RBSE Solutions for Class 11 Chemistry Chapter 8 Oxidation-Reduction Reactions

Rajasthan Board RBSE Class 11 Chemistry Chapter 8 Oxidation-Reduction Reactions

RBSE Class 11 Chemistry Chapter 8 Text Book Questions

RBSE Class 11 Chemistry Chapter 8 Multiple Choice Questions

Question 1.
Oxidation is a process :
(a) In which electrons are accepted
(b) In which electronegative element is added
(c) In which electro positive element is added
(d) In which oxygen is released
Answer:
(b) In which electronegative element is added

Question 2.
In which of the following pair of compounds, oxidation number of chromium is same
(a) K₂CrO₄ and KCrO₂
(b) KCrO₂ and Cr(CO)₆
(c) K₂Cr₂O₇ and Cr(CO)₄
(d) K₂Cr₂O₇ and K₂CrO₄
Answer:
(b) KCrO₂ and Cr(CO)₆

Question 3.
The oxidation number of carbon in diamond is :
(a) Zero
(b) +4
(c) -4
(d) +2
Answer:
(a) Zero

Question 4.
In the following reaction 2FeCl₃ +SnCl₂ ➝ 2FeCl₂ +SnCl₄
The substance which get reduced is :
(a) Sn⁺²
(b) Fe⁺²
(c) Sn⁺⁴
(d) Fe⁺³
Answer:
(d) Fe⁺³

Question 5.
Which of the following is a redox reaction ?
(a) 2FeCl₃ + SnCl₂ ➝ 4 2FeCl₂ + SnCl₄
(b) ASNO₃ + HCl ➝ AsCl + HNO₃
(c) 2Kl + Pb(NO)3 ➝ 2KNO₃ +Pbl₂
(d) BaCl₂ + H₂SO₄ ➝ BaSO4 + 2HCl
Answer:
(a) 2FeCl₃ + SnCl₂ ➝ 4 2FeCl₂ + SnCl₄

RBSE Class 11 Chemistry Chapter 8 Very Short Answer Type Questions

Question 6.
What is cell potential ?
Answer:
The cell potential is the measure of potential difference between two half cell in an electrochemical cell. It is represent by the symbol E-cell.

Question 7.
Which is of two F or I shows both positive and negative oxidation states ?
Answer:
I (iodine) shows both positive and negative oxidation states.

Question 8.
Calculate the value of x in the following equation :
CIO^– + H₂O + xe^– ➝ + Cl + 20H^–
Answer:
The value of x is 2 in the given reaction :
CIO^– +H₂O + 2e^– ➝ Cl^– + 20H^–
gain of two electrons in this reaction :

Question 9.
Identify the reducing agent in the following reaction:
5H₂O₂ +Br₂ ➝ 2HBrO₃ +4H₂0
Answer:
Br₂ acts as reducing agent in the given reaction, because it reduces hydrogen peroxide into water.
H₂O₂ + 2H⁺ + 2e^– ➝ 2H₂O (reduction)

Question 10.
In the given structure the electronegativity of Y is more then Z. Find the oxidation Number of Z.

Answer:
In this structure, the oxidation number of Z is zero.

Question 11.
Write the oxidation number of nitrogen in ammonium nitrate.
Answer:
NH₄NO₃
x + 4 × (+1) + (-1)= 0 × + 4 – 1= 0
x = – 4 + 1
x= – 3
∴ The oxidation number of nitrogen in ammonium nitrate is + 5 and -3.

Question 12.
Arrange the following in increasing order of oxidation states of Mn,
K₂MnO₄ , MnO₂ , KMnO₄
Answer:
Oxidation state of Mn in K₂MnO₄
1(2) + x + 4 (-2) = 0
2 + x – 8 = 0
x = 8 – 2
x = +6
Oxidation state of Mn in MnO₂
x + 2 (-2) =0
x – 4 = 0
x= +4
Oxidation state of Mn in KMnO₄
1 + x + 4 (-2) = 0
1 + x – 8 = 0
x – 7 = 0
x = + 7
∴ Increasing order of oxidation states of Mn,
MnO₂ < K₂MnO₄ < KMnO₄

Question 13.
What do you mean by redox reaction ?
Answer:
Redox reaction is the reaction in which both oxidation and reduction take place.

Question 14.
What is the oxidation number of Li i LiAlH₄ ?
Answer:
x + 3 + 4 (-1) = 0
x + 3 – 4 = 0
x – 1 = 0
x = 1
∴ The oxidation number of Li : LiAlH₄ is + 1.

Question 15.
Calculate the oxidation number of Fe in K₄ [Fe(CN)₆],
Answer.
4 (+1) + x + 6 (-1) – 0
4 + x – 6 = 0
x – 2 = 0
x= + 2
∴ The oxidation number Fe in K₄ [Fe (CN)₆] is + 2.

Question 16.
What is the maximum oxidation state of sulphur ?
Answer.
The maximum oxidation state of sulphur is +6.

