Chapter 7

PARAMOUNT SCHOOL SYSTEM

Subject: Chemistry – I

Unit 7: Electrochemistry

CONCEPT ASSESSMENT EXERCISE 7.1

Identify elements undergoing oxidation and reduction in the following reactions.

i) N₂ + 3H₂ → 2NH₃

N-atoms undergo reduction because they gain hydrogen.

H-atoms undergo oxidation.

ii) 2H₂ + O₂ → 2H₂O

H-atoms undergo oxidation because they gain oxygen.

O-atoms undergo reduction because they losses oxygen.

iii) Fe₂O₃+ 3CO → 2Fe + 3CO₂

C-atoms undergo oxidation because they gain oxygen.

Fe-atoms undergo reduction because they losses oxygen.

iv) 4 Al + 3O₂ → 2Al₂O₃

Al-atoms under oxidation because they gain oxygen.

O-atoms undergo reduction.

CONCEPT ASSESSMENT EXERCISE 7.2

In the following reactions, identify which element is oxidized and which element is reduced?

4Na + O₂ → 2Na₂O

Na belongs to Group IA and loses one electron, so it undergoes oxidation.

O belongs to Group VIA and gain two electrons, so it undergoes reduction.

2Al + 3Cl₂ → 2AlCl₃

Al belongs to Group III A and loses three electrons, so it undergoes oxidation. Cl belongs to Group VII A and gains one electron, so it undergoes reduction.

Mg + Cl₂ → MgCl₂

Mg belongs to Group IIA and loses two electrons, so it undergoes oxidation. Cl belongs to Group VIIA and gain one electron so undergo reduction.

CONCEPT ASSESSMENT EXERCISE 7.3

One major problem of air pollution is the formation of acid rain. Air pollutants such as SO2 and NO₂ combine with oxygen and water vapours in the air to form H₂SO₄ and HNO₃. These acids fall to the ground with the rain, making the rain acidic. Clouds can also absorb the acids and carry them hundreds of kilometer away from where the pollutants are released. Determine the oxidation number of N is NO₂ and HNO₃, S is SO₂ and H₂SO₄.

The sum of oxidation number must be zero

Oxidation number of N in NO₂:

x + 2 (-2) = 0

x – 4 = 0

x = 4

Thus, the oxidation state for N in NO₂ is + 4

Oxidation number of N in HNO₃

1 + x + 3(-2) = 0

1 + x – 6 = 0

x – 5 = 0

x = 5

Thus, the oxidation state for N is HNO₃ is + 5.

Oxidation number of S in SO₂:

X + 2 (-2) = 0

X – 4 = 0

X = 4

Thus, oxidation state for S in SO₂ is + 4

Oxidation number of S in H₂SO₄:

2(+1) + x +4(-2) = 0

2 + x – 8 = 0

X – 6 = 0

X = 6

The oxidation state for S in H₂SO₄ is + 6

CONCEPT ASSESSMENT EXERCISE 7.4

Determine the oxidation state of

1. S in sulphate ion, SO₄^-2

x + 4(−2) = − 2

x – 8 = − 2

x = -2 + 8

x = + 6

So, the oxidation state of sulfur (S) in SO₄^-2 + 6.

2. P in phosphate ion, PO₄^-3

x + 4(−2) = −3

x – 8 = − 3

x = -3 + 8

x = + 5

So, the oxidation state of phosphorus (P) in PO₄^-3 + 5.

3. N in ammonium ion, NH₄⁺​

x+4(+1) = +1

x + 4 = 1

x = 1 – 4

x = -3

So, the oxidation state of Nitrogen (N) in NH₄⁺ is -3.

CONCEPT ASSESSMENT EXERCISE 7.5

1. Identify oxidizing and reducing agents in the following reactions.
2. 2S + Cl₂ → S₂ Cl₂

First assign oxidation number to each atom.

2+ →

Oxidation number of Cl is decreased, Cl is reduced so Cl₂ is oxidizing agent. Oxidation number of S is increased, S is oxidized. So S is reducing agent.

1. 2Na + Br₂ → 2NaBr

First assign oxidation number to each atom.

2+ → 2

Oxidation number of Br is decreased, Br is reduced, so Br₂ is oxidizing agent.

Oxidation number of Na is increased, Na is oxidized, So Na is reducing agent.

2. Differentiate between oxidizing and reducing agents.

Oxidizing Agent	Reducing Agent
Causes another substance to oxidize.	Causes another substance to be reduced.
Accepts electrons.	Donates electrons.
Is reduced by gaining electrons.	Is oxidized by losing electrons.
Example: Chlorine (Cl₂) is reduced in 2Na + Cl₂ → 2NaCl.	Example: Sodium (Na) is oxidized in 2Na + Cl₂ → 2NaCl.

Q. Will Fe oxidize in this region? (Corrosion)

Yes, iron (Fe) will oxidize in the region acting as the anode during the corrosion process.

