2Ag(s)+2HBr(aq)+C_6{H}_4{O}_2(aq)$

Answer :

Substance reduced/oxidizing agent- $AgBr$

Substance oxidized/reducing agent- $C_6{H}_6{O}_2]$

Question 7.13(b) Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions

$(b)HCHO(l)+2[Ag(N{H}_3){]}^{+}(aq)+3O{H}^{-}(aq)\to 2Ag(s)+HCO{O}^{-}(aq)+4N{H}_3(aq)+2H_{2}O(l)$

Answer :

Substance reduced/oxidising agent- $[Ag(N{H}_3{)}_2{]}^{+}$

Substance oxidised/reducing agent- $HCHO$

Question 7.13(d) Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions

$(d)N_2{H}_4(l)+2H_2{O}_2(l)\to N_2(g)+4H_{2}O(l)$

Answer :

Substance oxidized/reducing agent- $N_2{H}_4$

Substance reduced/oxidizing agent- $H_2{O}_2$

Question 7.13(e) Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions

$(e)Pb(s)+Pb{O}_2(s)+2H_{2}S{O}_4(aq)\to 2PbS{O}_4(s)+2H_{2}O(l)$

Answer :

Substance oxidized/reducing agent- $Pb$

Substance reduced/oxidizing agent- $Pb{O}_2$

Question 7.14 Consider the reactions :
$2S_2{O}_3^{2-}(aq)+I_2(s)\to S_4{O}_6^{2-}(aq)+2I^{-}(aq)$

$2S_2{O}_3^{2-}(aq)+2B{r}_2(l)+5H_{2}O(l)\to S{O}_4^{2-}(aq)+4B{r}^{-}(aq)+10H^{+}(aq)$

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

Answer :

$F_2>C{l}_2>B{r}_2>I_2$ oxidizing power order

Bromine is a stronger oxidizing agent than iodine. So in the case of bromine (avg. oxidation number of sulphur is changed from +2 to +6)

and in case of iodine it (+2 to +2.5). So that's why thiosulphate reacts differently with bromine and iodine.

Question 7.15 Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

Answer :

part(i)

Fluorine can oxidize other halogen ions. On the other hand $B{r}_2,I_2,C{l}_2$ cannot oxidize $2F^{-}\to F_2$

$$\begin{gathered} F_2+2I^{-}\to 2F^{-}+I_2 \\ {F}_2+2B{r}^{-}\to 2F^{-}+B{r}_2 \\ {F}_2+2C{l}^{-}\to 2F^{-}+C{l}_2 \end{gathered}$$

And hence we say that fluorine is the better oxidant among halogen.

part(ii)

$HI$ & $HBr$ are able to reduce $H_{2}S{O}_4\to S{O}_2$ but $HCl,HF$ are unable to reduce sulphuric acid.

So here we can say that $HI$ & $HBr$ are better reductant than $HCl,HF$ .

Again $I^{-}$ can only able to reduce $C{u}^{2+}\to C{u}^{+}$ but $B{r}^{-}$ cannot.

$4I^{-}+2C{u}^{2+}\to C{u}_2{I}_2+I_2$

Hence among hydrohalic compound hydroiodic acid is the best reductant.

Question 7.16 Why does the following reaction occur?

$Xe{O}_6^{4-}(aq)+2F^{-}(aq)+6H^{+}(aq)\to Xe{O}_3(g)+F_2(g)+3H_{2}O(l)$

What conclusion about the compound $N{a}_{4}Xe{O}_6$ (of which $Xe{O}_6^{4-}$ is a part) can be drawn from the reaction ?

Answer :

we conclude that the oxidation state of Xenon changes from +8 to +6

$Xe{O}_6^{4-}(+8)\to Xe{O}_3(+6)$

and oxidation state of F changes from -1 to 0

$F^{-}(-1)\to F_2(0)$

$N{a}_{4}Xe{O}_6$ is reduced by accepting an electron.

It is a strong oxidizing agent than F

Question 7.20 What sorts of information can you draw from the following reaction?

$(CN{)}_2(g)+2O{H}^{-}(aq)\to C{N}^{-}(aq)+CN{O}^{-}(aq)+H_{2}O(l)$

Answer :

Carbon shows different oxidation states according to the compound formula.

here we can clearly say that Carbon is in its +3 oxidation state.

$$\begin{gathered} (CN{)}_2(+3)\to C{N}^{-}(+2) \\ (CN{)}_2(+3)\to CN{O}^{-}(+4) \end{gathered}$$

The oxidation state of carbon is increased(oxidized) and decreased(reduced) as well in the product side. So it is a redox reaction and more specifically we can say it disproportion redox reaction.

Question 7.21 The $M{n}^{3+}$ ion is unstable in solution and undergoes disproportionation to give $M{n}^{2+}$ , $Mn{O}_2$ and $H^{+}$ ion. Write a balanced ionic equation for the reaction

Answer :

The base equation

$M{n}^{3+}\to M{n}^{2+}+Mn{O}_2+H^{+}$