RBSE Class 11 Chemistry Chapter 8 Short Answer Type Questions

Question 17.
How to balance a reaction taking place in basic medium by ion electron method ? Explain.
Answer:
There are following steps used-to balance a reaction taking place in basic medium by ion electron method :
Step I Write the skeletal equation and indicate the oxidation number of all the elements.
Step II Find out the species which are oxidised and which are reduced.
Step III Split the equation in two half reactions i.e. oxidation half reaction and reduction half reaction
Step IV Balance the two half reactions separately by using following steps :

• First of all, balance the atoms of the elements which have undergone a change in oxidation number
• Add electrons to whatever side in necessary to make up the difference in oxidation number in each half reaction.
• Balance change by adding OH^–.
• Balance oxygen atoms by adding required number of H₂O molecules to the side deficient in O-atoms.
• H-atoms are balanced by adding H₂O molecules equal in number to the deficiency of H atoms and equal number of OH^– ions are included in opposite side of the equation.

Step V The two half reactions are multiplied by suitable integers to balance the number of electrons. The half reaction, are then added up.

Question 18.
Explain the difference between oxidation state and valency of an element.
Answer:

	Oxidation state		Valency
(1)	In an atom or ion of a substance, the number of effective charge present on an element is called oxidation state of an element.		The valency of an element is a measure of its power with other atoms.
(2)	The value of oxidation state may be positive, negative or zero.		Valency is always +ve. Inert gases have valency due to complete octet zero.
(3)	In covalent compounds, oxidation state of an element is variable		In covalent compounds,          the valency of an element is generally constant.

Question 19.
2KI + Cl₂ ➝ 2KCl + I₂ , Identify the reducing agent in the reaction.
Answer:
2KI + Cl₂ ➝ 2KCl + I₂
Oxidation state of of I in KI => 1 + x = 0
x = -l
Oxidation state of I in I₂= 0
Oxidation state of Cl in Cl₂ = 0
Oxidation state of Cl in KC1 => 1 + x = 0
x = -l

Since, in this reaction Cl₂ is reduced to KCl which is due to KI So, KI is reducing agent in the reaction.

Question 20.
Calculate the oxidation number :
(i) Mo in (NH₄ )₂Mo O₄
(ii) Ni in [Ni (CN)₄] ^2-
Answer:
(i) Suppose oxidation number of Mo in (NH₄ )₂ MoO₄ = x
2 (+1) + x+ 4 (-2) = 0
2 + x – 8 = 0
x – 6 = 0
x = + 6

(ii) Suppose, oxidation number of Ni in [Ni(CN)₄ ]^2- = + x
x + 4 (-1) = -2
x – 4 = – 2
x = -2 + 4 = +2
∴ Oxidation number of Ni in [Ni(CN)₄ ]² = + 2

Question 21.
Arrange the following metals in the order in which they displace each other from the solution of their salts – Cr, Cu, Mg, Zn, Fe, Al.
Answer:
More negative the value of standard electrode potential of a metal, more active will be the metal and a good reducing agents, This is the reason when a more active metal is placed in a solution of less active metal from the solution.
Metal ➝ Cr Cu Mg Zn Fe Al standard electrode
potential – 0.74, + 0. 34, – 2. 36, – 0.76, – 0.44, – 1.66, so, the order of metals in which they displace each other from the solution of their salts in :
Mg, Zn,                Cr, Fe, Cu
(More active)     (Least active)

Question 22.
Represent an electrochemical cell showing indirect redox reaction between Zn and ZnSO₄ + Cu solution.
Answer:
Zn + CuSO₄ ➝ ZnSO₄ + Cu
The electrochemical cell of this reaction is represented as follows :
Anodic representation – Zn/ Zn²⁺(1M), which is written on the left side.
Cathodic representation- Cu²⁺ (1M)/Cu, which is written on the right side.

Salt bridge is represented by two line between the two half reactions.

Question 23.
Define oxidation and reduction :
(a) On the basis of electronic concept
(b) On the basic of oxidation number.
Answer:
(a) On the basic of electronic, oxidation may be defined as a reaction in which one or more electrons is lost by atom, ion or molecule.
The reduction may be defined as a reaction in which one or more electrons is gained by an atom, ion or molecule,

(b) On the basis of oxidation number, oxidation may be defined as a reaction in which the oxidation number of an element in an atom, molecule or ion increase. Reduction may be defined as a reaction in which the oxidation number of an element in an atom, ion or molecule decreases.

Question 24.
Identify oxidising agent and reducing agent in the following reactions :
H₂O₄ + O₃ ➝ H₂O + 2O₂
2Na₂S₂O₃ + I₂ ➝ Na₂S₄O₆ + 2Na I
Answer:
In a reaction

H₂O₂ is reduced to H₂O and oxidised to oxygen. So, H₂O₂ acts as oxidising agent as well as reducing agent.
In a reaction

I₂ acts as oxidising agent, which oxidise Na₂S₂O₃ into Na₂S₄O₆ and Na₂S₂O₃ acts as reducing agent, which reduces I₂ into NaI.

