AP Inter 1st Year Chemistry Study Material Chapter 5 Stoichiometry

Andhra Pradesh BIEAP AP Inter 1st Year Chemistry Study Material 5th Lesson Stoichiometry Textbook Questions and Answers.

AP Inter 1st Year Chemistry Study Material 5th Lesson Stoichiometry

Very Short Answer Questions

Question 1.
How many number of moles of glucose are present in 540 gms of glucose ?
Answer:
Formulae:
No. of moles = \(\frac{\text { Weight }}{\text { G.M. Wt }}\)
= \(\frac{540}{180}\)
= 3
Given
Weight of glucose = 540 gms
G.M. wt = 180 (C₆H₁₂O₆)
No. of moles of glucose in 540 gms = 3 moles.

Question 2.
Calculate the weight of 0.1 mole of sodium carbonate.
Answer:
Formulae:
No. of moles = \(\frac{\text { Weight }}{\mathrm{GMWt}}\), 0.1 = \(\frac{\text { Weight }}{106}\)
Weight = 10.6 × 0.1 = 10.6 gms
Given
No. of moles = 0.1
G.M. Wt of Na₂CO₃ = 106

Question 3.
How many molecules of glucose are present in 5.23g of glucose (Molecular weight of glucose 180 u).
Answer:
Formulae:
No. of molecules = No. of moles × Avagadro no
No. of moles = \(\frac{\text { Weight }}{\text { GM Wt }}\) = \(\frac{5.23}{180}\) = 0.02906 moles
No. of molecules = 0.02906 × 6.023 × 10²³
= 1.75 × 10²² molecules
Given
Wt of glucose = 5.23 gms
G.M. Wt = 180

Question 4.
Calculate the number of molecules present in 1.12 × 10^-7 c.c. of a gas at STP (c.c. cubic centimeters = cm³).
Answer:
1 mole of any gas occupies 22400 cc of volume
1 mole of any gas contains 6.023 × 10²³ molecules
∴ 22400 cc of gas at STP contains 6.023 × 10²³ molecules
1.12 × 10^-7 cc of a gas at STP contains ——–?
\(\frac{1.12 \times 10^{-7} \times 6.023 \times 10^{23}}{22400}\) = 3.015 × 10¹² molecules

Question 5.
The empirical formula of a compound is CH₂O. Its molecular weight is 90. Calculate the molecular formula of the compound. (A.P. Mar. ‘16) (Mar. 13)
Answer:
Formulae:

∴ Molecular formula = 3(CH₂O) = Cr (NO₃)_3(aq) + Pb_(s).
Given
Molecular wt = 90
Emperical formula = CH₃O
∴ Emperical wt = 30

Question 6.
Balance the following equation by the oxidation number method
Cr_(S) + Pb(NO₃)_2(aq) → Cr (NO₃)_3(aq) + Pb_(s).
Answer:

So the balanced equation is 2Cr + 3Pb(NO₃)₂ → 2Cr(NO₃)₃ + 3Pb

Question 7.
What volume of H₂ at STP is required to reduce 0.795 g of CuO to give Cu and H₂O.
Answer:
Balance equation is
CuO + H₂ → Cu + H₂O
79.5 gms → 1 mole H₂ gas for reduction → 22.4 lit of volume at STP
∴ 79.5 gms of CuO require 22.4 lit of H₂ at STP
Then 0.795 gms of CuO require ——-?
\(\frac{0.795 \times 22.4}{79.5}\) = 0.224 lit.

AP Inter 1st Year Chemistry Study Material Chapter 5 Stoichiometry

Question 8.
Calculate the volume of O₂ at STP required to completely burn 100 ml of acetylene.
Answer:
Balanced chemical equation for combustion of acetylene is
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
2 moles of C₂H₂ require 5 moles of O₂ for complete combustion at STP
2 × 22400 ml of C₂H₂ require 5 × 22400 ml of O₂ at STP
Then 100 ml of C₂H₂ require ————?
\(\frac{100 \times 5 \times 22400}{2 \times 22400}\) = \(\frac{500}{2}\) = 250 ml

Question 9.
Now-a-days it is thought that oxidation is simply decrease in electron density and reduction is increase in electron density. How would you justify this?
Answer:
Oxidation : The process of removing electrons from an element is called oxidation. It is also called
as de-electronation.

• Oxidation means decrease in electron density
  e.g. : 2Cl^– → Cl₂ + 2\(e^{\ominus}\)
• Oxidation process occurs at anode in the electrolysis process.
• Reduction : The process of addition of electrons to an element is called reduction. It is also called as electronation.
• Reduction means increase in electron density.
  e.g.: Mg²⁺ + 2e^– → Mg
• Reduction process occurs of cathode in the electrolysis process.

Question 10.
What is a redox concept? Give an example.
Answer:
Redox reactions are the reactions in which reduction and oxidation both takes place.
e.g.:

Question 11.
Calculate the mass per cent of the different elements present in sodium sulphate (Na₂SO₄).
Answer:
Given compound is sodium sulphate (Na₂SO₄)
Molecular weight of the compound = 2(23) + 1(32) + 4(16)
= 46 + 32 + 64
= 142

Step – I:
Mass percent of Na
142 gms of Na₂SO₄ → 46 gms of Na
100 gms of Na₂SO₄ → ?
\(\frac{100 \times 46}{142}\) = 32.39%

Step – II:
Mass percent of ‘S’
142 gms of Na₂SO₄ → 32 gms of ‘S’
100 gms of Na₂SO₄ → ?
\(\frac{100 \times 32}{142}\) = 22.53%

Step-III:
Mass percent of ‘0’
142 gms of Na₂SO₄ → 64 gms of oxygen
100 gms of Na₂SO₄ → ?
\(\frac{100 \times 64}{142}\) = 45.07%
∴ Mass percents of Na, S, O are 32.39, 22.53, 45.07.

Question 12.
What do you mean by significant figures?
Answer:
The meaningful digits which are known with certainty are called significant figures.

• The uncertainty in the experimental (or) calculated values is indicated by mentioning the number of significant figures.

Question 13.
If the speed of light is 3.0 × 10⁸ ms^-1. Calculate the distance covered by light in 2.00 ns.
Answer:
Given the speed of light 3 × 10⁸ m/sec.
i.e., 1 sec → 3 × 10⁸ m
In 2 nano seconds → ?
2 × 10^-9 sec
\(\frac{2 \times 10^{-9} \times 3 \times 10^8}{1}\) = 6 × 10^-1 = 0.6 m

Short Answer Questions

Question 1.
The approximate production of sodium carbonate per month is 424 × 10⁸ g, while that of methyl alcohol is 320 × 10⁶ g. Which is produced more in terms of moles?
Answer:
Given that weight of Na₂CO₃ produced per month = 424 × 10⁶ gms
No. of moles of Na₂CO₃ = \(\frac{w t}{\text { GMwt }}\) = \(\frac{424 \times 10^6}{106}\) = 4 × 10⁶ moles per month
Given weight of CH₃OH produced per month = 320 × 10⁶ gms
No. of moles of CH₃OH = = \(\frac{320 \times 10^6}{32}\) = 10⁷ moles per month
Methyl Alcohol is produced more (per month) in terms of moles.

Question 2.
How much minimum volume of CO at STP is needed to react completely with 0.112 L of O₂ at 1.5 atm pressure and 127°C to give CO₂.
Answer:
Formula \(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\)
At STP
P₁ = 1 atm
V₁ = ?
T₁ = 0°C = 273 K
Given P₂ = 1.5 at m
V₂ = 0.112 lit
T₂ = 127°C = 400 K
\(\frac{1 \times V_1}{273}\) = \(\frac{1.5 \times 0.112}{400}\)
V₁ = \(\frac{1.5 \times 0.112 \times 273}{400}\)
= 0.1147 lit
Chemical equation
2CO + O₂ → 2CO₂
2 moles of CO → 1 mole of O₂ at STP
2 × 22.4 lit → 1 × 22.4 lit
? ← 0.1147 lit
\(\frac{0.1147 \times 2 \times 22.4}{22.4}\) = 0.2294 lit
The volume of CO required at STP = 0.2294 lit = 229.4 ml

Question 3.
Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present, carbon = 10.06%, hydrogen = 0.84%. chlorine = 89.10%. Calculate the empirical formula of the compound.
Answer:

∴ Emperical formula of given compound = C₁H₁Cl₃
= CHCl₃

Question 4.
A carbon compound on analysis gave the following percentage composition, carbon 14.5%, hydrogen 1.8%, chlorine 64.46%, oxygen 19.24%. Calculate the empirical formula of the compound.
Answer:

∴ The formula of the compound = C₁ H_1.5 Cl_1.5 O₁
Emperical formula of the compound = C₂ H₃ Cl₃ O₂

Question 5.
Calculate the empirical formula of a compound having percentage composition : Potassium (K) = 26.57, chromium (Cr) = 35.36; oxygen (0) 38.07. (Given the atomic weights of K. Cr and O as 39; 52 and 16 respectively).
Answer:

Formula of the compound = K₁ Cr₁ O_3.5
Emperical formula of the compound = K₂ Cr₂ O₇

Question 6.
A carbon compound contains 12.8% carbon, 2.1% hydrogen, 85.1% bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula.
Answer:

∴ Emperical formula of the compound = C₁ H₂ Br
Molecular formula = n (Emperical formula)

