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

$(a) NaH_{2}\bar{P}O_{4}$

$(b)NaH\bar{S}O_{4}$

$(c) H_{4}\bar{P_{2}}O_{7}$

$(d) K_{2}\bar{Mn}O_{4}$

$(e) Ca\bar{O_{2}}$

$(f) Na\bar{B}H_{4}$

$(g) H_{2}\bar{S_{2}}O_{7}$

$(h) KAl(\bar{S}O_{4})_{2}.12H_{2}O$

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

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

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

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

(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

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

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

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

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

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

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

$(a) K\underline{I}_{3}$

$(b) H_{2}\underline{S_{4}}O_{6}$

$(c) \underline{Fe}_{3}O_{4}$

$(d) \underline{C}H_{3}\underline{C}H_{2}OH$

$(e) \underline{C}H_{3}\underline{C}OOH$

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)

$(a) K\underline{I}_{3}$

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 $KI_{3}$

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

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.)

$(b) H_{2}\underline{S_{4}}O_{6}$

Assume O.N of S is x

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

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]

$(c) \underline{Fe}_{3}O_{4}$

If you calculate the oxidation number of Fe in $Fe_{3}O_{4}$ it would be 8/3 and however, O.N cannot be in fractional.

$Fe_{3}O_{4} \:\:\: is \:an\:equimolar \:mixture\: of \:(FeO)\: and \:(Fe_{2}O_{3})$ Here one iron atom has +2 O.N and the other two are of +3 O.N.

$(d) \bar{C}H_{3}\bar{C}H_{2}OH$

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.

$(e) \bar{C}H_{3}\bar{C}OOH$

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

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

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

$\overset{\:\:2+\:\:-2}{CuO}\:+\:\overset{0}{H_{2}}\rightarrow \overset{0}{Cu}\:+ \:\overset{+1\:-2}{H_{2}O}$

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

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

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

$\overset{\:\:+3\:\:-2}{Fe_{2}O_{3}}\:+ 3\overset{+2\:-2}{CO}\rightarrow 2\overset{0}{Fe}\:\:+$ $3\overset{+4\:\:-2}{CO_{2}}$

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

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

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

$4\overset{+3\:\:-1}{BCl_{3}}\:\:+\:3\overset{+1\:\:+3\:\:-1}{LiAlH_{4}}\:\rightarrow\:2\overset{-3\:\:+1}{B_{2}H_{6}}\:\:+\:\:3\overset{+1\:-1}{LiCl}\:\:+\:3\overset{+3\:-1}{AlCl_{3}}$

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 $LiAlH_{4}$ 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

$2 K_{(s)}+F_{2(g)} \rightarrow 2 K^{+} F_{(s)}^{-}$

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

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

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

and $O(0)\rightarrow O(-2)$ 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

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

Justify that this reaction is a redox reaction

Answer : $F_{2}$ $HF$ $HOF$

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

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 $H_{2}SO_{5}$ , $Cr_{2}O_{7}^{2-}$ and $NO_{3}^{-}$ . Suggest the structure of these compounds. Count for the fallacy.

Answer :

(i) $H_{2}SO_{5}$ let the oxidation number of sulphur be x

So,

$\\2*1+x+5(-2)= 0\\ x= +8$

There is a fallacy Sulphur cannot have +8 oxidation state because it has maximum +6 oxidation number, not more than that.The structure of $H_{2}SO_{5}$ is shown as follows:

$2(H)+1(S)+3(O)+2(O \:\:in \:peroxy\: linkage)$

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

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

(ii) $Cr_{2}O_{7}^{2-}$

let the oxidation number of chromium be x

now

$\\2x+7*(-2)=-2\\ x= (-2+14)/2\\ x= +6$

There is no fallacy here

(iii) $NO_{3}^{-}$

let assume oxidation number of N is x

Now, $\\x+(-2)*3= -1\\(x-6)=-1\\x= +5$

here is no fallacy about the O.N of N in $NO_{3}^{-}$

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

Mercury(II) chloride

Answer : $HgCl_{2}$

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 : $NiSO_{4}$

sulphate has $-2$ oxidation state

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

Tin(IV) oxide

Answer : $SnO_{2}$

Oxygen has $-2$ oxidation state

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

Thallium(I) sulphate

Answer : $Tl_{2}SO_{4}$

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

Iron(III) sulphate

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

$Fe_{2}(SO_{4})_{3}$

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

Chromium(III) oxide

Answer :