Question 25.
When an iron rod is dipped in silver nitrate solution, it became silverish. Explain the reason.
Answer:
The standard electrode potential of iron (Fe) is -0.44V and silver (Ag) is 0.80V. So, iron is more active metal which displaces silver from its solution, which gets deposited on iron rod so, iron rod becomes silverish when it is dipped in silver nitrate solution.
Fe + 2 Ag NO₃ ➝ Fe(NO₃ )₂ + 2Ag

Question 26.
What is the oxidation number of oxygen in KO₂ ?
Answer:
Lets the oxidation number of oxygen is KO₂ = x
1+ 2 x = 0
2x = -1
x = -1/2
So, the oxidation number of oxygen on KO₂ is – 1/2

Question 27.
During test of nitrate, a complex [Fe (H₂O)₅ (NO)(SO₄ )] is formed. What the oxidation number of Fe in this complex ?
Answer:
[Fe(H₂O)₅(NO) (SO₄)]
x + 0 + 0 – 2 = 0
x – 2 = 0
x= + 2
∴ Oxidation number of Fe in complex = + 2

Question 28.
Arrange the following in increasing order of oxidation states of iodine :
I₂ , HI, HIO₄, ICl
Answer:
Oxidation state of I in I₂ => 0
Oxidation state of I in HI => 1 + x = 0, ⇒  x = -1
Oxidation state of I in HIO₄ => 1 + x + 4(-2) = 0
1 + x – 8 = 0
x – 7 = 0
x = + 7
Oxidation state of I in ICl => x – 1 = 0
x =+l
Increasing order of oxidation state of iodine.

RBSE Class 11 Chemistry Chapter 8 Long Answer Type Questions

Question 29.
What is oxidation number ? How oxidation and reduction can be identified on the basis of change in oxidation number ? Write the step of balancing equation by oxidation number method.
Answer:
The number of effective charge present on an element of atom, ion or molecule in called oxidation number of that element. If oxidation number of an element on ion, atom or molecule increases, then oxidation takes place. Whereas if the oxidation number of element decreases, reduction takes place. The following steps are used for balancing equation by oxidation number method:

Step I First write the oxidation number of each element above its symbol.
Step II Identify the oxidation number of each element above its symbol.
Step III Identify the elements which undergo change in oxidation number.
Step IV Calculate the increases in oxidation number per atom. If more than .one atom of the same element is involved, find out the total number of increase or decrease in oxidation number.
Step V Equate the increase in oxidation number with decrease in oxidation number on reactant side by multiplying the formula of oxidizing and reducing agent.
Step VI Balance all other atoms except H and O.

Question 30.
What is standard electrode potential ? What is the importance of standard electrode potential in chemical reactions ? Explain with example.
Answer:
Standard electrode potential is defined as the potential of each electrode if the concentration of each taking part in reaction is unity at 1 atmospheric and 298K temperature. Importance of standard electrode potential in chemical reactions :
1. The reduction potential of lithium is minimum. It has highest capacity to lose electrons, hence, it is the strongest reducing agent. Whereas the reduction potential of fluoride is higher, it as higher capacity to accept electrons and thus strongest oxidising agent.

2. More negative the value of standard electrode potential of metal, more active will be the metal and a good reducing agent. This is the reason when a more active metal is placed in a solution of less active metal, it displaces the less active metal from the solution.

3. The metal in the top of activity series are good reducing agents therefore, their cations cannot be reduced by chemical metals. This is the reason why alkali metals, alkaline earth metals and aluminum etc. are obtained by electrolytic reduction method. The value of E° will be negative.

4. The metals placed above the hydrogen in activity series reduces hydrogen ion into hydrogen. This is why metals react with acids and liberate hydrogen gas.
Zn + 2H⁺ -> Zn²⁺ + H₂
Mg + 2H⁺ -> Mg²⁺ +H₂

5. Any non-metal displace the anion of the non-metal placed above it in the series from its solution.
Cl₂ + 2Br^– —> 2Cl^– + Br₂

6. The element present below in the series have more capacity to accept electrons as compared to lose electrons. They are good oxidising agents. For example, their cations oxidised H₂ to H⁺ions
2Au⁺ + 3H₂ —> 2Au + 6H⁺

7. Gold and Silver can be precipitated from their salt solutions by the metals more electronegative like zinc, aluminum etc.
2Na[Ag(CN₂)]+ Zn -> Na₂[Zn(CN)₄ ] + 2Ag ↓

Question 31.
Balance the following reaction by oxidation number method and also identify oxidising agent.
(i) H₂S + \(\mathrm{MnO}_{4}^{-}\) ➝ S + Mn²⁺ + H₂O (Acidic medium)
Cl₂O₇ + H₂O₂ ➝ \(\mathrm{ClO}_{2}^{-}\) + O₂₍gas) (Basic medium)
Answer:
(i)(a) Write unbalanced equation,

(b) Oxidation number of Mn decreases from + 7 to +2. So, \(\mathrm{MnO}_{4}^{-}\) is reduced to Mn⁺² and oxidation number of S increase from -2 to 0 so, H₂S is oxidised to S. Hence, writing partial equation as,

Oxidation number decreases 5 (Reduction)

Oxidation number increases by 2 (Oxidation)
Multiplying equation 2 by 2 and equation 3 by 5, we get
\(2 \mathrm{MnO}_{4}^{-}\) ➝ 2Mn⁺² … (4)
5H₂S ➝ 5S …(5)
Adding equations 4 and 5, we get
2MnO₄^– + 5H₂S ➝ 2Mn²⁺ + 5S … (6)
Change (-1) (0)  (+4) (0)

(c) Addition of water molecules in the direction of deficiency of oxygen.
2MnO₄ + 5H₂ S ➝ 2Mn²⁺ + 5S + 2Mn²⁺+5S + 8H₂O

(d) To balance hydrogen atom, H⁺ions are added to the left hand side of the reaction.
5H₂S + \(2 \mathrm{MnO}_{4}^{-}\) + 6H⁺ ➝ 5S + 2Mn²⁺ + 8H₂O
This is balanced equation.