Given molecular wt = 187.9
Emperical wt = 94 (CH₂ Br)
∴ Molecular formula = 2 (CH₂Br)
= C₂H₄Br₂

Question 7.
0.188 g of an organic compound having an empirical formula CH₂Br displaced 24.2 cc. of air at 14 °C and 752 mm pressure. Calculate the molecular formula of the compound. (Aqueous tension at 14°C is 12mm).
Answer:
Given Emperical formula CH₂Br
Wt. of compound = 0.188 gms
Volume of displaced air = 24.2 CC
Temperature = 14°C = 287 K
Pressure = 752 mm.
STP conditions
P₁ = 760 mm
v₁ = ?
T₁ = 273 K
Given conditions
P₂ = Pressure – aqueous tension
= 752 – 12mm = 740 mm
V₂ = 24.2K
T₂ = 287 K

Formula

\(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\) ⇒ \(\frac{760 \times V_1}{273}\) = \(\frac{740 \times 24.2}{287}\)
V₁ = \(\frac{740 \times 24.2}{287} \times \frac{273}{760}\) = 22.414ml (or) 22.414 CC

∴ 0.188 gms of organic compound displaced 22.414 CC of air gms of organic compound displaced 22400 CC of air
\(\frac{22400}{22.414} \times 0.188\) = 188 ∴ Gram molecular wt of compound = 188
Molecular formula = n (Emperical formula)
= n (CH₂Br)
n = \(\frac{\text { Molecular wt }}{\text { Emperical wt }}\) = \(\frac{188}{94}\) = 2 ∴ Molecular formula = C₂ H₄ Br₂

Question 8.
Calculate the amount of 90% H2S04 required for the preparation of 420 kg HCl.
2 NaCl + H₂SO₄ → Na₂SO₄ + 2 HCl
Answer:
Given equation is
2NaCl + H₂SO₄ → 4 Na₂SO₄ + 2HCl
1 Mole H₂SO₄ → 2 Moles HCl
98 gms of H₂SO₄ → 2 × 36.5 = 73 gms of HCl
? → 420 × 10³ gms of HCl
\(\frac{420 \times 10^3 \times 98}{73}\) = 563.84 × 10³ gms
For 100% H₂SO₄ → 563.84 × 10³
For 90% H₂SO₄ → 563.84 × 10³ × \(\frac{100}{90}\)
= 626.5 × 10³ gms of H₂SO₄
∴ 626.5 × 10³ gms of 90% H₂SO₄ is required to produce 420 × 10³ gms of HCl

Question 9.
An astronaut receives the energy required in his body by the combustion of 34g of sucrose per hour. How much oxygen he has to carry along with him for his energy requirement in a day ?
Answer:
Given sucrose quantity required per hour = 34 gms
Sucrose quantity required per a day = 34 × 24 gms
Chemical equation of combustion of sucrose
C₁₂ H₂₂ O₁₁ + 12O₂ → 12CO₂ + 11 H₂O + Energy
1 mole sucrose → 12 moles of Oxygen
342 gms of sucrose → 12 × 32 gms of oxygen
34 × 24 gs of sucrose → ?
\(\frac{34 \times 24}{342}\) × 12 × 32 = 916.21 gms
∴ The astronaut require 916.21 gms of Oxygen per a day

Question 10.
What volume of CO₂ is obtained at STP by heating 4g of CaCO₃ ?
Answer:
Chemical Equation is
CaCO₃ CaO + CO₂
1 mole CaCO₃ → 1 mole CO₂ at STP
100 gms of CaCO₃ → 22.4 lit of CO₂ at STP
4gms of CaCO₃ → ?
\(\frac{4 \times 22.4}{100}\) = 0.896 lit

Question 11.
When 50g of a sample of sulphur was burnt in air 4% of the sample was left over. Calculate the volume of air required at STP containing 21% oxygen by volume.
Answer:
Given that when a 50 gms of sample of sulphur was burnt 4% of the sample was left over.
∴ 50 gms of S → 48 gms of ‘S’ burnt.
Chemical equation is S + O₂ → SO₂
32 gm of ‘S’ 22.4 lit of O₂ at STP
48 gms of S →
100 ml air contains 21 ml of oxygen
100 lit of air contains 21 lit of oxygen
33.6 lit of oxygen is present in \(\frac{33.6 \times 100}{21}\) = 160 lit
∴ The volume of air needed for combustion = 160 lit

Question 12.
Calculate the volume of oxygen gas required at STP conditions for the complete combustion of 10 cc of methane gas at 20°C and 770 mm pressure.
Answer:
Given 10 cc of methane (CH₄) undergoes combustion at 20°C and 770 mm pressure
STP Conditions
P₁ = 760 mm
V₁ = ?
T₁ = 273 k
Given conditions
P₂ = 770 mm
V₂ = 10 cc
T₂ = 20°C = 293 k

Formulae
\(\frac{P_1 V_1}{T_1}\) = \(\frac{\mathrm{P}_2 \mathrm{~V}_2}{\mathrm{~T}_2}\)
\(\frac{760 \times V_1}{273}\) = \(\frac{770 \times 10}{293}\)
V₁ = \(\frac{770 \times 10 \times 273}{293 \times 760}\) = 9.44 cc ∴ Volume of methane (CH₄) at STP = 9.44 cc.

Chemical equation is
CH₄ + 2O₂ → CO₂ + 2H₂O
1 mole CH₄ → 2 moles of oxygen
22,400 cc of CH₄ → 2 × 22400 cc of oxygen
9.44 cc of CH₄ → ?
\(\frac{9.44}{22400}\) × 2 × 22400 = 18.88 cc ∴ Volume of oxygen gas required STP = 18.88 cc.

Question 13.
Calculate the volume of H₂ liberated at 27°C and 760 mm of Hg pressure by action by 0.6g of magnesium with excess of dil HCl.
Answer:
Chemical equation is
Mg + 2HCl → MgCl₂ +H₂
24 gms. of Mg → 1 mole of H₂ at STP
= 22.4 lit at STP
0.6 gms. of Mg → ?
\(\frac{0.6 \times 22.4}{24}\) = 0.56 lit = 560 ml

Formula
\(\frac{P_1 V_1}{T_1}\) = \(\frac{P_2 V_2}{T_2}\)
\(\frac{760 \times V_1}{300}\) = \(\frac{760 \times 560}{273}\)
V₁ = \(\frac{560 \times 300}{273}\) = 615.4 ml
∴ volume of H₂ liberated at 27°C and 760 mm of presence = 615.4 ml = 0.6154 Lit.

Given conditions
P₁ = 760 mm
V₁ = ?
T₁ = 27° mm

STP conditions
P₂ = 760 mm
V₂ = 560 ml
T₂ = 0° C = 273 k

Question 14.
Explain the role of redox reactions in titrimetre processes and galvanic cells.
Answer:
a) Redox reactions in titrimetric quantitative analysis : In titrimetric analysis the substance of known concentration is called the titrant and the substance being titrated is called the titrand. The standard solution is generally added from a long graduated tube called burette. The process of adding the standard solution until the reaction is just complete is called titration. The substance to be estimated is titrated. The point at which the titrand just completely reacts is called the equivalence point (or) the theoretical point (or) stoichiometric end point. In the redox reactions the completion of the titration is detected by a suitable method like

1. Observing a physical change, for example, the light pink colour of KMnO₄ titrations.
2. By using a reagent known as indicator which gives a clear visual change like colour change, formation of turbidity etc. The point at wh’ch this is observed is called the end point of the titration which should coincide with theoretical end point.

Examples:

1. In Cr₂\(\mathrm{O}_7^{2-}\) titrations dippenyl amine is used as indicator and at the end poirtt it produces intense blue colour due to oxidation by Cr₂\(\mathrm{O}_7^{2-}\).
2. In the titration of Cu⁺² with I^–
   2Cu²⁺ (aq) + 4I^– (aq) → Cu₂I₂ (s) + I₂ (aq)
   A redox reaction I₂ gives a deep blue colour with starch solution.
3. In the titration of I₂ (aq) S₂\(\mathrm{O}_3^{-2}\) as per the stoichiometric reduce equation.
   In this way redox reactions are taken as the basis for titrimetric analysis with Mn\(\mathrm{O}_4^{-}\), Cr₂\(\mathrm{O}_7^{-2}\) etc., as oxidising agents and S₂\(\mathrm{O}_3^{-}\) etc., as a reducing agents,

b) Redox Reactions – Galvanic cells : Redox reaction (i.e) cell reaction that takes place in gavamic cell is

The process of transfer of electrons from Zn(s) to Cu⁺² take place directly. To make this transfer indirectly, Zn rod is kept in ZnSO₄ solution in one beaker and in the other beaker CuSO₄ solution is taken and a copper rod is dipped in it. Now the redox reaction takes place in either of the beakers. Each beakers contains both oxidised and reduced forms of the respective species in the beakers containing CuSO₄ solution & Cu rod at the interface Cu and Cu⁺² and in the other beaker at the interface Zn and Zn⁺². The two forms i.e., oxidised and reduced forms of a species participating in oxidation and reduction half reactions is called redox couple. Both the beakers contains each a redox couple. The oxidised form and reduced form are separated by a vertical line (or) a slash that represent an interface, e.g.: Zn (s) / Zn⁺² (aq)

In the above arrangement, the two redox couples are represented by Zn⁺² / Zn and Cu⁺² / Cu.
halvamic cell is represented as : Zn/Zn²⁺//Cu⁺²/Cu.

