NCERT Solutions for Class 11 Chemistry Chapter 7 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 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 a 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 7.3(a) Justify that the following reactions are redox reaction

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

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 oxidised to H₂O. Hence, it is a redox reaction.

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

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

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(d) Justify that the following reactions are redox reaction

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

Answer: We know that oxidation = loss of electrons by an 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 reactions

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

Answer: Here, N (−3) → N (+2) oxidation reaction

and O (0) → 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 a 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

Answer :

$\begin{array}{rcc}{F}_2& \mathrm{HF}& \mathrm{HOF}\\ \text{ Oxidation state of }F\to 0& -1& +1\end{array}$

Here, F is oxidised 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 a 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

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

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

There is no fallacy here

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

Let's assume the oxidation number of N is x

Now, $x+(-2)\ast 3=$

$$\begin{gathered} -1 \\ (x-6) \end{gathered}$$

$$\begin{gathered} =-1 \\ x=+5 \end{gathered}$$

There 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:

NiSO₄

Sulphate has a −2 oxidation state

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

Tin(IV) oxide

Answer:

SnO₂

Oxygen has a −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 :

The Formula of the Chromium(III) oxide compounds is

$C{r}_2{O}_3$

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 (SO₂) 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 an oxidant and a reductant.
• 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 oxidising and reducing properties.
• For $HN{O}_3$, Nitrogen has a +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 the only oxidant.
• And in case of $O_3$, the oxidation state of O is zero(0), and the range of oxidation number of O is 0 to -2. It only decreases in this case also so it acts as only an oxidant.

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 KMnO₄ both are polar in nature and alcohol is homogenous with toluene because both are organic compounds. So the reaction is faster in the homogeneous medium than in the 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 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 $S{O}_2$ that's why we get colourless pungent 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 the evolution of a red vapour 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/oxidising agent- $AgBr$

Substance oxidised/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 oxidised/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 oxidised/reducing agent- Pb

Substance reduced/oxidising agent- PbO₂

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 oxidise other halogen ions. On the other hand $B{r}_2,I_2,C{l}_2$ cannot oxidize

$2F^{-}\to F_2$

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

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

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 reducing agents 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 compounds, hydroiodic acid is the best reducing agent.

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(oxidised) and decreased(reduced) as well on 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

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

write the oxidation half with their oxidation state

$M{n}^{3+}(+3)\to Mn{O}_2(+4)$

Balance the charge on $Mn$ by adding 1 $e^{-}$ on the RHS side. To balance charge, add $H^{+}$ ions on the RHS side and then for oxygen balance, add $H_{2}O$ molecule on the LHS side.

$M{n}^{3+}+2H_{2}O\to Mn{O}_2+1e^{-}+4H^{+}$

reduction half

$M{n}^{3+}(+3)\to M{n}^{2+}(+2)$

balancing the reduction half by adding 1 $e^{-}$ on LHS side

$M{n}^{3+}+1e^{-}\to M{n}^{2+}$

Add both balanced reduction half and oxidation half

$2M{n}^{3+}+2H_{2}O\to M{n}^{2+}+Mn{O}_2+4H^{+}$

Question 7.22 Consider the elements: Cs, Ne, I and F
(a) Identify the element that exhibits only a negative oxidation state.

(b) Identify the element that exhibits only a positive oxidation state

(c) Identify the element that exhibits both positive and negative oxidation states

(d) Identify the element which exhibits neither the negative nor the positive oxidation state.

Answer :

(a) F (because of its highly electronegative in nature)

(b) Cs (highly electropositive)

(c) I (has empty d orbitals)

(d) Ne (inert gas)

Question 7.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating it with sulphur dioxide. Present a balanced equation for this redox change taking place in water

Answer :

Base equation- $S{O}_2+C{l}_2\to C{l}^{-}+S{O}_4^{2-}$ --------------(have to remember)

Now we have to balance the oxidation half and the reduction half.

oxidation half = $S{O}_2\to S{O}_4^{2-}$

balancing - Oxygen is balanced by adding water molecule, Hydrogen is balanced by $H^{+}$ ion and for charge add $e^{-}$ (electron) $S{O}_2+2H_{2}O\to S{O}_4^{2-}+4H^{+}+2e^{-}$

Reduction hallf = $C{l}_2\to C{l}^{-}$

Balancing - to balance charge add an electron $C{l}_2+2e^{-}\to 2C{l}^{-}$

Now add both balanced oxidation half and reduction half, we get

$C{l}_2+S{O}_2+2H_{2}O\to 2C{l}^{-}+S{O}_4^{2-}+4H^{+}$

Question 7.24 Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non-metals that can show a disproportionation reaction.

(b) Select three metals that can show a disproportionation reaction

Answer :

(a) Phosphorus, sulphur and chlorine can show disproportionation reaction.

(b) Manganese, copper, indium and gallium can show disproportionation reaction.