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Have you ever wondered how the process of rusting of iron or metabolism in our body takes place? Redox reactions help in understanding such important phenomena. Various industrial processes, workings of batteries, combustion of fuels, etc., are very well explained by redox reaction concepts. Chapter 7 of Class 11 Chemistry NCERT, Redox Reactions, provides detailed descriptions of various topics, such as the process of oxidation and reduction, the change in oxidation state of elements during chemical reactions, and different types of oxidation reactions, among others.
In this Chapter, there are 30 questions in the exercise. The NCERT solutions for class 11 Chemistry Chapter 7 Redox Reactions are prepared and designed by experts. Chemistry is a scoring subject for competitive exams. These NCERT Solutions will help you in your preparation for the class 11 final examination as well as in the various competitive exams like JEE, NEET, etc.
Also Read :
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
$\therefore \:\:1*1+1*3+[x+(-2)*4]*2 +12*[1*2+(-2)] = 0\Rightarrow x = +6$
$(a) K\underline{I}_{3}$
$(b) H_{2}\underline{S_{4}}O_{6}$
$(c) \underline{Fe}_{3}O_{4}$
$(d) \underline{C}H_{3}\underline{C}H_{2}OH$
$(e) \underline{C}H_{3}\underline{C}OOH$
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.
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
|
O.N. Of C
|
---|---|
$CH_{4}$
|
-4
|
$C_{2}H_{6}$
|
-3
|
$C_{2}H_{4}\: or\: CH_{3}Cl$
|
-2
|
$C_{2}H_{2}$
|
-1
|
$CH_{2}Cl_{2}$
|
0
|
$C_{6}Cl_{6}$
|
+1
|
$CHCl_{3}$
|
+2
|
$(COOH)_{2}$
|
+3
|
$CO_{2} \: or\: CCl_{4}$
|
+4
|
substance for Nitrogen-
$NH_{3}$
|
-3
|
$N_{2}H_{4}$
|
-2
|
$N_{2}H_{2}$
|
-1
|
$N_{2}$
|
0
|
$N_{2}O$
|
+1
|
$NO$
|
+2
|
$N_{2}O_{3}$
|
+3
|
$N_{2}O_{4}$
|
+4
|
$N_{2}O_{5}$
|
+5
|
Substance
|
O.N. Of N
|
---|
Answer :
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)
Answer :
These can be understood by the following examples-
$P_{4}+10Cl_{2}(Excess)\rightarrow PCl_{5}$ [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
$C+O_{2}\rightarrow CO_{2}$ (O is in excess) [O.N of C +4]
$C+O_{2}\rightarrow CO$ (C is in excess) [O.N of C +2]
$K+O_{2}\rightarrow K_{2}O$ (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 :
$2NaCl\:+\:2H_{2}SO_{4}\rightarrow \: 2NaHSO_{4}\:+\: 2HCl$
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.
$2NaBr\:+\:2H_{2}SO_{4}\rightarrow \: 2NaHSO_{4}\:+\: 2HBr$
$2HBr + H_{2}SO_{4}\:\rightarrow \:Br_{2} ( Red \:Vapour)\:+\:SO_{2}\:+\:2H_{2}O$
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.
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 7.17(a) Consider the reactions:
(a) $H_{3}PO_{2}(aq)+4AgNO_{3}(aq)+2H_{2}O(l)\rightarrow H_{3}PO_{4}(aq)+4Ag(s)+4HNO_{3}(aq)$
$(b)H_{3}PO_{2}(aq)+2CuSO_{4}(aq)+2H_{2}O(l)\rightarrow H_{3}PO_{4}(aq)+2Cu(s)+4H_{2}SO_{4}(aq)$
What inference do you draw about the behaviour of $Ag^{+}$ and $Cu^{+}$ from these reactions ?
Answer :
In the first reaction, we can see that $Ag^{+}$ oxidizes the phosphorus from (+1 $\rightarrow$ +5) also in second, we clearly see that $Cu^{+}$ oxidize the phosphorus from (+1 $\rightarrow$ +5).
Both are oxidizing agents.
Question 7.17(b) Consider the reactions:
$(d)C_{6}H_{5}CHO(l)+2Cu^{2+}(aq)+5OH^{-}(aq)\rightarrow$ No change observed
Answer :
Here, by looking at the reaction, we conclude that $Ag^{+}$ oxidises $C_{6}H_{5}CHO$ and in the second reaction $Cu^{+}$ not able to oxidise. So we can say that $Ag^{+}$ is stronger oxidizing agent than $Cu^{+}$ .
Question 7.18(a) Balance the following redox reactions by ion – electron method
$MnO_{4}^{-}(aq)+ I^{-}(aq)\rightarrow MnO_{2}(s)+ I_{2}(s)$ (In basic medium)
Answer :
reduction half reaction
$MnO_{4}^{-}\rightarrow MnO_{2}$ (+7 to +4)
Add 3 electron on LHS side and after that to balance charge add OH ions. And to balance O atom add water molecule on whichever side it needed
balance it
$MnO_{4}^{-}+2H_{2}O+3e^{-}\rightarrow MnO_{2} +4OH^{-}$
oxidation half
$I\rightarrow I_{2}$
balance it
$2I^{-}\rightarrow I_{2}+2e^{-}$
equalising the no. of electrons by multiplying the oxidation half by 3 and reduction half by 2 and then add it.