Question 15.
Define and explain molar mass.
Answer:
Molar mass : The mass of one mole of any substance in gms is called its molar mass,
e.g. :

1. Molar mass of sulphuric acid = 98 g
2. Molar mass of hydrogen is one gram for gram atomic mass and two grams for gram molecular mass.

Question 16.
What are disproportionation reactions ? Give example. (T.S. Mar. ’16)
Answer:
Disproportionation Reaction : Reactions in which same element in the given form to undergo both oxidation and reduction simultaneously.

Here Cl₂ undergoes both oxidation and reduction reactions. So, above reaction is disproportionation reaction.

Question 17.
What are comproportionation reactions ? Give example.
Answer:
Comproportionation reactions : In these reactions, two species with the same element in two different oxidation states form a single product in which the element is in an intermediate oxidation state. Reverse of disproportionation is comproportionation.
e.g.:

Question 18.
Determine the empirical formula of an oxide of iron which has 69.9% iron and 30.1% dioxygen by mass.
Answer:

∴ Formula of compound = Fe₁O_1.5
Emperical formula = Fe₂O₃

Question 19.
Calculate the mass of sodium acetate (CH₃COONa) required to make 500 ml of 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol^-1.
Answer:
Formula
Molarity = \(\frac{\mathrm{Wt}}{\mathrm{GMwt}}\) × \(\frac{1000}{V(m l)}\)
Molarity = 0.375M
V = 500 ml
GMwt of CH₃COONa = 82.0245
0.375 = \(\frac{W t}{82.0245}\) × \(\frac{1000}{500}\)
Wt = \(\frac{82.0245 \times 0.375}{2}\)
= 15.3795 gms
∴ Mass of CH₃COONa required = 15.3795 gms

Question 20.
What is the concentration of sugar (C₁₂H₂₂O₁₁) in mol L^-1 if 20g are dissolved in enough water to make a final volume up to 2L ?
Answer:
Formula
Molarity = \(\frac{W t}{G M W t} \times \frac{1}{V(\text { lit })}\)
= \(\frac{20}{342} \times \frac{1}{2}\)
= 0.02924 M
Wt = 20 gms
GMWt = 342
[C₁₂H₂₂O_H]
V = 2 lit

Question 21.
How many significant figures are present in the following ?

1. 0.0025
2. 208
3. 5005
4. 126,000
5. 500.0
6. 2.0034

Answer:

1. 0.0025 has 2 significant figures
2. 208 has 3 significant figures
3. 5005 has 4 significant figures
4. 126000 has 3 significant figures
5. 500.0 has 4 significant figures
6. 2.0034 has 5 significant figures

Question 22.
Round up the following upto three significant figures :

1. 34.216
2. 10.4107
3. 0.04597
4. 2808

Answer:

1. 34.216 becomes 34.2
2. 10.4107 becomes 10.4
3. 0.04597 becomes 0.046
4. 2808 becomes 281

Question 23.
Calculate the molarity of a solution of ethanol in water in which the mole fraction of ethanol is 0.040 (assume the density of water to be one)
Answer:
The relation between Molarity and mole fraction is given by

Given data, x₂ = 0.040, M = ?
M₁ = 18(H₂O), M₂ = 46 (C₂H₅OH), Cl = 1
0040 = \(\frac{M \times 18}{M(18-46)+1000 \times 1}\) = \(\frac{M \times 18}{-28 M+1000}\)
18 M = -0.040 × M × 28 + 0.040 × 1000
18 M = -1.12M + 40
(18 + 1.12)M = 40
19.12 M = 40
M = \(\frac{40}{19.12}\) = 2 09 M ∴ Molarity of Ethanol = 2.09 M

Given data

Question 24.
A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g. calculate
(i) empirical formula,
(ii) molar mass of the gas and
(iii) molecular formula.
Answer:
Assuming 1 gm. of gas is burnt
The weight % of ‘C’ = \(\frac{3.38 \times 12}{44}\) × 100 = 92.18%
The weight % of ‘H’ = \(\frac{0.69 \times 2}{18}\) = 7.67%
i)

Experical formula of compound = C₁H₁

ii) Given 10 lit of gas at STP weighs – 11.6 gas
22.44 lit of gas at STP weights = \(\frac{22.4 \times 11.6}{10}\) = 25.984
∴ Molecular weight of given gas = 25.984

iii) Molecular formula = n(emperical formula)
n = \(\frac{\text { Mol. wt }}{\text { Emp. wt }}\) = \(\frac{25.984}{13}\) = 2 ∴ Molecular formula = 2(CH) = C₂H₂.

Question 25.
Calcium carbonate reacts with aqueous HCl to give CaCl₂ and CO₂ according to the reaction CaCO₃₍₅₎ + 2 HCl_(aq) → CaCl_2(aq) + CO_2(g) + H₂O (L). What mass of CaCO₃ is required to react completely with 25 mL of 0.75 M HCl?
Answer:
Given chemical reaction
CaCO_3(s) + 2HCl_(aq) → CaCl_2(aq) + CO_2(g) + H₂O_(l)
Molarity = \(\frac{\mathrm{Wt}}{\mathrm{GMWt}} \times \frac{1000}{\mathrm{~V}(\mathrm{~m})}\)
0.75 = \(\frac{\mathrm{Wt}}{36.5} \times \frac{1000}{25}\)
M = 0.75 M
W = ?
Wt = \(\frac{0.75 \times 36.5 \times 25}{1000}\)
= 0.6844 gms of HCl
GMwt = 36.5
V = 25 ml
From the equation
1 mole of CaCO₃ – 2 moles of HCl
100 gms of CaCO₃ – 2 × 36.5 gms of HCl
? – 0.6844 gms of HCl
∴ Mass of CaCO₃ = \(\frac{0.6844}{2 \times 36.5}\) × 100 = 0.937 5 gms

Question 26.
Chlorine is prepared in the laboratory by treating manganese dioxide (MnO₂) with aqueous hydrochloric acid according to the reaction
4 HCl (aq) + MnO₂(s) → 2 H₂O (l) + MnCl_2(aq) + Cl₂(g)
How many grams of HCl react with 5.0 g of manganese dioxide?
Answer:
Given chemical reaction is
4HCl_(aq) + MnO_2(s) → MnCl_2(aq) + 2H₂O_(l) + Cl_2(g)
4 moles of HCl – 1 mole of MnO₂
4 × 36.5 gms. of HCl – 87 gms. of MnO₂
∴ Weight of HCl required = \(\frac{5 \times 4 \times 36.5}{87}\) = 8.39 gms.

Question 27.
To 50 ml of 0.1 N Na₂CO₃ solution 150 ml of H₂O is added. Then calculate a normality of resultant solution.
Answer:
Formulae N₁V₁ = N₂V₂
0.1 × 50 = N₂ × 200
N₂ = \(\frac{0.1}{4}\)
= 0.025 N
N₁ = 0.1 N
V₁ = 50 ml
N₂ = ?
V₂ = 150 + 150 = 200 ml

Question 28.
Calculate the volume of 0.1 N H₂SO₄ required to neutralize 200 ml of 0.2 N NaOH solution. It is an acid base neutralisation reaction. Hence, at the neutralisation point. Number of equivalents of acid = Number of equivalents of base.
Answer:
Formula N₁V₁ = N₂V₂
0.1 × V₁ = 0.2 × V₂
V₂ = 2 × 200
= 400 ml
N₁ = 0.1 N
V₁ = ?
V₂ = 200 ml
N₂ = 0.2 N

Question 29.
Calculate normality of H₂SO₄ solutions if 50 ml of it completely neutralises 250 ml of 0.1 N Ba(OH)₂ solutions.
Solution:
Formulae N₁V₁ = N₂V₂
N × 50 = 0.1 × 250
N₁ = 0.5 N
N₁ = ?
V₁ = 50 ml
N₂ = 0.1 N
V₂ = 250 ml

Question 30.
Calculate the volume of 0.1 M KMnO₄ required to react with 100 ml of 0.1 M H₂C₂O₄.2H₂O solution in presence of H₂SO₄.
Answer:
Balanced chemical equation for the given data is
Formulae
\(\frac{M_1 V_1}{n_1}\) = \(\frac{M_2 V_2}{n_2}\)
\(\frac{0.1 \times V_1}{2}\) = \(\frac{0.1 \times 100}{8}\)
V₁ = 40 ml