(ii) (a) Write unbalanced equation,

(b) Oxidation number of Cl decrases from +7 to +3, so Cl₂O₇ is reduced to \(\mathrm{ClO}_{2}^{-}\) and oxidation number of O increases from -1 to 0. So, H₂O₂ is oxidised to O₂. Hence, writing partial equations,

Oxidation number decreases by 4 (Reduction)

Multiplying equation 2 by 1 and equation 3 by 4 and also balancing chlorine in ag (2)
Cl₂O₇ ➝ \(2 \mathrm{ClO}_{2}^{-}\) …(4)
4H₂O₂ ➝ 4O₂ •••(5)
Adding equation (4) and (5), we get
C1₂O₇ + 4H₂O₂ ➝ 2ClO₂ + 4O₂
Change (0) (0)  (-1)0

(c) Addition of water molecular in the direction of deficiency of oxygen and addion of OH^– ions to left hand side
Cl₂O₇ + 4H₂O₂ + 2OH^– ➝ 2ClO₂^– + 4O₂ + 5H₂O
This is balanced equation.

Question 32.
Write the step of balancing equation by Ion Electron method and balance the following reactions :
(i) Al + NO₃^– ➝ Al(OH)₄^– + NH₃ (Basic)
(ii) MnO₄^– + Br^– ➝ Mn²⁺ + Br₂ (Acidic)
(iii) Cr₂O₇^2- + Fe³⁺ ➝ Cr³⁺ + H₂O + Fe³⁺(Acidic)
Answer:
(i) (a) The unbalanced equation is :

First half reaction
A1 ➝ Al(OH)₄^– (Oxidation)
(a) To balance oxygen atoms, add 4H₂O to left hand side
Al + 4H₂O ➝  Al(OH)₄^–

(b) To balance hydrogen atoms, add H+ to right side
A1 + 4 H₂O➝ Al (OH)₄^– + 4H⁺
Since, reaction is in basic medium, OH^– is added to left side
Al + 4H₂O ➝ 4 Al(OH)₄^– + 4H₂O

(c) To balance change, electrons are added to right side
Al + 4H₂O+ 4OH^– ➝ Al(OH)₄^– + 4H₂O+3\(\overline{e}\) …(1)

(a) To balance oxygen atoms, add 3H₂O to right side
NO₃^– ➝ NH₃ + 3H₂O

(b) NO₃^– + 9H₂O ➝ NH₃ + 3H₂O+ 9 OH^–

(c) To balance charge, add electron to left side
NO₃^–+ 9H₂O+ 8\(\overline{e}\) ➝ NH₃ + 3H₂O+ 9OH^– …(2)
Multiply eq. (1) By 8 and equation 2 by 3 and add, we get equation.

(ii) \(\mathrm{MnO}_{4}^{-}\) + Br^– ➝ Mn²⁺ + Br₂ (Acidic medium)

To balance oxygen atoms, H₂O is added on right side.
\(\mathrm{MnO}_{4}^{-}\) ➝ Mn²⁺ + 4H₂O

(b) To balance hydrogen, H+ ions are added to left side
\(\mathrm{MnO}_{4}^{-}\) + 8H⁺ ➝ Mn²⁺ + 4H₂O

(c) To balance charge, electrons are added left side
\(\mathrm{MnO}_{4}^{-}\) + 8H⁺ + 5e^– ➝ Mn²⁺ + 4 H₂O …(1)

or 2Br^– ➝ Br₂
To balance charge, 2\(\overline{e}\) are added right hand side
2Br^– ➝ Br₂ + 2\(\overline{e}\) …(2)
Now adding the eq. (1) and (2)

This is balanced equation.

(iii) (a) Balancing Cr, \(\mathrm{CrO}_{7}^{2-}\) ➝ 2Cr³⁺

(b) to balance oxygen atoms, 7H₂O is added to the right side.
\(\mathrm{CrO}_{7}^{2-}\) ➝ Cr³⁺ + 7H₂O

(c) To balance H atom H⁺ are added to the left side.
\(\mathrm{CrO}_{7}^{2-}\) + 14H⁺ ➝ 2Cr³⁺ + 7H₂O

(d) To balance charge, electrons are added to the left
side. \(\mathrm{CrO}_{7}^{2-}\) + 14H⁺\(6 \overline{e}\) ➝ 2Cr³⁺+7H₂O …(1)

To balance charge, electrons are added to the right side.
Fe²⁺ ➝ Fe³⁺ + e^– …(2)
Multiply equation (1) by 1 and equation 2 by 6 and add, we get.