Review Questions

1. Encircle the correct answer.

1. In which of the following changes the nitrogen atom is reduced?

a. N₂ to NO b. N₂ to NO₂ c. N₂ to NH₃ d. N₂ to HNO₃

1. Which of the following change’s reaction is an example of oxidation?

a. Chlorine molecule to chloride ion b. Silver atoms to silver (I) ion

c. Oxygen molecule to oxide ion d. Iron (III) ion to iron(II) ion

1. Which of the following elements in the given reaction is reduced?

ZnO + H₂ Zn + H₂O

a. H₂ b. ZnO c. Zn d. O

1. Consider the following reaction:

H₂S +Cl₂

2HCl + S

In this reaction what does H₂S behave as?

a. Reducing agent b. Oxidizing agent c. Catalyst d. Electrolyte

1. The oxidation state of Cr in K₂Cr₂O₇ is

a. + 12 b. + 6 c. + 3 d. – 6

2. Give short answers

(i) What is oxidation state?

Oxidation state or oxidation number is defined as the number of apparent charges that an atom will have in the molecule.

The elements that show an increase in oxidation number are oxidized. The elements that show a decrease in oxidation number are reduced.

(ii) What is the oxidation number of Cr in chromic acid (H₂CrO₄)?

The sum of oxidation state of all the atoms is zero.

2×1 + x + 4 (-2) = 0

2 + x -8 = 0

x – 6 = 0

x = + 6

(iii) Identify reducing agent in the following reaction.

CuO + H₂ Cu + H₂O

First assign oxidation number to each atom.

+2 -2 0 0 2(+1) -2

CuO + H₂ Cu + H₂O

The oxidation number of Cu decreases, Cu is reduced so CuO₃ is an oxidizing agent. Similarly, the oxidation number of H increases, H is oxidized. Therefore, H₂ is reducing agent.

(iv) Why tin plated steel is used to make food cans?

Tin plated steel is used to make food cans. This is because the components of food beverages and the preservatives contain organic acids or their salts. They may form toxic substances by reacting with iron. These acids and salts are corrosive. Tin plating is non-poisonous and prevents corrosion.

(vi) Explain one example from daily life which involves oxidation-reduction?

Examples from daily life which involves oxidation-reduction is rusting of iron. Oxygen and water are necessary for iron to rust. A region of metal surface that has relatively less moisture, act as anode.

Fe _(s) → Fe⁺² _(aq) + 2ē

Another region on the surface of metal that has relatively more moisture act as cathode. The electrons released in the oxidation process reduce atmospheric oxygen to hydroxyl ions.

O₂ + 2H₂O +4ē → 4OH^–

The Fe⁺² ions formed at the anodic regions flow to the cathodic regions through the moisture on the surface. Fe⁺² ions further react with oxygen to form rust, Fe₂O₃. xH₂O

3. Compare and contrast oxidation and reduction.

Oxidation	Reduction
Gain of oxygen	Loss of oxygen
Loss of hydrogen	Gain of hydrogen
Loss of electrons	Gain of electrons
Increase in oxidation number	Decrease in oxidation number

4. Define oxidation and reduction in terms of loss or gain of electrons.

Oxidation:

The process that involves the loss of electrons by an element is called oxidation.

Example:

Fe (s) → Fe⁺²(aq) +2e⁻

Iron (Fe) loses two electrons to form Fe⁺².

Reduction:

A process that involves the gain of electrons by a substance is called reduction.

Example:

O₂​ + 4e⁻ + 2H₂​O → 4OH⁻

Oxygen (O₂) gains four electrons to form hydroxide ions (OH⁻).

5. Explain how food and beverage industries deal with corrosion.

Tin plated steel is used to make cans. Food and beverages industries use tinplated steel cans. This is because the components of food beverages and the preservatives contain organic acids or their salts. They may form toxic substances by reacting with iron. These acids and salts are corrosive. Tin plating is non-poisonous and prevents corrosion.

6. State the substances which are oxidized or reduced. Give reason for your answer.

(a) N₂ + 3H₂ → NH₃

0 0 -3 +3

N₂ + 3H₂ → NH₃

There is a decrease in oxidation state of N. Therefore N^– atoms undergo reduction.

There is an increase in oxidation state of H. Therefore H-atoms undergo oxidation.

(b) CO₂ + 2Mg → 2MgO + C

Ans. +4 2(-2) 0 +2 -2 0

CO₂ + 2Mg → 2MgO + C

There is a decrease in oxidation state of O. Therefore O-atoms undergo reduction.

There is a increase in oxidation state of Mg. Therefore Mg-atom undergo oxidation.

(c) Mg + H₂O → MgO + H₂

+ → +

There is a decrease in oxidation state of H. Therefore H-atoms undergo reduction.

There is an increase in oxidation state of Mg. Therefore Mg-atoms undergo oxidation.

(d) H₂S + Cl₂ → 2HCl + S

+ → 2 +

There is a decrease in oxidation state of Cl. Therefore Cl-atoms undergo reduction.

There is an increase in oxidation state of S. Therefore S-atoms undergo oxidation.

(e) 2NH₃ + 3CuO → 3Cu + N₂ + 3H₂O

2 + 3 → 3+ +

There is a decrease in oxidation state of Cu. Therefore Cu-atoms undergo reduction.