KMnO₄
M₁ = 0.1 M
V₁ = ?
n₁ = 2

Oxalic acid

M₂ = 0.1 M
V₂ = 100 ml
n₂ = 5

Question 31.
Assign oxidation number to the underlined elements in each of the following species:
a) NaH₂PO₄
b) NaHSO₄
c) H₄P₂O₇
d) K₂MnO₄
e) CaO₂
f) NaH₄
g) H₂S₂O₇
h) KAl(SO₄)₂.12 H₂O
Answer:
a) NaH₂PO₄
1(+1) + 2(+1) + x + 4(-2) = 0
1 + 2 + x – 8 = 0
x – 5 = 0
x= + 5
Oxidation no. of ‘P in NaH₂PO₄ = + 5

b) NaHSO₄
1(+1) + 1(+1) + x + 4(-2) = 0
1 + 1 + x – 8 = 0
x = +6
Oxidation no. of ‘S in NaHSO₄ = + 6

c) H₄P₂O₇
4(+1) + 2x + 7(-2) = 0
4 + 2x – 14 = 0
2x – 10 = 0.
x = +5
Oxidation no. of ‘P’ in H₄P₂O₇ is +5

d) K₂MnO₄
2(+1) + x + 4(-2) = 0
2 + x – 8 = 0
x = + 6
Oxidation no of Mn is K₂MnO₄ = +6

e) CaO₂
+ 2 + 2x = 0
2x = -2
x = -1
Oxidation no. of oxygen in CaO₂ = -1

f) NaBH₄
1(+1) + x + 4 (-1) = 0
1 + x – 4 = 0
x = +3
Oxidation no. of ‘B’ in NaBH₄ = + 3
But B’ most probably exhibits – 3 oxidation state.

g) H₂S₂O₇
2(1) + 2x + 7(-2) = 0
2 + 2x – 14 = 0
2x – 12 = 0
x = +6
Oxidation state of ‘S’ in H₂S₂O₇ = + 6

h) K Al(SO₄)₂ 12H₂O :
General formula of above compound is K₂SO₄ Al₂ (SO₄)₃ 2 + H₂O (Potash alum)
Consider Al₂(SO₄)₃ from the above double salt
2x + 3 (-2) = 0
2x – 6 = 0
x = + 3

Question 32.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?
a) KI₃
b) H₂SO₄O₆
c) Fe₃O₄
Answer:
a) KI₃
It is formed by the combining KI, I₂
∴ Oxidation no. of ‘I’ in KI = – 1
[1 + x = 0
x = – 1]
In I₂ oxidation no. of. I = 0

b) H₂S₄O₆
According to H₂S₄O₆ structure

2(1) + 4x + 6(-2) = 0 ⇒ 4x – 10 = 0
x = 2.5
‘S’ average Ox. No = 2.5

c) Fe₃O₄
3x + 4(-2) = 0
3x – 8 = 0
x = \(\frac{8}{3}\)
In general Fe₃O₄ obtained by FeO + Fe₂O₃
∴ In FeO → x = + 2 [x – 2 = 0 ⇒ x = + 2]
In Fe₂O₃ → x = + 3 [2x – 6 = 0 ⇒ x = + 3]

Question 33.
Justify that the following reactions are redox reactions :
Answer:
a) CuO(s) + H₂(g) → Cu(s) + H₂O(g)
b) Fe₂O₃(s) + 3CO(g) → 2 Fe(s) + 3 CO₂(g)
c) 4 BCl₃(g) + 3 LiAlH₄(s) → 2 B₂H₆(g) + 3 LiCl(s) + 3 AlCl₃(s)
d) 2 K(s) + F₂(g) → 2 K⁺F^–(s)
e) 4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(g)
a)

Hence it is a Redox reaction.

b)

Hence it is a Redox reaction.

c) \(4 \mathrm{BCl}_{3_{(g)}}\) + \(3 \mathrm{LiAlH}_{4_{(s)}}\) → \(2 \mathrm{~B}_2 \mathrm{H}_{6_{(g)}}\) + \(3 \mathrm{LiCl}_{(s)}\) + \(4 \mathrm{AlCl}_{3_{(\mathrm{s})}}\)
Here in the above reaction the oxidation states of all elements doesnot changed. So it is not a redox reaction.

d)

Hence it is a redox reaction.

e)

Hence it is a redox reaction.

Question 34.
Fluorine reacts with ice and results in the change
H₂O(S) + F₂(g) → HF(g) + HOF(g)
Justify that this reaction is a redox reaction.
Answer:
Given chemical equation

Hence it is a redox reaction.

Question 35.
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂\(O_7^{-2}\) and \(\mathrm{NO}_3^{-}\). Suggest structure of these compounds.
Answer:
a) H₂SO₅ Structure:

2(1) + x + 2(-1) + 3(-2) = 0 (One peroxy linkage)
2 + x – 2 – 6 = 0
x = +6

b) Cr₂\(\mathrm{O}_7^{-2}\) Structure:

2(x) + 7(-2) = -2
2x – 14 = -2
x = +6

c) \(\mathrm{NO}_3^{-}\) Structure:

x + 3(-2) = -1
x – 6 = -1
x = +5

Question 36.
Write formulas for the following compounds: .
a) Mercury (II) chloride
b) Nickel (II) sulphate
c) Tin (IV) oxide
d) Thallium (I) sulphate
e) Iron (III) sulphate
f) Chromium (III) oxide
Answer:
Formula
a) Mercury (II) Chloride — HgCl₂
b) Nickel (II) Sulphate — NiSO₄
c) Tin (IV) Chloride — SnCl₄
d) Thallium (I) Sulphate — Tl₂SO₄
e) Iron (III) Sulphate — Fe₂(SO₄)₃
f) Chromium (III) Oxide — Cr₂O₃

Question 37.
Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.
Answer:

Question 38.
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:
SO₃ and H₂O₂ both acts as oxidising as well as reducing agents.

• Ozone acts as strong oxidising agent in acidic medium (Ozone acts as reducing agent also in some cases).
• HNO₃ acts as strong oxidising agent because in HNO₃. He H⁺ ion readily released and the oxidation state of ‘N’ is +5 in HNO₃. So it is a good oxidising agent.

Question 39.
Consider the reactions :
a) 6 CO₂(g) + 6 H₂O (l) → C₆H₁₂O₆ (aq) + 6 O₂(g)
b) O₃ (g) + H₂O₂ (l) → H₂O () + 2 O₂ (g)
Why it is more appropriate to write these reactions as :
a) 6 CO₂ (g) + 12 H₂O(l) → C₆H₁₂O₆(aq) + 6 H₂O(l) + 6 O₂(g)
b) O₃(g) + H₂O₂(l) → H₂O₂(l) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer:
a) 6CO₂ + 12 H₂O → C₆H₁₂O₆ + 6 H₂O + 6 O₂
This equation is more appropriate writing equation of photosynthesis.
Because of evolution of oxygen from H₂O but not from CO₂.

b) O_3(g) + H₂O_2(l) → H₂O_(l) + O_2(g) + O_2(g)
This is the more appropriate equation to write because in this reaction clearly mentioned that which is oxidised and which is reduced.

Question 40.
The compounds AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?
Answer:

• AgF₂ is an unstable compound.
• If suppose it is formed it acts as good oxidising agent.

Reason :
AgF₂ releases fluorine gas which is a powerful oxidising agent.
∴ AgF₂ is a good oxidising agent.

Question 41.
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:

The above equations illustrates the given statement.

Question 42.
How do you count for the following observations ?
a) Though alkaline potassium permagnate 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.
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 ?
Answer:
a) Balanced equation of KMn04/H+ (Acidic medium)
Mn\(\mathrm{O}_4^{-}\) + 8 H⁺ + 5 e^– → Mn⁺² + 4 H₂O
Balanced equation of KMnO₄/ H⁺ (Basic medium)
Mn\(\mathrm{O}_4^{-}\) + 2 H₂O + 3 e^– → MnO₂ + 4 OH^–

• Toluene is oxidised to benzoic acid in alcoholic medium.
• Toluene is easily oxidised in presence of alcoholic KMnO₄
  

b) When Conc. H₂SO₄ reacts with NaCl then HCl vapours are evolved.
2 NaCl + H₂SO₄ → Na₂SO₄ + 2 HCl
When Conc. H₄SO₄ reacts with KBr than HBr vapours are evolved which on further oxidation gives reddish brown Br₂ vapours.
2 KBr + H₂SO₄ → Na₂SO₄ + 2 HBr
2MBr + H₂SO₄ → 2 H₂O + SO₂ + Br₂ (Reddish brown)

Question 43.
Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions :
a) 2 AgBr (s) + C₆H₆O₂ (aq) → 2 Ag (s) + 2 HBr (aq) + C₆H₆O₂ (aq)
b) HCHO(l) + 2[Af(NH₃)₂]⁺(aq) + 3 OH^–(aq) → 2 Ag(s) + HCOO^–(aq) + 4 NH₃(aq) + 2 H₂O(l)
c) HCHO(l) + 2 Cu²⁺(aq) + 5 OH^–(aq) → Cu₂O(s) + HCOO^–(aq) + 3 H₂O(l)
d) N₂H₄ (l) + 2 H₂O₂(l) → N₂(g) + 4 H₂O(l)
e) Pb(s) + PbO₂(s) + 2 H₂SO₄(aq) → 2 PbSO₄(s) + 2 H₂O(l)
Answer:
a) Given equation is
2 AgBr_(s) + C₆H₆O_2(aq)

• 2 Ag_(s) + 2 HBr_(aq) + C₆H₄O₂ (aq)
• C₆H₆O₂ oxidised to C₆H₄O₂
• Ag⁺Br^– reduced to Ag
• Oxidising agent is Ag⁺
• Reducing agent is C₆H₆O₂

b) Given equation is HCHO_(l) + \(2\left[\mathrm{Ag}\left(\mathrm{NH}_3\right)_2\right]_{(\mathrm{aq})}^{+}\) + \(3 \mathrm{OH}_{(\mathrm{aq})}^{-}\)