$\begin{aligned} & 6 \mathrm{I}^{-} \rightarrow 3 \mathrm{I}_2+6 e^{-} \\ & 2 \mathrm{MnO}_4^{-}+4 \mathrm{H}_2 \mathrm{O}+6 e^{-} \rightarrow 2 \mathrm{MnO}_2+8 \mathrm{OH}^{-}\end{aligned}$
final answer-
$2MnO_{4}^{-}+4H_{2}O+6I^{-}\rightarrow 2MnO_{2}+3I_{2}+8OH^{-}$
Question 7.18(b) Balance the following redox reactions by ion – electron method
$MnO_{4}^{-}(aq)+SO_{2}(g) \rightarrow Mn^{2+}(aq)+HSO_{4}^{-}$
(In an Acidic medium)
Answer -
oxidation half-reaction
$SO_{2}+2H_{2}O\rightarrow HSO_{4}^{-}+3H^{+}+2e^{-}$
reduction half reaction
$MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O$
Balancing the reaction
multiply the oxidation half by 5 and reduction half by 2 and then add these two reactions
$2MnO_{4}^{-}+5SO_{2}+2H_{2}O+H^{+}\rightarrow 2Mn^{2+}+5HSO_{4}^{-}$
Question 7.18(c) Balance the following redox reactions by ion – electron method
(c) $H_{2}O_{2}(aq)+Fe^{3+}(aq)\rightarrow Fe^{3+}(aq)+H_{2}O(l)$
in acidic medium
Answer :
In acidic medium
oxidation half-reaction-
$Fe^{2+}\rightarrow Fe^{3+}+e^{-}$
reduction half-reaction-
$H_{2}O_{2}+2H^{+}+2e^{-}\rightarrow 2H_{2}O$
Balancing the reaction
multiply by 2 on the oxidation half-reaction then add it with the reduction half-reaction
$H_{2}O_{2}+2Fe^{2+}+2H^{+}\rightarrow 2Fe^{3+}+2H_{2}O$
Question 7.18(d) Balance the following redox reactions by ion – electron method
$Cr_{2}O_{7}^{2-}+SO_{2}(g) \rightarrow Cr^{3+}(aq)+SO_{4}^{2-}(aq)$
in acidic medium
Answer :
Half-reaction
oxidation half $SO_{2}+2H_{2}O\rightarrow SO_{4}^{2-}+4H^{+}+2e^{-}$
reduction half $Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+3SO^{2-}+H_{2}O$
balancing them by multiplying oxidation half by 3 and adding the reaction
$Cr_{2}O_{7}^{2-}+3SO_{2}+2H^{+}\rightarrow 2Cr^{3+}+3SO_{4}^{2-}+H_{2}O$
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.
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
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.
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 7.26 Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible
(a) $Fe^{3+}(aq)$ and $I^{-}(aq)$
(b) $Ag^{+}(aq)$ and Cu(s)
(c) $Fe^{3+}$ (aq) and Cu(s)
(d) Ag(s) and $Fe^{3+}$ (aq)
(e) $Br_{2}(aq)$ and $Fe^{2+}(aq)$ .
Answer :
If $E^{0}$ for the overall reaction is positive $\rightarrow$ feasible
negative $\rightarrow$ not feasible
(a) $[Fe^{3+}+ e^{-}\rightarrow Fe^{2+} ]*2: E^{0} = 0.77V$
$2I^{-}\rightarrow I_{2}+2e^{-}: E^{0} = -0.54V$
---------------------------------------------------------------------------------------
$2Fe^{3+}+2I^{-}\rightarrow Fe^{2+}+I_{2}: E^{0} = +0.23V$
(b) $Ag^{+}+e^{-}\rightarrow Ag(s)]*2: E^{0} = +0.80V$
$Cu\rightarrow Cu^{2+}+2e^{-}:E^{0} =-0.34V$
---------------------------------------------------------------------------------------
$2Ag^{+}+Cu\rightarrow 2Ag+Cu^{2+}: E^{0}=+0.46V$
(c) $Fe^{3+}+e^{-}\rightarrow Fe^{2+}]*2 :E^{0}=+0.77V$
$2Br^{-}\rightarrow Br+2e^{- }:E^{0}=-1.09V$
-------------------------------------------------------------------------
$2Fe^{3+}+2Br^{-}\rightarrow 2Fe^{3+}+Br_{2}:E^{0}=-0.32$
(d) $Ag\rightarrow Ag^{+}+e^{-}:E^{0}=-0.80V$
$Fe^{3+}+e^{-}\rightarrow Fe^{2+}:E^{0}==0.77V$
------------------------------------------------------------------------------
$Ag+Fe^{3+}\rightarrow Ag^{+}+Fe^{2+}:E^{0}=-0.03V$
(e) $Br_{2}+2e^{-}\rightarrow 2Br^{-}:E^{0}=+1.09V$
$Fe^{2+}\rightarrow Fe^{3+}+e^{-}]*2:E^{0}=-0.77V$
-------------------------------------------------------------------------------
$Br_{2}+2Fe^{2+}\rightarrow 2Br^{-}+2Fe^{3+}:E^{0}=+0.32V$
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- $Cu^{2+}+2e^{-}\rightarrow Cu$
Answer :
Answer-
In order to displace a metal from its metal salt is done only when the other metal has higher electrode potential.