This is balanced equation.

Question 33.
Draw the diagram of Galvanic cell which represents the following reaction :
Zn + 2Ag⁺ ➝  Zn²⁺+ 2 Ag
Answer the following question :
(a) Give equation for the reaction taking place at each electrode.
(b) In which direction electrons in external circuit flows ?
(c) Identify amode and cathode.
(d) If E° Zn²⁺/Zn = 0.76 an E° Ag⁺/ Ag = +0. 80 volt, then whether E° =: 1.56 volt ?
Ans.

(a) Reaction taking place at anode :
Zn ➝> Zn²⁺ + 2e^–
Reaction taking place at cathode-
2Ag⁺ + 2e^– ➝ 2Ag

(b) Electrons in external circuit flow anode to cathode.

(c) Zn in anode and Ag in cathode.

(d) E°cell = E°cathode – E° anode
= E° Ag⁺ / Ag = E° Zn²⁺ / Zn
= 0. 80 – (- 0.76)
= 0.80 + 0.76 = 1.56 volt

Question 34.
What is electro chemical series ? The electrode potential of elements A, B, C and D are + 0. 79, -0. 74, + 1.08 and – 0.31 volts. Arrange them in increasing order of reactivity. To store 1 m HCl, which one aluminum or silver is suitable to use ? (E°Al³+ /Al = – 1.66V and E° Ag / Ag = +0.80V)
Answer:
Electro chemical series : The arrangement of elements in order of decreasing reduction potential values is called electro chemical series.

Element	Electrode potential (V)
A	+ 0.79
B	-0.74
C	+1.08
D	-0.31

Increasing order of reactivity of these elements :
C < A < D < B
Least reactive Most reactive
The reactivity of elements depends on value of standard electrode potential more negative the value of standard electrode potential of element, more active will be the element.

Aluminum is suitable to store lm HCl because its standard electrode potential is more negative than silver.

Question 35.
Balance the following reaction using Ion electron method :
(i) MnO₄^– + SO₃^2- -> Mn²⁺ + SO₄^2-
(ii) Cr₂O₇^2- + H⁺ + I^– -> Cr³⁺+ H₂O +I₂ (Acidic medium)
(iii) Cl₂ + OH^– -> Cl^– + ClO₃^–
(iv) N₂H₄ + ClO₃^– —> NO + Cl^– (Basic medium)
Answer:
(i) MnO₄^– + SO₃^2- —> Mn²⁺ + SO₄^2- (Acidic medium)

(a) To balance oxygen atoms, addH₂Oto right side.
MnO₄^– ➝ Mn²⁺ + 4H₂O

(b) To balance hydrogen atoms, add H⁺ ions to left side.
MnO₄^– + 8H⁺ ➝ Mn²⁺ + 4H₂O

(c) To balance charge, add electrons to left side.
MnO₄^–+ 8H⁺ + +5e^– ➝ Mn²⁺ + 4H₄O …(1)

(a) To balance oxygen atom, add H₂O to left side.
SO₃^2- + H₂O ➝ SO₄^2-

(b) To balance hydrogen atom, add H+ ions to right side.
SO₃^2- + H₂O ➝ SO₄^2- + 2H⁺

(c) To balance charge, add electron to right side.
SO₃^2- + H₂O ➝ SO₄^2- + 2H⁺ + 2e^– …(2)
Multiply equation 1 by 2 and equation 2 by 5, and add, we get.

This is balanced equation.

(ii) Cr₂O₇^2- + H⁺ + I^– ➝ Cr³⁺ + H₂O+ I₂

Balancing Cr, Cr₂O₇^2-  ➝ 2Cr³⁺

(a) To balance oxygen atoms, add H₂O to right side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) ➝ 2Cr³⁺ + 7H₂O

(b) To balance hydrogen atoms, add H⁺ ions to left side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) + 14H⁺ ➝ 2Cr³⁺ + 7H₂O

(c) To balance change, add electrons to left side.
\(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) + 14H⁺ + 6e^–➝ 2Cr³⁺ +7H₂O …(1)

(a) Balancing I, 2I^– ➝ I₂

(b) To balance charge, electron is added to right side.
2I^– ➝ I₂ + 2e^– …(2)
Multiply equation (1) by 1 and equation 2 by 3 and add, we get

This is balanced equation.

(iii) Cl₂ + OH^– ➝ Cl^– + ClO₃^–

Balancing Cl, Cl₂ ➝ 2ClO₃

(a) To balancing oxygen atoms, H20 as added to left side.
Cl₂ +6H₂O ➝ 2ClO₃^–

(b) To balance hydrogen atom, H⁺ are added to right side.
Cl₂ + 6H₂O ➝ 2CIO₃^– + 12H⁺
∴ Reaction is in basic medium OH^– are added to left side
∴ Cl₂+6H₂O + 120H^– ➝ 2ClO₃^– + 12H₂O

(c) To balance charge, electrons are added to right side.
Cl₂ +6H₂O + 120H^– ➝ 2CO₃^– + 12H₂O + 10e^– …(1)

Balancing Cl, Cl₂ ➝ 2C1^–

(a) To balance charge, electrons are added to left side.
Cl₂ + 2e^– ➝ 2Cl^– …(2)
Multiply eq (1) by 1 and eq. (2) by 2 and add. we get

This is balanced equation.