• 2 Ag_(s) + \(3 \mathrm{OH}_{\text {(aq) }}^{-}\) + 4NH_3(aq) + 2 H₂O_(l)
• HCHO oxidised to HCOO^–
• [Ag(NH₃)₂]⁺ reduced to Ag
• [Ag(NH₃)₂]⁺ is oxidising agent
• HCHO is reducing agent

c) Given equation is
HCHO_(l) + \(2 \mathrm{Cu}_{(\mathrm{aq})}^{+2}\) + 5OH^– → Cu₂O_(s) + \(\mathrm{HCOO}_{(a q)}^{-}\) + 3H₂O_(l)

• HCHO oxidised to HCOO^–
• Cu⁺² reduced to Cu⁺(in Cu₂O)
• Oxidising agent is Cu⁺² ions (In Fehling’s reagent)
• Reducing agent is HCHO

d) Given equation is
N₂H_4(l) + 2K₂O_2(l) → N_2(g) + 4H₂O_(l)

• \(N^{-2}\) oxidised to N₂
• \(\mathrm{O}_2^{-2}\) reduced to O^-2
• Oxidising agent is H₂O₂
• Reducing agent is N₂H₄
  Pb_(s) + pbO_4(s) + 2H₂SO_4(aq)

e)

• 2PbSO_4(s) + 2H₂O
• Pb oxidised to Pb⁺²
• PbO₂ reduced to Pb⁺²
• Oxidising agent is PbO₂
• Reducing agent is Pb

Question 44.
Consider the reactions :
2S₂\(\mathrm{O}_3^{-2}\) _(aQ) + I_{2 (s)} → S₄\(\mathrm{O}_6^{-2}\) (aQ) + 2I^– (aQ)
S₂\(\mathrm{O}_3^{-2}\) (aQ) + I_2(s) + 5H₂O_(l) → 2S\(\mathbf{O}_4^{-2}\) (aq) + 4\(\mathrm{Br}_{(\mathrm{aq})}^{-}\)(aq) + 10\(H_{(aq)}^{+}\)
Why does the same reductant, thiosulphate react differently with iodine and bromine ?
Answer:

• Thio sulphate ion is not a strong reducing agent.
• I₂ is not a strong oxidising agent.
• The reaction between I₂ and thio sulphate form tetrathionate (S₄\(\mathrm{O}_6^{-2}\)) ion.
  
• The above reaction has high rate of reaction.
• The reaction between Thiosulphate and Bromine involves the formation of sulphate ion.
  
• Br₂ is some what better oxidising agent than I₂.
  Hence the difference observed in the above reactions.

Question 45.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
a)

• The oxidising capacity of any substance depends on the net result of several contributing energy factors like enthalpy change in a reaction, standard Electrode potential.
• With reference to Born-Haber type energy cycle the change in enthalpy value is greater for fluorine.
• Greater the magnitude of negative change in enthalpy greater is the oxidising power.

Supporting reaction :

• Fluorine reacts with carbon while the other elements of group do not combine even under drastic conditions.
  C + 2 F₂ → CF₄
• Fluorine is also called as superhalogen.

b)

• Hydrohalic compounds acts as reducing agents. Their stability order is
  HF >> HCl > HBr > HI
• HI is least stable and acts as strons reducing agent among these halides.
  H₂ + I₂ \(\rightleftharpoons\) 2 HI

Question 46.
Why does the following reaction occur ?

What conclusion about the compound Na₄XeO₆(of which  is a part) can be drawn from the reaction.
Answer:
Given equation is

• In the above reaction ‘Xe’ under go reduction (+8 to +6)
• F^– oxidised to F₂ here perxenate ion acts as powerful oxidising agent.
• Perxenates are stable in alkali solutions.
• Na₄XeO₆ is a powerful oxidant.

Question 47.
Consider the reactions :
(a) H₃PO₂(aq) + 4 AgNO₃(aq) + 2 H₂O(1) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)
(b) H₃PO₂(aq) + 2 CuSO₄(aq) + 2 H₂O(1) → H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)
(c) C₆H₅CHO(l) + 2[Ag (NH₃)₂]⁺(aq) + 3OH^–(aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^–(aq) → No Change observed.
what inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions ?
Answer:

• H₃PO₂ is a strong reducing agent hence it reduce Ag⁺ to Ag and Cu⁺² to Cu.
• C₆H₅CHO is also a reducing agent it just reduced.
  Ag⁺ to Ag in tollen’s reagent but it does not reduced Cu⁺² in alkaline solution.

Question 48.
Balance the following redox reactions by ion – electron method :
(a) Mn\(\mathrm{O}_4^{-}\) (aq) + I^– (aq) → MnO₂(s) + I₂ (s) (in basic medium)
(b) Mn\(\mathrm{O}_4^{-}\) (aq) + SO₂ (g) → Mn²⁺ (aq) + HSO₄ (aq) (in acidic solution)
(c) H₂O₂ (aq) + Fe²⁺ (aq) → Fe³⁺ (aq) + H₂O(l) (in acidic solution)
(d) Cr₂\(\mathrm{O}_7^{2-}\)SO₂ (g) → Cr³⁺ (aq) + S\(\mathrm{O}_4^{2-}\) (aq) (in acidic solution)
Answer:
a) Mn\(\mathrm{O}_{4 \text { (aq) }}^{-}\) + \(\mathbf{I}_{(\mathrm{aq})}^{-}\) → MnO_2(s) + I_2(s) (Basic medium)

Reduction half cell
\(\mathrm{MnO}_4^{-}\) → MnO₂
\(\mathrm{MnO}_4^{-}\) → MnO₂ + 2H₂O (Oxygens balanced)
\(\mathrm{MnO}_4^{-}\) + 4H₂O → MnO₂ + 2H₂O + 4OH^–
(Hydrogens balanced)
\(\mathrm{MnO}_4^{-}\) + 2H₂O → MnO₂ + 4OH^–
\(\mathrm{MnO}_4^{-}\) + 2H₂O + 3e^– → MnO₂ +4OH^–]
(Charge balanced)
2 × [Mn\(\mathrm{MnO}_4^{-}\) + 2H₂O + 3e^– → MnO₂ + 4OH^-1]
3 × [2 I^– → I₂ + 2e^–]
Oxidation half cell
I^– → I₂
2 I^– → I₂ (Iodines balanced)
2 I^– → I₂ + 2 e^– (Charge balanced)

The above equation is balanced equation.

b)

Above equation is balanced chemical equation.

c)

d)

Question 49.
Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducting agent
(a) P₄(s) + OH^–(aq) → PH₃(g) + HP\(\mathrm{O}_2^{-}\)
(b) N₂H₄(l) + Cl\(\mathrm{O}_3^{-}(\mathrm{aq})\)(aq) → NO(g) + Cl^–(g)
(c) Cl₂O₇ (g) + H₂O₂ (aq) → ClO₂ (aq) + O₂ (g) + H⁺
Answer:
a) P₄ + OH^– → PH₃ + HP\(\mathrm{O}_2^{-}\)
Reduction half cell
P₄ → PH₃
P₄ → 4PH₃
P₄ + 12H₂O → 4PH₃ + 12 OH^– (H balanced)
P₄ + 12H₂O + 12e^– → 4PH₃ + 12 OH^–

Oxidation half cell
P₄ → H₂P\(\mathrm{O}_2^{-}\)
P₄ → 4H₂P\(\mathrm{O}_2^{-}\) (‘P’ balanced)
P₄ + 8H₂O → 4H₂\(\mathrm{PO}_2^{-}\) (Oxygen balanced)
P₄ + 8H₂O + 8 OH^– → 4H₂\(\mathrm{PO}_2^{-}\) + 8H₂O
(Hydrogen balanced)
P₄ + 8OH^– → 4H₂\(\mathrm{PO}_2^{-}\) + 4e^–
(Charge balanced)
Here P₄ is oxidising agent as well reducing agent.

b)

\(\mathrm{ClO}_3^{-}\) is oxidant
N₂H₄ is reductant

c)

Question 50.
What sorts of informations can you draw from the following reaction ?
(CN)₂(g) + 20H(aq → CN^–(aq) + CNO(aq) + H₂O(1)
Answer:
(CN)_2(g) + 2(OH^–) → \(\mathrm{CN}_{(\mathrm{aq})}^{-}\) + \(\mathrm{CNO}_{(\mathrm{aq})}^{-}\) + \(\mathrm{H}_2 \mathrm{O}_{(l)}\)
(CN)₂ + 2e^– → 2 CN^– (Reduction)
(CN)₂ + 2H₂O → 2 CNO^– + 4H⁺ + 2e^– (Oxidation)
Here (CN₂) undergo oxidation as well as reduction.
It is a dis proportionation reaction.

Question 51.
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.
Answer:

• Above equation is a dis proportionation reaction

Question 52.
Consider the elements :
Cs, Ne, I and F
(a) Identify the element that exhibits only negative oxidation state.
(b) Identify the element that exhibits only positive oxidation state.
(c) Identify the element that exhibits both postive and negative oxidation states.
(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.
Answer:
a) ‘F’ Exhibits negative oxidation state i.e. ‘-1’
b) ‘Cs’ Exhibits only positive oxidation state
c) ‘I’ Exhibits positive as well as negative oxidation states
d) ‘Ne’ doesnot exhibit any oxidation state in ground state
(Neither positive nor negative)

Question 53.
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.
Answer:
Balanced equations for the given data is
Cl₂ + SO₂ + H₂O → SO₃ + 2H⁺ + Cl^–

Question 54.
Refer to the periodic table given in your book and now answer the following questions :
a) Select the possible non metals that can show disproportionation reaction.
b) Select three metals that can show disproportionation reaction.
Answer:
a) Non metals
Chlorine, Bromine, Oxygen, Sulphur, Phosphorous, Iodine undergo disproportionations
b) Metals Cr, Mn and pb undergo disproportionation.

