NCERT Solutions for Class 11 Chemistry Chapter 8
Redox Reactions Class 11
Chapter 8 Redox Reactions Exercise Solutions
Exercise : Solutions of Questions on Page Number : 272
Assign oxidation numbers to the underlined elements in each of the following species:
(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4
(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2O
(a)
Let the oxidation number of P be x. We know that,
Oxidation number of Na = +1 Oxidation number of H = +1 Oxidation number of O = –2
Hence, the oxidation number of P is +5.
Then, we have
Hence, the oxidation number of S is + 6.
Hence, the oxidation number of P is + 5.
Then, we have
Hence, the oxidation number of Mn is + 6.
Then, we have
Hence, the oxidation number of O is – 1.
Then, we have
Hence, the oxidation number of B is + 3.
Hence, the oxidation number of S is + 6.
Then, we have
Or,
Hence, the oxidation number of S is + 6.
What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH
(a) KI3
In KI3, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3to find the oxidation states.
Hence, in a KI3molecule, the O.N. of the two I atoms forming the I2molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.
(b)
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 S atoms is +5 and the O.N. of the other two S atoms is 0.
(c)
(d)
2 (x) + 6 (+1) + 1 (-2) = 0
or, 2x + 4 = 0 or, x = -2
Hence, the O.N. of C is –2.
(e)
2 (x) + 4 (+1) + 2 (-2) = 0
or, 2x = 0 or, x = 0
Justify that the following reactions are redox reactions:
(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)
(a)
Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H2 to +1 in H2O i.e., H2 is oxidized to H2O. Hence, this reaction is a redox reaction.
Here, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe i.e., Fe2O3 is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO2 i.e., CO is oxidized to CO2. Hence, the given reaction is a redox reaction.
The oxidation number of each element in the given reaction can be represented as:
In this reaction, the oxidation number of B decreases from +3 in BCl3 to –3 in B2H6. i.e., BCl3 is reduced to B2H6. Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in B2H6 i.e., LiAlH4 is oxidized to B2H6. Hence, the given reaction is a redox reaction.
In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F2 to –1 in KF i.e., F2 is reduced to KF.
Hence, the above reaction is a redox reaction.
Here, the oxidation number of N increases from –3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2decreases from 0 in O2 to –2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.
Fluorine reacts with ice and results in the change:
Justify that this reaction is a redox reaction.
Here, we have observed that the oxidation number of F increases from 0 in F2 to +1 in HOF. Also, the oxidation number decreases from 0 in F2 to –1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.
Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, –. Suggest structure of these compounds. Count for the fallacy.
However,the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6. The structure of H2SO5is shown as follows:
Therefore, the O.N. of S is +6.
Here, there is no fallacy about the O.N. of Cr in . The structure of is shown as follows:
Here, each of the two Cr atoms exhibits the O.N. of +6.
Here, there is no fallacy about the O.N. of N in . The structure of is shown as follows:
Write the formulae for the following compounds:
(c) Tin(IV) oxide (d) Thallium(I) sulphate
Answer :
(a) Mercury (II) chloride:
HgCl2
(b) Nickel (II) sulphate:
NiSO4
(c) Tin (IV) oxide:
SnO2
(d) Thallium (I) sulphate:
Tl2SO4
(e) Iron (III) sulphate:
Fe2(SO4)3
(f) Chromium (III) oxide: Cr2O3
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
The substances where carbon can exhibit oxidation states from –4 to +4 are listed in the following table.
Substance |
O.N. of carbon |
CH2Cl2 |
0 |
ClC≡CCl |
+1 |
HC≡CH |
–1 |
CHCl3, CO |
+2 |
CH3Cl |
–2 |
Cl3C – CCl3 |
+3 |
H3C – CH3 |
–3 |
CCl4, CO2 |
+4 |
CH4 |
–4 |
The substances where nitrogen can exhibit oxidation states from –3 to +5 are listed in the following table.
Substance |
O.N. of nitrogen |
|
N2 |
0 |
|
N2O |
+1 |
|
N2H2 |
–1 |
|
NO |
+2 |
|
N2H4 |
–2 |
|
N2O3 |
+3 |
|
NH3 |
–3 |
|
NO2 |
+4 |
|
N2O5 |
+5 |
|
|
|
|
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?
In sulphur dioxide (SO2), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to -2.
Therefore, SO2can act as an oxidising as well as a reducing agent.
In hydrogen peroxide (H2O2), the O.N. of O is -1 and the range of the O.N. that O can have is from 0 to -2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H2O2can act as an oxidising as well as a reducing agent.
In ozone (O3), the O.N. of O is zero and the range of the O.N. that O can have is from 0 to -2. Therefore, the O.N. of O can only decrease in this case. Hence, O3acts only as an oxidant.
In nitric acid (HNO3), the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to -3. Therefore, the
Consider the reactions:
Why it is more appropriate to write these reactions as:
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
(a) The process of photosynthesis involves two steps.
Now, the net reaction of the process is given as:
Itis 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 H2O18in place of H2O.