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

$Cr_{2}O_{3}$

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 $(SO_{2})$ 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 $HNO_{3}$ , 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:

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

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

Why it is more appropriate to write these reactions as :

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

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 $H_{2}$ --> $2H_{2}O\rightarrow O_{2}+2H_{2}$

step-2 The $H_{2}$ produced in above reduces the $CO_{2}$ into glucose $(C_{6}H_{12}O_{6})$ and water $(H_{2}O)$

$6CO_{2}+12H_{2}\rightarrow C_{6}H_{12}O_{6}+6H_{2}O$

So, the final net reaction is

$2H_{2}O\rightarrow CO_{2}+2H_{2}]*6\\+6CO_{2}+12H_{2}\rightarrow C_{6}H_{12}O_{6}+6H_{2}O\\------------------\\6CO_{2}+12H_{2}O\rightarrow C_{6}H_{12}O_{6}+6H_{2}O$

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)

$\\O_{3}(g)\rightarrow O_{2}(g)+O(g)\\ H_{2}O_{2}(l)+O(g)\rightarrow H_{2}O(l)+O_{2}(g)\\ ---------------\\ H_{2}O_{2}(l)+O_{3}(g)\rightarrow H_{2}O(l)+O_{2}(g)+O_{2}(g)$ (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 $O^{18}_{3}/ H_{2}O^{18}$ .

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 $Cl_{2}$ is an oxidizing agent

$P_{4}+10Cl_{2}(Excess)\rightarrow PCl_{5}$ [O.N of phosphorus +5] $\Rightarrow$ Higher O.S of P

$P_{4}(Excess)+6Cl_{2}\rightarrow 4PCl_{3}$ [O.N of phosphorus +3] $\Rightarrow$ Lower O.S of P

• is an $O_{2}$ is a reducing agent and $C$ oxidising agent

$C+O_{2}\rightarrow CO_{2}$ (O is in excess) [O.N of C +4]

$C+O_{2}\rightarrow CO$ (C is in excess) [O.N of C +2]

• $K$ is a reducing agent and $O_{2}$ is an oxidizing agent

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

$K+O_{2}\rightarrow K_{2}O_{2}$ (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 $KMnO_{4}$ 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-

$C_{6}H_{6}CH_{3}+MnO_{4}^{-}(alc.)\rightarrow C_{6}H_{6}COO^{-}+MnO_{2}+H_{2}O(l)+OH^{-}(aq)$

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-

$2NaCl\:+\:2H_{2}SO_{4}\rightarrow \: 2NaHSO_{4}\:+\: 2HCl$

HCl is a weak reducing agent and it cannot reduce $H_{2}SO_{4}$ to $SO_{2}$ thats why we get colourless pungent smelling gas HCl.

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

$2NaBr\:+\:2H_{2}SO_{4}\rightarrow \: 2NaHSO_{4}\:+\: 2HBr$

$2HBr + H_{2}SO_{4}\:\rightarrow \:Br_{2} ( Red \:Vapour)\:+\:SO_{2}\:+\:2H_{2}O$

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 $H_{2}SO_{4}$ to $SO_{2}$ 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

$(a)2AgBr(s)+C_{6}H_{6}O_{2}(aq)\rightarrow 2Ag(s)+2HBr(aq)+C_{6}H_{4}O_{2}(aq)$

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

$(b) HCHO(l)+2[Ag(NH_{3})]^+(aq) +3OH^-(aq)\rightarrow 2Ag(s)+HCOO^- (aq)+4NH_{3}(aq) +2H_{2}O(l)$

Answer :

Substance reduced/oxidising agent- $[Ag(NH_{3})_{2}]^+$

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

$(d) N_{2}H_{4}(l)+2H_{2}O_{2}(l)\rightarrow N_{2}(g)+4H_{2}O(l)$

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

$(e)Pb(s)+PbO_{2}(s)+2H_{2}SO_{4}(aq)\rightarrow 2PbSO_{4}(s)+2H_{2}O(l)$

Answer :

Substance oxidized/reducing agent- $Pb$

Substance reduced/oxidizing agent- $PbO_{2}$

Question 7.14 Consider the reactions :
$2S_{2}O_{3}^{2-}(aq)+I_{2}(s)\rightarrow S_{4}O_{6}^{2-}(aq)+2I^{-}(aq)$