Question 55.
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:
Chemical reaction is
4NH₃ + 5O₂ → 4NO + 6H₂O
4 moles NH₃ – 5 moles of O₂
Given 10 gms of NH₃
∴ No. of moles = \(\frac{10}{17}\) = 0.588 moles
Given 20 gms of NH₃
∴ No. of moles = \(\frac{20}{32}\) = \(\frac{5}{8}\) = 0.4
4 moles of ammonia reacts with 5 moles of O₂
0.588 moles of NH₃ …. ?
\(\frac{0.588}{4}\) × 5 = 0.735 moles
5 moles of O₂ reacts with 4 moles of NH₃
0.4 moles of O₂…. ?
\(\frac{0.4}{5}\) × 4 = \(\frac{1.6}{5}\) = 0.32

• Here O₂ is not present in sufficient amount NH₃ has sufficient in amount

5 moles of O₂ → 4 moles of NO.
5 × 32 gms of O₂ → 4 × 30 gms of NO.
20 gms of O₂ → ?
\(\frac{20 \times 4 \times 30}{5 \times 32}\) = \(\frac{20 \times 24}{32}\) = \(\frac{120}{8}\) = 15 gms

Question 56.
Arrange the following metals in the order in which they displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Answer:
According to electro chemical series electrode potentials of given elements as follows
Al – -1.66 V
Cul – + 0.34 V
Fe – -0.40 V
Mg – -2.37 V
Zn – -0.76 V
∴ The order of the above metals in which they displace each other from the solution of their salts. Mg > AZ > Zn > Fe > Cu

Long Answer Questions

Question 1.
Wtire the balanced ionic equation which represents the oxidation of iodide (I-) ion by permanganate ion in basic medium at give iodine (I) and managanese dioxide (MnO₂)
Answer:
Basic equation is


The above equation is balanced equation.

Question 2.
Write the balanced equation for the oxidation of suiphite ions to sulphate ions in acid medium by permanganate ion.
Answer:

Question 3.
Oxalic acid is oxidised by permanganate ion is acid medium of Mn²⁺ balance the reaction by ion — electron method.
Answer:

Question 4.
Phosphorus when heated with NaOH solution gives phosphine (PH₃) and H₂P\(\mathrm{O}_2^{-}\). Give the balanced equation.
Answer:

Here P₄ is oxidising agent as well reducing agent.

Question 5.
Balance the following equation

Answer:
Oxidation half reaction Cr(OH)₃ → \(\mathrm{CrO}_4^{2-}\)
(The ox. no of Cr changes from +3 to +6)
Reduction half reaction \(\mathrm{IO}_3^{-}\) → I^– (The ox. no. of ‘I’ changes from +5 to —1)

Question 6.
Balance the following equation by the oxidation number method.
Mn\(\mathrm{O}_4^{-2}\) + Cl₂ → Mn\(\mathrm{O}_4^{-2}\) + Cl^–
Answer:

Question 7.
Exaplain the different types of redox reactions.
Answer:
Redox reaction : “The reaction that involves loss of electrons is called an oxidation reaction and that involving gain of electrons is called a reduction reaction. The overall reaction is called as oxidation – reduction reaction” or simply ‘redox reaction’.

Types of redox reactions:

a) Chemical combination reactions: In these reactions one species combine with another species to form product. In this conversion one species undergo oxidation and other species undergo reduction.
e.g.:

As the oxidation state of carbon increases from ‘0’ to ‘+4’ in the reaction. So, carbon undergoes oxidation.
Oxygen changes from ‘0’ to ‘-2’ oxidation state. So, it undergoes reduction. Therefore, the above overall combination reaction is redox reaction.

b) Decomposition reactions : Chemical compounds chemically split into two or more simpler substances during decomposition reactions. These are again redox reactions.

Here, water decomposes. H₂ undergoes reduction from +1 to 0 in the reaction. Oxygen on the other hand changes its oxidation state from -2 to 0 (or) it undergoes oxidation.
Therefore, decomposition of H20 is a redox reaction.

c) Displacement reactions : In these reactions the place of one species in its compound is taken up by other species.

‘Zn’ displaces ‘Cu’ from CuSO₄ solution. In this reaction Zn undergoes oxidation and Cu undergoes reduction. Therefore overall reaction is a redox reaction.

d) Disproportionation reactions : These reactions involve the same element in the given form to undergo both oxidation and reduction simultaneously.

In the above reaction ‘Cl₂‘ undergoes both oxidation and reduction. It is a special type of redox reaction.

e) Comproportionation reactions : Reverse of disproportionation is comproportionation. In comproportionation reactions, two species with the same element in two different oxidation states form a single product in which the element is in an intermediate oxidation state.
e.g.:

Question 8.
Calculate the amount of carbon dioxide that could be produced when
(i) 1 mole of carbon is burnt in air.
(ii) 1 mole of carbon is burnt in 16 g of dioxygen,
(iii) 2 moles of carbon are burnt in 16 g of dioxygen.
Answer:
i) C + O₂ → CO₂
1 mole of ‘C’ → 1 mole of CO₂
∴ 44 gms of CO₂

ii) C + O₂ → CO₂
Given 1 mole of O₂ burnt in 16 gms of di oxygen

iii) C + O₂ → CO₂
2 moles of ‘C’ burnt in 16 gms of O₂
∴ 1 mole of O₂ → 44 gms of CO₂
32 gms of O₂ → 44 gms of CO₂
16 gms of O₂ → 22 gms of CO₂

Question 9.
Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation :
N₂(g) + H₂(g) → 2NH₃ (g)
(i) Calculate the mass of ammonia produced if 2.00 × 10³ g dinitrogen reacts with 1.00 × 10³ g of dihydrogen.
(ii) Will any of the two reactants remain unreacted ?
(iii) If yes, which one and what would be its mass ?
Answer:
i) N₂ + 3H₂ → 2NH₃
Given Nitrogen = 2 × 10³ gms
No. of moles = \(\frac{2000}{28}\) = 71.4285
Given hydrogen = 1 × 10³ gms
No. of moles = \(\frac{1000}{2}\) = 500 moles
N₂ is limiting agent
∴28 gms of N₂ → 2 × 17gms NH₃
2000 gms of N₂ →
\(\frac{2000 \times 17}{28}\) = 2428.57 gms

ii) Used amount of hydrogen
28 → 6 gms
100 gms → ?
\(\frac{1000 \times 6}{28}\) = 214.285 gms

iii) Remaining amount of hydrogens
= 1000 – 2140.285
= 785.715 gms

Question 10.
Assign oxidation number to the underlined elements in each of the following epecies :
(a) NaH₂PO₄
b) NaHSO₄
c) H₄P₂O₇
d) K₂MnO₄
(e) CaO2
(f) NaBH₄
(g) H₂S₂O₇
(h) KA1(SO₄)₂.12 H₂O
Answer:
(a) NaH₂PO₄
1(+1) + 2(+1) + x + 4 (-2) = 0
1 + 2 + x – 8 = 0
x – 5 = 0
x = +5
Oxidation no. of ‘P’ in NaH₂PO₄ = + 5

b) NaHSO₄
1(+1) +1(+1) + x + 4(-2) = 0
1 + 1 + x – 8 = 0
x = + 6
Oxidation no. of ‘S’ in NaHSO₄ = + 6

c) H₄P₂O₇
4(+1) + 2x + 7(-2) = 0
4 + 2x – 14 = 0
2x – 10 = 0
x = + 5
Oxidation no. of ‘P’ in H₄P₂O₇ is +5

d) K₂MnO₄
2(+1) + x + 4(-2) = 0
2 + x + 8 = 0
x = + 6
Oxidation no of Mn is K₂MnO₄ = + 6

e) CaO₂
+ 2 + 2x = 0.
2x = -2
x = -1
Oxidation no. of oxygen in CaO₂ = -1

f) NaBH₄
1(+1) + x+ 4(-1) = 0
1 + x – 4 = 0
x = +3
Oxidation no. of ‘B’ in NaBH₄ = +3
But ‘B’ most probably exhibits -3 oxidation state.

g) H₂S₂O₇
2(1) + 2x + 7(-2) = 0
2 + 2x – 14 = 0
2x – 12 = 0
x = +6
Oxidation state of ‘S’ in H₂S₂O₇ = + 6

h) k Al(SO₄)₂ 12H₂O :
General formula of above compound is
k₂SO₄ Al₂ (SO₄) 24 H₂O (Potash alum)
Consider Al₂(SO₄)₃ from the above double salt
2x + 3(-2) = 0
2x – 6 = 0
x = +3

Question 11.
What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?
(a) H₂S₄O₆
(b) Fe₃O₄
(c) CH₃CH₂OH
(d) CH₃COOH
Answer:
a) 45 (b)

b) 45 (c)

c) CH₃ CH₂ – OH
C₂H₆O
2x + 6(1) + (-2) = 0
2x + 6 – 2 = 0
2x + 4 = 0
x = -2

d) CH₃COOH
C₂H₄O₂
2x + 4(+1) + 2(-2) = 0
2x + 4 – 4 = 0
x = 0

Solved Problems

Question 1.
Calculate molecular mass of glucose (C₆H₁₂O₆) molecule.
Solution:
Molecular mass of glucose (C₆H₁₂O₆)
= 6(12.011 u) + 12(1.008 u) + 6(16.00.u)
= (72.066 u) + (12.096 u) + (96.00 u)
= 180.162 u

Question 2.
A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulae ?
Solution:
Step 1.
Conversion of mass per cent to grams :
Since we are having mass percent, it is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07 g hydrogen; 24.27 g carbon; and 71.65 g chlorine are present.