There is an increase in oxidation state of N. Therefore N-atoms undergo oxidation.

7. (a) Define oxidation number or oxidation state.

Oxidation state or oxidation number is defined as the number of charges an atom will have in a molecule or a compound.

The elements that show an increase in oxidation number are oxidized.

The elements that show a decrease in oxidation number are reduced.

(b) Find the oxidation state of Nitrogen in the following compounds.

(i) NO₂

The sum of oxidation state of all the atoms is zero

x + 2 (-2) = 0

x – 4 = 0

x -4 = 0

x = 4

(ii) N₂O

The sum of oxidation state of all the atoms is zero

2x + (-2) = 0

2x – 2 = 0

2x = 2

2x = 4

2 2

X = 1

(iii) N₂O₃

The sum of oxidation state of all the atoms is zero

2x + 3 (-2) = 0

2x –6 = 0

2x = 6

2x = 6

2 2

X = 3

(iv) HNO₃

The sum of oxidation state of all the atoms is zero

+1 +x + 3 (-2) = 0

1 + x – 6 = 0

x -5 = 0

x = 5

8. Find the oxidation state of S in the following compound

(a) H₂S

The sum of oxidation state of all the atoms is zero

2(+1) x = 0

2 + x = 0

x = -2

Oxidation number for N in H₂S is -2.

(b) H₂SO₃

The sum of oxidation state of all the atoms is zero.

2(+1) + x + 3(-2) = 0

2 + x – 6 = 0

x – 4 = 0

X = 4

Oxidation number for S in H₂SO₃ is +4.

(c) Na₂S₂O₃

The sum of oxidation state of all the atoms is zero

2(+1) + 2x + 3(-2) = 0

2 + 2x – 6 = 0

2x -4 = 0

2x = 4

2x = 4

2 2

X = 2

Oxidation number for S in Na₂S₂O₃ is +2

9. Define oxidizing and reducing agents.

(a) Oxidizing agents:

An oxidizing agent is the reactant containing the element that is reduced (gains electrons) in a chemical reaction.

Example:

2Na + Cl₂​ → 2NaCl

Chlorine (Cl₂) is Oxidizing Agent, because it gains electrons to form chloride ions (Cl⁻).

Reducing agents:

A reducing agent is the reactant containing the element that is oxidized (loses electrons) in the chemical reaction.

Example:

Sodium (Na) is Reducing Agent, because it loses electrons to form sodium ions (Na⁺).

(b) Identify the oxidizing agents and reducing agents in the following reactions.

(i) H₂S + Cl₂ → 2HCl + S

First assign oxidation number to each atom.

2(+1) -2 0 +1 -1 0

H₂S + Cl₂ → 2HCl + S

The oxidation number of Cl decreases, Cl is reduced so Cl₂ is an oxidizing agent.

Similarly, the oxidation number of S increases, S is oxidized, therefore S is reducing agent.

ii) 2 FeCl₂ + Cl₂ → 2FeCl₃

First assign oxidation number to each atom

+2 2(-1) 0 +3 3(-1)

2 FeCl₂ + Cl₂ → 2FeCl₃

The oxidation number of Cl decreases, Cl is reduced so Cl₂ is an oxidizing agent. Similarly, the oxidation number of Fe increases, Fe is therefore oxidized. FeCl₂ is reducing agent.

(iii) 2KI + Cl₂ → 2KCI + I₂

First assign oxidation number to each atom.

2+ → 2 +

The oxidation number of Cl decreases, Cl is reduced. So Cl₂ is an oxidizing agent. Similarly, the oxidation number of I increases, I is oxidized. Therefore, KI is reducing agent.

(iv) Mg + 2HCl → MgCl₂ + H₂

First assign oxidation number to each atom.

+ 2 → +

The oxidation number of Cl decreases, Cl is reduced. So Cl₂ is oxidizing agent.

Similarly, the oxidation number of Mg increases, Mg is oxidized. Therefore, Mg is reducing agent.

10. Hydrogen peroxide reacts with silver oxide and lead (II) sulphide according to the following equations.

(i) H₂O₂ + Ag₂O → 2Ag + H₂O + O₂

(ii) 4H₂O₂ + PbS → PbSO₄ + 4H₂O

Is hydrogen peroxide an oxidizing or reducing agent in these reactions? Give your reasons.

2(+1) 2(-1) 2(+1) -2 0 +2 -2 0

i) H₂O₂ + Ag₂O → 2Ag + H₂O + O₂

Hydrogen peroxide is reducing agent. Because the oxidation number of Ag decreases, Ag is reduced. So Ag₂O is oxidizing agent.

Similarly, the H is oxidized. Therefore, H₂O₂ is reducing agent.

ii) 2(+2) 2(-2) +2 -2 +2 +6 4(-2) +2 -2

4H₂O₂ + PbS → PbSO₄ + 4H₂O

Hydrogen per oxide is an oxidizing agent. Because oxidation number of S increases, S is oxidized. So PbS is reducing agent.

Similarly, oxidation number of H decreases, H is reduced. Therefore, H₂O₂ is oxidizing agent.