$2S_{2}O_{3}^{2-}(aq)+2Br_{2}(l)+5H_{2}O(l)\rightarrow SO_{4}^{2-}(aq)+4Br^{-}(aq)+10H^{+}(aq)$

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

Answer :

$F_{2}>Cl_{2}>Br_{2}>I_{2}$ 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 $Br_{2},I_{2},Cl_{2}$ cannot oxidize $2F^{-} \rightarrow F_{2}$

$F_{2}+2I^{-}\rightarrow 2F^{-}+I_{2} \\F_{2}+2Br^{-}\rightarrow 2F^{-}+Br_{2} \\F_{2}+2Cl^{-}\rightarrow 2F^{-}+Cl_{2}$

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

part(ii)

$HI$ & $HBr$ are able to reduce $H_{2}SO_{4}\rightarrow SO_{2}$ 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 $Cu^{2+}\rightarrow Cu^{+}$ but $Br^{-}$ cannot.

$4I^{-}+2Cu^{2+}\rightarrow Cu_{2}I_{2}+I_{2}$

Hence among hydrohalic compound hydroiodic acid is the best reductant.

Question 7.16 Why does the following reaction occur?

$XeO_{6}^{4-}(aq)+2F^{-}(aq)+6H^{+}(aq)\rightarrow XeO_{3}(g)+F_{2}(g)+3H_{2}O(l)$

What conclusion about the compound $Na_{4}XeO_{6}$ (of which $XeO_{6}^{4-}$ is a part) can be drawn from the reaction ?

Answer :

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

$XeO_{6}^{4-} (+8)\rightarrow XeO_{3}(+6)$

and oxidation state of F changes from -1 to 0

$F^{-}(-1)\rightarrow F_{2}(0)$

$Na_{4}XeO_{6}$ 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?

$(CN)_{2}(g)+2OH^{-}(aq)\rightarrow CN^{-}(aq) +CNO^{-}(aq)+H_{2}O(l)$

Answer :

Carbon shows different oxidation states according to the compound formula.

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

$\\(CN)_{2}(+3)\rightarrow CN^{-}(+2)\\ (CN)_{2}(+3)\rightarrow CNO^{-}(+4)$

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 $Mn^{3+}$ ion is unstable in solution and undergoes disproportionation to give $Mn^{2+}$ , $MnO_{2}$ and $H^{+}$ ion. Write a balanced ionic equation for the reaction

Answer :

The base equation

$Mn^{3+}\rightarrow Mn^{2+}+MnO_{2}+H^{+}$

write oxidation half with their oxidation state

$Mn^{3+}(+3)\rightarrow MnO_{2}(+4)$

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

$Mn^{3+}+2H_{2}O\rightarrow MnO_{2}+1e^{-}+4H^{+}$

reduction half

$Mn^{3+}(+3)\rightarrow Mn^{2+}(+2)$

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

$Mn^{3+}+1e^{-}\rightarrow Mn^{2+}$

Add both balanced reduction half and oxidation half

$2Mn^{3+}+2H_{2}O\rightarrow Mn^{2+}+MnO_{2}+4H^{+}$

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 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- $SO_{2}+Cl_{2}\rightarrow Cl^{-}+SO_{4}^{2-}$ --------------(have to remember)

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

oxidation half = $SO_{2}\rightarrow SO_{4}^{2-}$

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

Reduction hallf = $Cl_{2}\rightarrow Cl^{-}$

Balancing - to balance charge add an electron $Cl_{2}+2e^{-}\rightarrow 2Cl^{-}$

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

$Cl_{2}+SO_{2}+2H_{2}O\rightarrow 2Cl^{-}+SO_{4}^{2-}+4H^{+}$

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.

Question 7.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Answer :

Answer-

we have,

number of moles(n) = given mass/ molecular mass ------------------------------(eq.1)

No. of moles of ammonia $(NH_{3})$ = 10/17 = 0.588

No. of moles of oxygen $(O_{2})$ = 20/32= 0.625

Balanced Reaction $4NH_{3}+5O_{2}\rightarrow 4NO + 6H_{2}O$

Here we see that 4 moles of ammonia required 5 moles of oxygen. So

0.588 moles of ammonia = $(5/4)*0.588 = 0.735$ moles of $(O_{2})$ . But we have only 0.625 moles of $(O_{2})$.