Step 2.
Convertion into number of moles of each element:
Divide the masses obtained above by respective atomic masses of various elements.
Moles of hydrogen = \(\frac{4.07 \mathrm{~g}}{1.008 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 4.04
Moles of carbon = \(\frac{24.27 \mathrm{~g}}{12.01 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 2.021 Moles of chlorine = \(\frac{71.65 \mathrm{~g}}{35.453 \mathrm{~g} \mathrm{~mol}^{-1}}\) = 2.021

Step 3.
Divide the mole value obtained above by the smallest number:
Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:Cl.

In case the ratios are not whole numbers, then they may be converted into whole number by multiplying by the suitable coefficient.

Step 4.
These numbers indicate the relative number of atoms of the elements. Write empirical formula by mentioning the numbers after writing the symbols of respective elements :
CH₂Cl is thus, the empirical formula of the above compound.

Step 5.
Writing molecular formula :

a) Determine empirical formula mass.
Add the atomic masses of various atoms present in the empirical formula.
For CH₂Cl, empirical formula mass is 12.01 + 2 × 1.008 + 35.453 = 49.48 g

b) Divide Molar mass by empirical formula mass

= 2 = (n)

c) Multiply empirical formula by n obtained above to get the molecular formula.
Empirical formula = CH₂Cl, n = 2. Hence molecular formula is C₂H₄Cl₂.

Question 3.
Calculate the amount of water (g) produced by the combustion of 16 g of methane.
Solution:
The balanced equation for combustion of methane is :
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)

1. 16 g of CH₄ corresponds to one mole.
2. From the above equation, 1 mol of CH₄(g) gives 2 mol of H₂O(g).
   2 mol of water (H₂O) = 2 × (2 + 16)
   = 2 × 18 = 36 g

1 mol H₂O = 18g H₂O ⇒ \(\frac{18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}\) = 1
Hence 2 mol H₂O × \(\frac{18 \mathrm{~g} \mathrm{H}_2 \mathrm{O}}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{O}}\)
= 2 × 18 g H₂O = 36 g H₂O

Question 4.
How many moles of methane are required to produce 22 g CO₂(g) after combustion ?
Solution:
According to the chemical equation
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
44 g CO₂ (g) is obtained from 16 g CH₄(g).
[∵ 1 mol CO₂(g) is obtained from 1 mol of CH₄(g)]
Mole of CO₂(g)
= 22 g CO₂(g) × \(\frac{1 \mathrm{~mol} \mathrm{CO}_2(\mathrm{~g})}{44 \mathrm{gCO}_2(\mathrm{~g})}\)
= 0.5 mol CO₂(g)
Hence 0.5 mol CO₂(g) would be obtained from 0.5 mol CH₄(g) or 0.5 mol of CH₄(g) would be required to produce 22g CO₂(g).

Question 5.
50.0 kg of N₂(g) anfd 10.0 kg of H₂(g) are mixed to produce NH₃(g). Calculate the NH₃ (g) formed. Identify the limiting reagent in the production of NH₃ in this situation.
Solution:
A balanced equation for the above reaction is written as follows :
Calculation of moles:
N₂ + 3H₂(g) ⇌ 2HN₃(g)
Moles of N₂

According to the above equation, 1 mol N₂(g) requires 3 mol H₂(g), for the reaction. Hence, for 17.86 × 10² mol N₂, the moles of H₂(g) required would be
17.86 × 10² mol N₂ × \(\frac{3 \mathrm{~mol} \mathrm{H}_2(\mathrm{~g})}{1 \mathrm{molN}_2(\mathrm{~g})}\)
= 5.36 × 10³ mol H₂

But we have only 4.96 × 10³ mol H₂. Hence, dihydrogen is the limiting reagent in
this case. So NH₃(g) would be formed only from that amount of available dihydrogen i.e.,
4.96 × 10³ mol.
Since 3 mol H₂(g) gives 2 mol NH₃(g)
4.96 × 10³ mol H₂ (g) × \(\frac{2 \mathrm{~mol} \mathrm{NH}_3(\mathrm{~g})}{3 \mathrm{molH}_2(\mathrm{~g})}\)
= 3.30 × 10³ mol NH₃ (g)
3.30 × 10³ mol NH₃ (g) is obtained.

If they are to be converted to grams, it is done as follows :
1 mol NH₃(g) = 17.0 g NH₃(g)
3.30 × 10³ mol NH₃ (g) × \(\frac{17.0 \mathrm{~g} \mathrm{NH}_3(\mathrm{~g})}{1 \mathrm{~mol} \mathrm{NH}_3(\mathrm{~g})}\)
= 3.30 × 10³ × 17 g NH₃ (g)
= 56.1 × 10³ g NH₃ (g)
= 56.1 kg NH₃

Question 6.
A solution ¡s prepared by adding 2 g of a substance A to 18 g water. Calculate the mass percent of the solute.
Solution:

Question 7.
Calculate the molarity of NaOH in the solution prepared by dissolving 4 g in enough water to form 250 mL of the solution.
Solution:
Since molarity

Note that molarity of a solution depends upon temperature because volume of a solution is temperature dependent.

Question 8.
The density of 3 M solution of NaCl is 1.25 g mL^-1. Calculate molality of the solution.
Solution:
M = 3 mol L^-1
Mass of NaCl in 1 L solution
= 3 × 58.5 = 175.5 g
Mass of
1 L solution = 1000 × 1.25 = 1250 g
(Since density = 1.25 gmL^-1)
Mass of water of solution = 1250 – 175.5
= 1074.5 g = 1.0745 kg

Often in a chemistry laboratory, a solution of a desired concentration is prepared by diluting a solution of known higher concentration. The solution of higher concentration is also known as stock solution. Note that molality of a solution does not change with temperature since mass remains unaffeoted with temperature.

Question 9.
Calculte the normality of oxalic acid solutions containing 6.3g of H₂C₂O₄.2H₂O in 500 ml of solutions.
Solution:
Weight of solute = 6.3 g
GEW of solute = \(\frac{126}{2}\) = 63
∴ Normality (N) = \(\frac{\omega}{\mathrm{GEW}} \times \frac{1000}{\mathrm{~V}(\mathrm{~mL})}\)
Normality (N) = \(\frac{6.3}{63} \times \frac{1000}{500}\) = 0.2 N

Question 10.
Calculate the mass of Na₂CO₃ required to prepare 250 ml of 0.5 N solution.
Solution:
Normality of required solution = 0.5 N
Volume of required solution = 250 ml
Equivalent wt.of. Na₂CO₃ = \(\frac{106}{2}\) = 53
Normality (N) = \(\frac{\mathrm{W}}{\mathrm{GEW}} \times \frac{1000}{\mathrm{~V}(\mathrm{~mL})}\)
wt. of solute = N × GEW × \(\frac{V(\mathrm{ml})}{1000}\)
= 0.5 × 53 × \(\frac{250}{1000}\) = \(\frac{53}{8}\) = 6.62 g

Question 11.
In the reactions given below, identify the species undergoing oxidation and reduction :
(i) H₂S (g) + Cl₂ (g) → 2HC (g) + S (s)
(ii) 3Fe₃O₄ (s) + 8Al (s) → 9Fe (s) + 4Al₂O₃ (s)
(iii) 2Na (s) + H₂ (g) → 2NaH (s)
Solution:
(i) H₂S is oxidised because a more electronegative element, chlorine is added to hydrogen (or a more electropositive element, hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it.

(ii) Aluminium is oxidised because oxygen is added to it. Ferrous ferric oxide (Fe₃O₄) is reduced because oxygen has been removed from it.

(iii) This reaction is very interesting. It may be said from the above definitions that the reaction is only a reduction reaction as it involves the addition of sodium (electropositive metal) or hydrogen whereas sodium undergoes oxidation and hydrogen undergoes reduction. It is to say that the above definitions for oxidation and reduction can not explain this and they have limitations like this. Therefore a new concept has to be considered for the oxidation and the reduction.

Question 12.
Justify that the reaction :
2Na (s) + H₂ (g) → 2NaH (s) is a redox change.
Solution:
Since in the above reaction the compound formed is an ionic compound, which may also be represented as Na⁺H^– (s), this suggests that one half reaction in this process is :
2Na (s) → 2Na⁺ (g) + 2e^– and the other half reaction is H₂ (g) + 2e^– → 2H^– (g)
This splitting of the reaction under examination into two half reactions automatically reveals that here sodium is oxidised and hydrogen is reduced, therefore, the complete reaction is a redox change.