(b) O2is produced from each of the two reactants O3and H2O2. For this reason, O2is written twice.
The path of this reaction can be investigated by using or .
The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
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.
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. This can be illustrated as follows:
(i) P4and F2are reducing and oxidising agents respectively.
However, if P4is treated with an excess of F2, then PF5will be produced, wherein the O.N. of P is +5.
(ii) K acts as a reducing agent, whereas O2is an oxidising agent.
However, if K reacts with an excess of O2, then K2O2will be formed, wherein the O.N. of O is –1.
(iii) C is a reducing agent, while O2acts as an oxidising agent.
On the other hand, if C is burnt in an excess of O2, then CO2will be produced, wherein the O.N. of C is +4.
How do you count for the following observations?
(b) 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?
(a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of 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) KMnO4 and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence, in alcohol, KMnO4 and toluene can react at a faster rate.
The balanced redox equation for the reaction in a neutral medium is give as below:
(b)
But, when conc. H2SO4 is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H2SO4 to SO2.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:
(b) HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH-(aq) → 2Ag(s) + HCOO-(aq) + 4NH3(aq) + 2H2O(l)
(e) Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l)
(a) Oxidised substance →C6H6O2 Reduced substance →AgBr Oxidising agent →AgBr Reducing agent →C6H6O2 (b)Oxidised substance →HCHO
Reduced substance →
Oxidising agent →
Reducing agent →HCHO
(c) Oxidised substance →HCHO Reduced substance →Cu2+
Oxidising agent →Cu2+
Reducing agent →HCHO
(d) Oxidised substance →N2H4 Reduced substance →H2O2 Oxidising agent →H2O2 Reducing agent →N2H4
(e) Oxidised substance →Pb
Reduced substance →PbO2 Oxidising agent →PbO2 Reducing agent →Pb
Consider the reactions:
aq) + 2Br2(l) + 5 H2O(l) → (aq) + 4Br–(aq) + 10H+(aq)
Answer :
The average oxidation number (O.N.) of S in is +2. Being a stronger oxidising agent than I2,
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
F2 can oxidize Cl– to Cl2, Br– to Br2, and I– to I2 as:
On the other hand, Cl2, Br2, and I2 cannot oxidize F– to F2. The oxidizing power of halogens increases in the order of I2 < Br2 < Cl2 < F2. Hence, fluorine is the best oxidant among halogens.
Again, I– can reduce Cu2+ to Cu+, but Br– cannot.
Hence, hydroiodic acid is the best reductant among hydrohalic compounds.
Why does the following reaction occur?
What conclusion about the compound Na4XeO6(of which is a part) can be drawn from the reaction. Answer :
In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in to +6 in XeO3and the O.N. of F increases from –1 in F– to O in F2.
Consider the reactions:
(b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) → H3PO4(aq) + 2Cu(s) + H2SO4(aq)
(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH-(aq) → No change observed.
Answer :
Ag+ and Cu2+ act as oxidising agents in reactions (a) and (b) respectively.
.
In reaction (c), Ag+
oxidises
C6H5CHO to C6H5COO-,
but in reaction (d), Cu2+ cannot oxidise C6H5CHO. Hence, we can say that Ag+ is a stronger oxidising
agent than Cu2+
Balance the following redox reactions by ion-electron method:
(b) (aq) + SO2(g) → Mn2+(aq) + (aq) (in acidic solution)
(d) + SO2(g) → Cr3+(aq) + (aq) (in acidic solution) Answer :
(a) Step 1: The two half reactions involved in the given reaction are:
Oxidation half reaction:
Reduction half reaction:
Now, to balance the charge, we add 2 e–to the RHS of the reaction.
(b)Following the steps as in part (a), we have the oxidation half reaction as:
Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:
(c)
And the reduction half reaction as:
(d) Following the steps as in part (a), we have the oxidation half reaction as:
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
(a)
The O.N. (oxidation number) of P decreases from 0 in P4 to – 3 in PH3 and increases from 0 in P4 to + 2 in . Hence, P4 acts both as an oxidizing agent and a reducing agent in this reaction.
The P atom is balanced as:
The charge is balanced by adding 8OH– as:
The O and H atoms are already balanced. The reduction half equation is:
The O.N. is balanced by adding 12 electrons as:
The O and H atoms are balanced by adding 12H2O as:
The oxidation number of N increases from – 2 in N2H4 to + 2 in NO and the oxidation number of Cl decreases from
+ 5 in to – 1 in Cl–. Hence, in this reaction, N2H4 is the reducing agent and is the oxidizing agent.
The N atoms are balanced as:
The charge is balanced by adding 8 OH–ions as:
The reduction half equation is:
The charge is balanced by adding 6OH– ions as:
The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:
Total decrease in oxidation number of N = 2 × 4 = 8 Total increase in oxidation number of Cl = 1 × 6 = 6
The N and Cl atoms are balanced as:
This is the required balanced equation.
The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in and the oxidation number of O increases from – 1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidizing agent and H2O2 is the reducing agent.