It means oxygen is a limiting reagent and the maximum weight of nitric oxide $(NO)$ can be produced by 0.635 moles of $(O_{2})$

So, 5 moles of $(O_{2})$ produced 4 moles of C.

therefore 0.625 moles of $(O_{2})$ = $(4/5)*0.625 = 0.5$ moles of $(NO)$ .

from Eq. 1

mass of $(NO)$ = number of moles $(NO)$ * molecular weight $(NO)$

= $0.5* 30$

= 15 g

Alternate Method

directly consider the molecular weight

(17*4) g of NH3 required (5*32) g of O to produce (30*4) g of NO

So, 10g of NH3 required (5*32/17*4)*10 = 23.5g of O. But we have only 20g (means O is limiting reagent) whatever the max. NO produce is from 20g of O.

and we know that 5*32g of O produce 30*4 g of NO

So, 20g of O produce =(30*4/5*32)*20 g of NO = 15g of NO

Question 7.27(i) Predict the products of electrolysis in each of the following:

(i) An aqueous solution of $AgNO_{3}$ with silver electrodes

Answer :

Answer-

(i) $AgNO_{3}$ dissociate into $Ag^{+}$ and $NO_{3}^{-}$

Cathode - $Ag^{+}(aq)+e^{-}\rightarrow Ag(s)$ (reduction potential of silver is higher than $H_{2}O$ ) $E^{0} = 0.80V$

Anode - $Ag(s)\rightarrow Ag^{+}(aq)+e^{-}$ ( oxidation potential of silver is higher than water molecule.So silver electrode oxidized ) $E^{0} = -0.83V$

Question 7.27(ii) Predict the products of electrolysis in each of the following:

An aqueous solution $AgNO_{3}$ with platinum electrodes

Answer :

Answer-

(ii) since platinum $(Pt)$ electrode cannot easily oxidize. So at the anode $H_{2}O$ will oxidize and liberate oxygen and at the cathode $Ag$ will be deposited.

At cathode- $Ag^{+}+e^{+}\rightarrow Ag$

At anode- $H_{2}O\rightarrow O_{2}+4H^{+} +4e^{-}$

Question 7.27(iii) Predict the products of electrolysis in each of the following:

(iii) A dilute solution of $H_{2}SO_{4}$ with platinum electrodes

Answer :

Answer-

given sulphuric acid is dilute.

ionize into $H_{2}SO_{4}\rightarrow 2H^{+}+ SO_{4}^{2-}$

At cathode $H^{+}+e^{-}\rightarrow 1/2H_{2}(g)$

At anode, There will be -(liberation of oxygen gas)

$H_{2}O\rightarrow 1/2O_{2}+4H^{+}+4e^{-}$

Question 7.27(iv) Predict the products of electrolysis in each of the following:

An aqueous solution of CuCl2 with platinum electrodes

Answer :

Answer-

(iv) In aqueous solution $CuCl_{2}$ ionise into $Cu^{2+}$ and $2Cl^{-}$

At the cathode, the copper ion will be deposited because it has a higher reduction potential than the water molecule

At the anode, the lower electrode potential value will be preferred but due to the over potential of oxygen, chloride ion gets oxidized at the anode.

Anode- $2Cl^{-}\rightarrow Cl_{2}+2e^{-}$

cathode- $Cu^{2+}+2e^{-}\rightarrow Cu$

Question 7.29 Given the standard electrode potentials

$K^{+}/K = -2.93 V$ $Ag^{+}/Ag = 0.80V$

$Hg^{2+}/Hg= 0.79V$

$Mg^{2+}/Mg = -2.37V$ $Cr^{3+}/Cr = -0.74V$

arrange these metals in their increasing order of reducing power.

Answer :

Answer-

A negative electrode potential $(E^{0})$ means redox couple is a stronger reducing agent. So as per data the increasing order of the following is-

$Ag<Hg<Cr<Mg<K$

Question 7.30 Depict the galvanic cell in which the reaction

$Zn(s)+2Ag^{+}\rightarrow Zn^{2+}(aq)+2Ag(s)$
takes place, Further show:

(i) Which of the electrodes is negatively charged,

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.

Answer :

Answers-

(i) $Zn$ electrode is negatively charged because it loses electrons (acts as an anode)

(ii) electron flow from a negatively charged electrode to a positively charged electrode (anode to cathode) and the flow of current is just reversed. So current flows throughthe silver cathode to the zinc anode.

(iii) At Anode- $Zn\rightarrow Zn^{2+}+2e^{-}$

At Cathode $Ag^{+}+e^{-}\rightarrow Ag$