$Mg(-2.36),Al(-1.66),Zn(-0.76),Fe(-0.44),Cu(+0.34)$
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 $Ag^{+}+e^{-}\rightarrow Ag$
Question: Given below are two statements :
Statement I : Mohr's salt is composed of only three types of ions-ferrous, ammonium, and sulphate.
Statement II : If the molar conductance at infinite dilution of ferrous, ammonium and sulphate ions are $\mathrm{x}_1, \mathrm{x}_2$ and $\mathrm{x}_3 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$, respectively then the molar conductance for Mohr's salt solution at infinite dilution would be given by $\mathrm{x}_1+\mathrm{x}_2+2 \mathrm{x}_3$
In the light of the given statements, choose the correct answer from the options given below :
1)Both statements I and Statement II are false
2)Statement I is false but Statement II is true
3)Statement I is true but Statement II are false
4)Both statements I and Statement II are true
Answer:
Mohr's salt : $\mathrm{FeSO}_4 \cdot\left(\mathrm{NH}_4\right)_2 \mathrm{SO}_4 \cdot 6 \mathrm{H}_2 \mathrm{O}$
Using Kohlrousch law
$\lambda_{\mathrm{m}}^*(\text { Mohr's salt })=\mathrm{x}_1+2 \mathrm{x}_2+2 \mathrm{x}_3$
Hence, the correct answer is option (3).
Sometimes, problems related to Redox Reactions seem difficult, but once we understand the basic rules and strategy, it becomes very easy to solve all the questions related to Redox reactions.
We can follow the steps given below to solve the questions based on Redox Reactions :
1). First of all we need to understand the type of question asked, basically numerical of Redox Reactions can be divided into the following categories:
2). Assign the oxidation numbers by using the oxidation number rule to find the oxidation state of each element.
3). Then we have to identify the oxidising and the reducing agent.
4). The next step to solve questions is the balancing of the Redox Reactions.
5). For redox titrations we can use the n-factor to find equivalent
These are the steps that can be followed while solving problems related to redox reactions. If students have a basic understanding of rules then redox problems can be easily solved.
7.1 Classical Idea of Redox Reactions-Oxidation and Reduction Reactions |
7.2 Redox Reactions in Terms of Electron Transfer Reactions |
7.3 Oxidation Number |
7.4 Redox Reactions and Electrode Processes |
Concept | JEE | NCERT |
CLASSICAL IDEA OF REDOX REACTIONS – OXIDATION AND REDUCTION REACTIONS | ✅ | |
Oxidation Number | ✅ | ✅ |
Types of Redox Reactions | ✅ | |
Displacement Reaction | ✅ | ✅ |
Balancing of Redox Reaction: Ion Electrode Method | ✅ | ✅ |
Balancing of Disproportionation Redox Reaction: Ion Electrode Method | ✅ | ✅ |
Balancing of Redox Reaction: Oxidation Number Method | ✅ |
Redox Reaction is an important chapter from the point of view of competitive exams as the chapter is very scoring as well as it will form the basis of the chapters students are going to learn in future. Some of the topics like Redox balancing are frequently asked in JEE mains. The basics of Redox Reactions play a crucial role in Electrochemistry, Thermodynamics, Metallurgy, etc. Many JEE and NEET questions are directly asked from the concepts given in NCERT. Redox reactions act as the link between inorganic chemistry and Biology. Redox reactions develop problem-solving and analytical abilities in students.
NCERT Solutions for Class 11 Maths |
NCERT solutions for Class 11 Biology |
NCERT solutions for Class 11 Chemistry |
NCERT solutions for Class 11 Physics |
Also Check,
In order to balance redox reaction using oxidation number method, we have to follow the steps given below:
In order to identify the oxidizing and reducing agents in the reaction, first we need to assign oxidation numbers to all the involved elements.
Oxidation number | Valency |
1). Imaginary charge an atom have in an ionic compound | 1). It is the combining capacity of atoms to form bonds |
2). Oxidation number can be positive, negative and zero | 2). Always positive |
3). Varies for same elements in different compounds | 3). Constant for an element in most of the compounds |
4). Shows loss and gain of electrons during a reaction. | 4). Valence shows the number of electrons required to complete an octet. |
5). Example : Oxidation number of Fe in FeCl3 is +3. | 5). Valency of Fe is usually 2 or 3. |
Rules of assigning oxidation number are given below:
Redox reactions are important in the biological system because they are fundamental to metabolism and production of energy. They are used in cellular respiration, photosynthesis, detoxification, immune response and help in maintaining the cell balance in ions and molecules.
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As per latest syllabus. Physics formulas, equations, & laws of class 11 & 12th chapters
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