Question 13.
Using stock notation, represent the following compounds : HAuCl₄, Tl₂O, FeO, Fe₂O₃, Cul, CuO, MnO and MnO₂.
Solution:
By applying various rules of calculating the oxidation number of the desired element in a compound, the oxidation number of each metallic element in its compound is as follows:
HAuCl₄ → Au has 3
Tl₂O → Tl has 1
FeO → Fe has 2
Fe₂O₃ → Fe has 3
Cul → Cu has 1
CuO → Cu has 2
MnO → Mn has 2
MnO₂ → Mn has 4
Therefore, these compounds may be represented as :
HAu (III) Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O₂.

Question 14.
Justify that the reaction :
2Cu₂O(S) + Cu₂S(s) → 6Cu(s) + SO₂(g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.
Solution:
Let us assign oxidation number to each of the species in the reaction under examination. This results into :

We therefore, conclude that in this reaction copper is reduced from +1 state to zero oxidation state and sulphur is oxidised from -2 state to +4 state. The above reaction is thus a redox reaction.

Further, Cu₂O helps sulphur in Cu₂S to increase its oxidation number, therefore Cu (I) is an oxidant; and sulphur of Cu2S helps copper both in Cu₂S itself and Cu₂O to decrease its oxidation number ; therefore, sulphur of Cu₂S is reductant.

Question 15.
Which of the following species, do not show disproportionation reaction and why ? ClO^–, Cl\(\mathrm{O}_2^{-}\), ClO₃ and ClO₄. Also write reaction for each of the species that disproportionates.
Solution:
Among the oxoanions of chlorine listed above, ClO₄ does not disproportionate because in this oxoanion chlorine is present in its highest oxidation state that is, +7. The disproportionation reactions for the other three oxoanions of chlorine areas follows :

Question 16.
Suggest a scheme of classification of
the following redox reactions
(a) N₂ (g) + O₂ (g) → 2NO (g)
(b) 2Pb(NO₃)₂(s) → 2PbO(s) + 2NO₂(g) + 1/2 O₂(g)
(c) NaH(s) + H₂O (l) → NaOH (aq) + H₂(g)
(d) 2NO₂(g) + 2OH^–(aq) → N\(\mathrm{O}_2^{-}\) (aq) + N\(\mathrm{O}_3^{-}\) (aq) + H₂O(l)
Solution:
In reaction
(a) the compound nitric oxide is formed by the combination of the elemental substances, nitrogen and oxygen ; therefore, this is an example of combination redox reactions. The reaction.

(b) involves the breaking down of lead nitrate into three components, therefore, this is categorised under decomposition redox reaction. In reaction.

(c) hydrogen of water has been displaced by hydride ion into dihydrogen gas. There-fore, this may be called as displacement redox reaction.

(d) The reaction involves disproportionation of NO₂ (+4 state) into NO₂ (+3 state) and NO₃ (+5 state). Therefore reaction (d) is an example of disproportionation redox reaction.

Question 17.
Why do the following reactiolns proceed differently ?
Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O and
Pb₃O₄ + 4HNO₃ → 2Pb(NO₃)₃ + PbO₂ + 2H₂O
Solution:
Pb₃O₄ is actually a stoichiometric mixture of 2 mol. of PbO and 1 mol. of PbO₂. In PbO₂, lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO₂ thus can act as an oxidant (oxidising agent) and, therefore, can oxidise Cl ion of HCl into chlorine. We may also keep in mind that PbO is a basic oxide. Therefore, the reaction
Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O can be splitted into two reactions namely :
2 PbO + 4HCl → 2PbCl₂ + 2H₂O (acid base reaction)

Question 18.
Write the net ionic equation for the reaction of potassium dichromate (VI) K₂Cr₂O₇ with sodium sulphite, Na₂SO₃ in an acid solution to give chromium (III) ion and the sulphate ion.
Solution:
Step – 1 :
The skeletal ionic equation is :
Cr₂\(\mathrm{O}_7^{2-}\) (aq) + S\(\mathrm{O}_3^{-2}\) (aq) → Cr³⁺ (aq) + S\(\mathrm{O}_4^{2-}\) (aq)

Step – 2 :
Assign oxidation numbers for Cr and S

This indicates that the dichromate ion is the oxidant (it oxidises sulphite ion to sulphate ion) and the sulphite ion is the reductant. (it reduces dichromate ion to chromium (III).

Step – 3 :
Calculate the increase and decrease of oxidation numbers of respective species and make them equal.

As reduction is total 6 units due to two Cr³⁺ formed, oxidation also must be 6 units. This is obtained by multiplying S\(\mathrm{O}_3^{-2}\) with 3.

Step – 4 :
Adjust the coefficients of the products accordingly

Step – 5 :
a) Add H⁺ ions in acid medium or H₂O molecules in basic medium in the required number to hydrogen atoms deficient side.
b) Add H₂O molecules in acid medium or OH^– ions in basic medium in the required number to oxygen atoms diffident side, (a) and (b) may be repeated any number of times by hit and trial method until hydrogen and oxygen atoms are same in number on both the sides of the redox reaction.
The given reaction is in acid medium

Question 19.
Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.
Solution:
Step – 1 : The skeletal ionic equation is :
Mn\(\mathrm{O}_4^{-}\) (aq) + Br^– (aq) → MnO₂ (s) + Br\(\mathrm{O}_3^{-}\) (aq)

Step – 2 : Assign oxidation numbers for Mn and Br.

This indicates that permanganate ion is the oxidant and bromide ion is the reductant.

Step – 3 :
Calculate the increase or decrease in the oxidation number per atom and for the entire molecule/ion in which it occurs. The number units of oxidations change must be equal to the number units of reduction change. If this is not observed then multiply the oxidising agent with the number of units of oxidation and the reducing agent with the number of units of reduction i.e., multiply oxidant MnO₄ by 2 and reductant Br^– by I.

Know number of units oxidation number of units reduction.

Step – 4:
Adjust the coefficients of products accordingly.
2Mn\(\mathrm{O}_4^{-}\) (aq) + Br (aq) → 2MnO₂ (s) + BrO₃ (aq)

Step – 5 :
a) Add H⁺ ions in acid medium or H₂O molecules in basic medium in the required number to hydrogen atoms deficient side.
b) Add H₂O molecules in acid medium or OH^– ions in basic medium in the required number to oxygen atoms dificient side, (a) and (b) may be repeated any number of times by hit and trial method until hydrogen and oxygen atoms are same in number on both the sides of the redox reaction. This reaction is basic medium.
2Mn\(\mathrm{O}_4^{-}\) (aq) + Br (aq) + H₂O (I)

• 2MnO₂ (s) + Br\(\mathrm{O}_3^{-}\) (aq) + 2OH^– (aq)

Question 20.
Permanganate (VII) ion, MnO₄ in basic solution oxidises iodide ion. I to produce molecular iodine (I₂) and manganese (IV) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction.
Solution:
Step – 1 :
First we write the skeletal ionic equation, which is
Mn\(\mathrm{O}_4^{-}\) (aq) + I^– (aq) → MnO₂ (s) + I₂ (s)

Step – 2 :
The two half-reactions are :
-1        0
Oxidation half :

Reduction half : Mn\(\mathrm{O}_4^{-}\) (aq) → MnO2 (s)

Step – 3 :
To balance the I atoms in the oxidation half reaction, we rewrite it as:
2I^– (aq) → I₂ (s)

Step – 4:
As the reaction takes place to basic medium, to balance the 0 atoms in the reduction half reaction, we add OH^– ions in required number.
Mn\(\mathrm{O}_4^{-}\) (aq) → MnO₂ (s) + 2HO^– (I)
To balance the H atoms, we add two H₂O molecules on the left.
Mn\(\mathrm{O}_4^{-}\) (aq) + 2H₂O (aq) → MnO₂ (s) + 2 HO^– (I)
Balance H and 0 atoms by bit and trial method even if it requires morethan once we haye to do it.
The resultant equation is :
Mn\(\mathrm{O}_4^{-}\) (aq) + 2H₂O (I) → MnO₂ (s) + 4OH^– (aq)
Note : While balanceing H atom 0 atoms do not disturb the coefficients of other species (oxidant and reduCtant and the products).

Step-5 :
In this step we balance the charges of the two half-reactions in the manner depicted as:
2I^– (aq) → I₂ (s) + 2e^–
Mn\(\mathrm{O}_4^{-}\) (aq) + 2H₂O (I) + 3e^– → MnO₂ (s) + 4OH^– (aq)

Now to equalise the number of electrons, in the two half reactions we multiply the oxidation hatf-reaction by 3 and the reduction half-reaction by 2.
6I^– (aq) → 3I₂ (s) + 6e^–
2MnO₄ (aq) + 4H₄O (I) + 6e^– → 2MnO₂ (s) + 8OH^–(aq)

Step-6:
Add two half-reactions to obtain the net reaction and canceel the electrons on both sides.
6I^– (aq) + 2Mn\(\mathrm{O}_4^{-}\) (aq) + 4H₂O (I) → 3I₂ (s) + 2MnO₂ (s) + 8 OH^– (aq)

Step – 7:
A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides.

Question 21.
Calculate the normality of H₂So₄ solution. If 50 ml of it completely neutralised 250 ml of 0.2 N sodium hydroxide (NaoH) solution.
Answer:

∴ Normality of H₂So₄ solution = 1 N.