(iv) N₂H₄ + ClO₃^– ➝ NO + Cl^– (Basic medium)

Balancing N, N₂H₄ ➝ 2NO

(a) To balance oxygen atoms, add H₂O to left side.
N₂H₄ + 2H₂O ➝ 2NO

(b) To balance hydrogen atoms, add H⁺ ions to right side.
N₂H₄ + 2H₂O ➝ 2NO+ 8 H⁺
∴ Reaction is in basic medium, add OH^– to left side
N₂H₄ +2H₂O + 80H⁺ ➝ 2NO + 8H₂O + 8e^– …(1)

(a) To balance O-atoms, add H₂O to right side.
ClO₃^– ➝ Cl^– +3H₂O

(b) To balance hydrogen atoms, add H⁺ ions to left side.
ClO₃^– + 6H₃⁺ ➝ Cl^– + 3H₂O
∴ Reaction is in basic medium , add OH^– to right side
ClO₃^– + 6H₃O ➝ Cl^– + 3H₂O + 60H^–

(c) To balance charge, add electrons to left side.
ClO₃^– + 6H₂O + 6e ➝ Cl^– + 3H₂O + 60H^– …(2)
Multiply equation 1 by 3 equation 2 by 4 and add, we get.

This is balanced equation.

Question 36.
Balance the following reaction using Ion Electron method,
(i) \(\mathrm{NO}_{3}^{-}\) + H⁺ + I^–  ➝ NO+ H₂O + I₂
(ii) CrO₄^2- + SO₃^2- + OH^– ➝ CrO₂^2- + \(\mathrm{SO}_{4}^{2-}\)
(iii) \(\mathrm{MnO}_{4}^{-}\) + Fe₃O₄ + OH^– ➝ Fe₂O₃ + MnO₂
(iv) P₄ + OH^– ➝ PH₃ + \(\mathrm{HPO}_{2}^{-}\)
Answer:
(i)  \(\mathrm{NO}_{3}^{-}\) + H⁺ +I^– ➝ NO + H₂O +I₂

(a) To balance oxygen atoms add H₂O to right side.
\(\mathrm{NO}_{3}^{-}\) ➝ NO + 2H₂O

(b) To balance hydrogen atoms add H⁺ ions to left side.
\(\mathrm{NO}_{3}^{-}\) + 4H⁺ ➝ NO + 2H₂O

(c) To balance charge, add electrons to left side.
\(\mathrm{NO}_{3}^{-}\) + 4H⁺ + 3e^– ➝ NO + 2H₂O …. (1)

Balancing I  2I^– ➝ I₂

(a) To balancing charge, electrons are added to right side.
2I^– ➝ I₂ + 2e^–
Multiply eq (1) by 2 and eq. (2) by 3 and add, we get

This is balanced equation.

(ii)  \(\mathrm{CrO}_{4}^{2-}\) + SO₃^2- + OH^– ➝ \(\mathrm{CrO}_{2}^{2-}\) + SO₄^2-

(a) To balance oxygen atoms add H₂O to right side.
\(\mathrm{CrO}_{4}^{2-}\) ➝ \(\mathrm{CrO}_{2}^{2-}\) + 2H₂O

(b) To balance hydrogen atoms add H⁺ ions to left side.
\(\mathrm{CrO}_{4}^{2-}\) + 4H⁺ ➝ \(\mathrm{CrO}_{2}^{2-}\) + 2H₂O

(c) To balance charge, electrons are added to right side.
\(\mathrm{SO}_{3}^{2-}\) + 20H^–  ➝ \(\mathrm{SO}_{4}^{2-}\) + 2H₂O + 2e^– … (2)
Multiply eq 1 by 1 and eq. 2 by 2 and add, we get

This is balanced equation.

(iii) \(\mathrm{MnO}_{4}^{-}\) + Fe₃O₄ + OH^– ➝ Fe₂O₃ + MnO₂

(a) To balance oxygen atoms add H₂O to right side.
\(\mathrm{MnO}_{4}^{-}\) ➝ MnO₂ + 2H₂O
(b) To balance hydrogen atoms add H⁺ ions to left side.
\(\mathrm{MnO}_{4}^{-}\) + 4H⁺ ➝ MnO₂ + 2H₂O
∴ Reaction is in basic medium, OH^–  ions are added to right side
\(\mathrm{MnO}_{4}^{-}\) + 4H₂O ➝ MnO₂ + 2H₂O + 4OH^–
(c) To balance charge, electrons are added to right side.
\(\mathrm{MnO}_{4}^{-}\) + 4H₂O + 3e^– ➝  MnO₂ + 2H₂O + 4OH^– … (1)

Balancing Fe, 2Fe₃O₄ ➝ 3Fe₂O₃

(a) To balance oxygen atoms, add H₂O to left side.
2Fe₃O₄ + H₂O ➝ 3Fe₂O₃

(b) To balance hydrogen atoms, add H⁺ ions to right side.
2Fe₃O₄ + H₂O  ➝ 3Fe₂O₃ + 2H₂O
∴ Reaction is in basic medium, so add OH^– ions to left side.
2Fe₃O₄ + H₂O + 20H^–  ➝ 3Fe₂O₃ + 2H₂O

(c) To balance charge, add electrons to right side,
2Fe₃O₄ + H₂O + 20H^–  ➝ 3Fe₂O₃ + 2H₂O + 2e^– … (2)
Multiply eq. (1) by 2 and eq. (2) by 3 and add, we get

This is balanced equation.