The oxidation number is balanced by adding 2 electrons as:
The oxygen atoms are balanced by adding 2H2O as:
The Cl atoms are balanced as:
The charge is balanced by adding 6OH– as:
The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:
Total decrease in oxidation number of Cl2O7 = 4 × 2 = 8 Total increase in oxidation number of H2O2 = 2 × 1 = 2
The Cl atoms are balanced as:
The O atoms are balanced by adding 3H2O as:
This is the required balanced equation.
The oxidation numbers of carbon in (CN)2, CN– and CNO– are +3, +2 and +4 respectively. These are obtained as shown below:
Let the oxidation number of C be x. (CN)2
2(x – 3) = 0
∴x = 3
CN–
x – 3 = –1
∴x = 2
CNO–
x – 3 – 2 = –1
∴x = 4
It can be easily observed that the same compound is being reduced and oxidised simultaneously in the given equation. Reactions in which the same compound is reduced and oxidised is known as disproportionation reactions. Thus, it can be said that the alkaline decomposition of cyanogen is an example of disproportionation reaction.
The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.
The oxidation half equation is:
The charge is balanced by adding 4H+ ions as:
The reduction half equation is:
The balanced chemical equation can be obtained by adding equation (i) and (ii) as:
Consider the elements:
(a) Identify the element that exhibits only negative oxidation state.
(c) Identify the element that exhibits both positive and negative oxidation states.
Answer :
(a) F exhibits only negative oxidation state of -1.
(b) Cs exhibits positive oxidation state of +1.
(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of - 1, + 1, + 3, + 5, and + 7.
(d)
Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
The oxidation half reaction is:
The charge is balanced by adding 4H+ ions as:
The reduction half reaction is:
The oxidation number is balanced by adding electrons
Refer to the periodic table given in your book and now answer the following questions:
(b) Select three metals that can show disproportionation reaction.
In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states.
(a) P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states.
(b)
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?
Thus, 68 g of NH3 reacts with 160 g of O2.
Therefore, 10g of NH3 reacts with g of O2, or 23.53 g of O2. But the available amount of O2 is 20 g.
Therefore, O2 is the limiting reagent (we have considered the amount of O2 to calculate the weight of nitric oxide obtained in the reaction).
Now, 160 g of O2 gives 120g of NO.
Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:
(b) Ag+(aq) and Cu(s)
(d) Ag(s) and Fe3+(aq)
Answer :
(a)
E° for the overall reaction is positive. Thus, the reaction between and is feasible.
(b)
E° positive for the overall reaction is positive. Hence, the reaction between and is feasible.
(c)
E° positive for the overall reaction is positive. Hence, the reaction between and is feasible.
(d) The possible reaction between and is given by,
Here, E° for the overall reaction is negative. Hence, the reaction between and is not feasible.
(e)
Predict the products of electrolysis in each of the following:
(ii) An aqueous solution AgNO3 with platinum electrodes
(iv) An aqueous solution of CuCl2 with platinum electrodes.
(i) AgNO3 ionizes in aqueous solutions to form Ag+ and ions.
Hence, Ag+ ions are reduced at the cathode. Similarly, Ag metal or H2O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H2O molecules.
Therefore, Ag metal gets oxidized at the anode.
(ii) Pt cannot be oxidized easily. Hence, at the anode, oxidation of water occurs to liberate O2. At the cathode, Ag+ ions are reduced and get deposited.
(iii)
On electrolysis, either of H+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of H+ ions is higher than that of H2O molecules.
Hence, at the cathode, H+ ions are reduced to liberate H2 gas.
On the other hand, at the anode, either of ions or H2O molecules can get oxidized. But the oxidation of involves breaking of more bonds than that of H2O molecules. Hence, ions have a lower oxidation potential than H2O. Thus, H2O is oxidized at the anode to liberate O2 molecules.
(iv)
On electrolysis, either of Cu2+ ions or H2O molecules can get reduced at the cathode. But the reduction potential of Cu2+is more than that of H2O molecules.
Hence, Cu2+ ions are reduced at the cathode and get deposited.
But oxidation of H2O molecules occurs at a lower electrode potential than that of Cl– ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl– ions are oxidized at the anode to liberate Cl2 gas.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Answer :
A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt. The order of the increasing reducing power of the given metals is Cu < Fe < Zn < Al < Mg.
Hence, we can say that Mg can displace Al from its salt solution, but Al cannot displace Mg.
Thus, the order in which the given metals displace each other from the solution of their salts is given below: Mg>Al> Zn> Fe,>Cu
Given the standard electrode potentials, K+/K = -2.93V, Ag+/Ag = 0.80V,
Mg2+/Mg = -2.37V. Cr3+/Cr = -0.74V
Answer :
Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place, further show:
(ii) the carriers of the current in the cell, and
Answer :
(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+ and the leaving electrons accumulate on this electrode.
(ii) Ions are the carriers of current in the cell.
(iii) The reaction taking place at Zn electrode can be represented as:
And the reaction taking place at Ag electrode can be represented as:
.