(iv) P₄ + OH^–  ➝ PH₃ + HPO₂^–

Balancing P, P₄ ➝ 4PH₃
(a) To balance H-atoms, add H⁺ ions to left side.
P₄ + 12H⁺ ➝  4PH₃
∴ Reaction is in basic medium, add OH” ions to right side.
P₄ + 12H₂O ➝ 4PH₃ + 120H^–

(b) To balance charge, electrons add to left side.
P₄ +12H₂O + 12e^– ➝ 4PH₃ + 120H^– …( 1)

Balancing P, P₄ ➝ 4HPO₂^–

(a) To balance hydrogen atoms, add H₂O to left side.
P₄ + 8H₂O ➝ 4HPO₂^–

(b) To balance hydrogen atom, add H⁺ ions to right side.
P₄ + 8H₂O ➝ 4HPO₂^– + 12H⁺
∴ Reaction is in basic medium, add OH^– to left side
P₄ + 8H₂O + 120H^– ➝ 4 HPO₂^– +12 H₂O

(c) To balance charge, electrons are added to right side.
P₄ + 8H₂O + 120H^– ➝ 4HPO₂^– + 12H₂O + 8e^– …(2)
Multiply eq. (1) by 2 and eq (2) by 3, and add we get.

This is balanced equation.

Question 37.
Balance the following reactions using Ion Electron method :
(i) MnO₄^– + H₂O₂ ➝ MnO₂ + O₂ + OH^–
(ii) As O₃^3- + H₂O + I₂ ➝ ASO₄^3- + H⁺ + I^–
(iii) Cl₂O₇ +  H₂O₂ ➝ ClO₂^– + O₂ + H⁺
(iv) Cr₂O₇^2-+SO₂ ➝ Cr³⁺ + SO₄^2-
Answer:
(i) MnO₄^– + H₂O₂ ➝ MnO₂ + O₂ + OH^–

(a) To balance oxygen atoms, add H₂O to right side
MnO₄^– ➝ MnO₂ + 2H₂O

(b) To balance hydrogen atoms, add H⁺ ions to left side
MnO₄^– + 4H⁺ ➝ MnO₂ + 2H₂O
∵ The reaction is in basic medium, OH^– are added to right side.
MnO₄ + 4H₂O ➝ MnO₂ + 2H₂O + 4OH^–

(c) To balance charge, electrons are added to left side
MnO₄^– + 4H₂O➝ MnO₂ + 2H₂O + 4OH^– …(1)

(a) To balance hydrogen atoms, add H⁺ ions to right side
H₂O₂ ➝ O₂ + 2H⁺
∵ Reaction is in basic medium, add OH^– ions to left side
H₂O₂ + 20H^{– ➝} O₂ + 2H₂O

(b) To balance charge, electrons are added to right side
H₂O₂ + 20H^– ➝ O₂ + 2H₂O + 2e^– …(2)
Multiply equation 1 by 2 and equation 2 by 3, and add, we get

This is balanced equation.

(ii) AsO₃^3- + H₂O + I₂ —> AsO₄^3- + H⁺ + I^–

(a) To balance oxygen atom, add H₂O to left side
AsO₃^3- + H₂O ➝ AsO₄^3-

(b) To balance hydrogen atoms, add H⁺ ions to right side
AsO₃^3- + H₂O ➝ AsO₄^3- + 2H⁺

(c) To balance charge, electrons are added to right side
AsO₃^3- + H₂O ➝ AsO₄^3- + 2H⁺ + 2e^– …(1)

Balancing I, I₂ ➝ 2I^–

(a) To balance charge, electrons are added to leftside
I₂ + 2e^– ➝ 2I^– …(2)
Add eq. (1) and (2)

This is balanced equation.

(iii) Cl₂O₇ + H₂O₂ ➝ ClO₂ + O₂ + H⁺

Balancing Cl, Cl₂O₇ ➝ 2ClO₂^–

(a) To balance oxygen atoms, add H₂O to right side
Cl₂O₇ ➝ 2ClO₂^– + 3H₂O

(b) To balance hydrogen atoms add H⁺ ion to left side
Cl₂O₇ + 6H⁺ ➝ 2ClO₂^– + 3H₂O

(c) To balance charge, electrons are added to left side
Cl₂O₇ + 6H⁺ + 8e^– ➝ 2ClO₂^– + 3H₂O …(1)

(a) To balance hydrogen atoms, add H⁺ ions to right side
H₂O₂ ➝ O₂ + 2H⁺

(b) To balance charge, add electrons to right side
H₂O₂ ➝ O₂ + 2H⁺ + 2e^– …(2)
Multiply eq. (1) by 1 and eq. (2) by 4, and add, we get

This is balanced equation.

(iv) Cr₂O₇^2- + H₂O ➝ Cr³⁺ + SO₄^2-

Balancing Cr, Cr₂O₇^2-  ➝ 2Cr³⁺ 

(a) To balance oxygen atoms, add H₂O to right side
Cr₂O₇^2-  ➝ 2Cr³⁺ + 7H₂O

(b) To balance hydrogen atoms add H⁺ ion to left side
Cr₂O₇^2- + 14H⁺ ➝ 2Cr³⁺ + 7H₂O

(c) To balance charge, electrons are added to left side
Cr₂O₇^2- + 14H⁺ + 6e^– ➝ 2Cr³⁺ + 7H₂O …(1)

(a) To balance oxygen atoms, add H₂O to left side
SO₂ + 2H₂O ➝ SO₄^2-

(b) To balance hydrogen atoms add H⁺ ions to right side
SO₂ + 2H₂O ➝ SO₄^2- + 4H⁺
(c) To balance charge, add electrons to right side.
SO₂ + 2H₂O ➝ SO₄^2- + 4H⁺ + 2e^– … (2)
Multiply eq. (1) by 1 and eq (2) by 3, and add, we get

This is balanced equation.

Question 38.
Calculate oxidation number of the following compounds :
(i) Fe in FeSO₄
(ii) Fe in Fe(CO)₅
(iii) P in H₃PO₃
(iv) S in H₂S₂O₇
(v) C in C₁₂H₂₂O₁₁
(vi) P in NaH₂PO₂
Answer:
(i) Suppose oxidation number of Fe in FeSO₄ = x
x + 1(-2) = 0
x – 2 = 0
x = +2
∴ Oxidation number of Fe FeSO₄ = +2

(ii) Suppose, oxidation number of Fe in Fe(CO)₅ = x
x + 5(0) = 0
x + 0 = 0
x = 0
∴ Oxidation number of Fe Fe(CO)₅ = 0

(iii) Suppose, oxidation number of P in H₃PO₃ = x
3(+1) + x + 3(-2) = 0
3 + x – 6 = 0
x – 3 = 0
x = +3
∴ Oxidation number of P in H₃PO₃ = +3

(iv) Suppose, oxidation number of S in H₂S₂O₇ = x
2(+1) + 2x + 7(-2) = 0
2 + 2x – 14 = 0
2x – 12 = 0
2x = 12
x\(=\frac{12}{2}\) = 6
∴ Oxidation number of S in H₂S₂O₇ = +6

(v) Suppose, oxidation number of C in C₁₂H₂₂O₁₁ = x
12x + 22(+1) + 11(-2) = 0
12x + 22 – 22 = 0
12x + 0 = 0
12x = 0
x = 0
∴ Oxidation number of C in C₁₂H₂₂O₁₁ = 0

(vi) Suppose, oxidation number of C in P in NaH₂PO₂ = x
1(+1) + 2(+1) + x + 2(-2) = 0
1 + 2 + x – 4 = 0
x – 1 = 0
x = +1
∴ Oxidation number of C in P in NaH₂PO₂ = +1

Question 39.
Identify oxidising and reducing agent in the following:
(i) 3I₂ + NaOH ➝ NaIO₃ + 5NaI + 3H₂O
(ii) AlCl₃ + 3K ➝ Al + 3KCl
(iii) SO₂ + 2H₂S ➝ 3S + 2H₂O
(iv) SnCl₂ + 2FeCl₃ ➝ SnCl₄ + 2FeCl₂
(v) H₂O₂ + H₂O₂ ➝ 2H₂O + O₂
Answer:

In this reaction, iodine (I₂) is oxidised to NaIO₃ and also reduced to Nal. So, it acts as both oxidising and reducing agent.

In this reaction, potassium (K) reduces AlCl₃ to Al, So it acts as reducing agent and AlCl₃ oxidises K to KCl, so it acts as oxidising agent
Oxidising agent = AlCl₃
Reducing agent = K

In this reaction, SO₂ oxidises H₂S to S, so it acts as oxidising agent and H₂S reduces SO₂ to S, so it acts as reducing agent.
SO₂ = Oxidising agent
H₂S = Reducing agent

In this reaction, FeCl₃ oxidises SnCl₂ to SnCl₄, so it acts as oxidising agent and SnCl₂ reduces FeCl₃ to FeCl₂, so it acts as reducing agent.
FeCl₃ = Oxidising agent
SnCl₂ = Reducing agent

In this reaction, one molecule of H₂O₂ is oxidised to O₂ and another molecule of H₂O₂ is reduced to H₂O. So, H₂O₂ acts as both oxidising and reducing agent.

Question 40.
In which of the following reaction, H₂O₂ acts as oxidising agent and in which it acts as reducing agent?
(i) 2KI+ H₂O₂ ➝ 2KOH + I₂
(ii) Cl₂ + H₂O₂ ➝ 2HCl + O₂
Answer:

In this reaction, H₂O₂ oxidises KI to I₂ so it acts as oxidising agent.

In this reaction, H₂O₂ reduces Cl₂ to HCl, so it acts as reducing agent.

RBSE Solutions for Class 11 Chemistry