Chapter 8.18 - More Solved Examples Related to Redox Reactions

In the previous section, we some solved examples related to redox reactions. In this section, we will see a few more solved examples.

Solved example 8.18
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
Solution:
(a) Mercury(II) chloride
1. This compound contains mercury (Hg) and chlorine (Cl)
2. Given that, Hg has an oxidation state of +2. So it has lost two electrons.
   ♦ Those two electrons must be accepted by Cl
3. One Cl can accept only one electron. So there must be two Cl atoms in the compound.
4. Thus the formula of this compound is HgCl₂

(b) Nickel(II) sulphate
1. This compound contains Nickel (Ni) and sulphate (SO₄^2-)
2. Given that, Ni has an oxidation state of +2. So it has lost two electrons.
   ♦ Those two electrons must be accepted by the sulphate group.
3. During the reaction, one Ni atom donates two electrons. Those two electrons are used to form one SO₄^2- group.
• So one Ni²⁺ and one SO₄^2- combine to form a single molecule.
4. Thus formula of that compound will be NiSO₄

(c) Tin(IV) oxide
1. This compound contains Tin (Sn) and oxygen (O)
2. Given that, Sn has an oxidation state of +4. So it has lost four electrons.
   ♦ Those four electrons must be accepted by O
3. One O can accept only two electron. So there must be two O atoms in the compound.
4. Thus the formula of this compound is SnO₂

(d) Thallium(I) sulphate
1. This compound contains Thallium (Tl) and sulphate (SO₄^2-)
2. Given that, Tl has an oxidation state of +1. So it has lost one electron.
   ♦ This one electron must be accepted by the sulphate group.
3. During the reaction, one Tl atom donates one electron. But one electron is not sufficient to form the sulphate group.
• So there must be two Tl atoms to supply the two electrons
   ♦ Those two electrons are used to form one  SO₄^2- group.
• So two Tl⁺ and one SO₄^2- combine to form a single molecule.
4. Thus formula of that compound will be Tl₂SO₄

(e) Iron(III) sulphate
1. This compound contains Iron (Fe) and sulphate (SO₄^2-)
2. Given that, Fe has an oxidation state of +3. So it has lost three electrons.
   ♦ Those three electrons must be accepted by the sulphate group.
3. The Fe donates 3 electrons. But the sulphate group can accept only 2 electrons.
4. So we must take the LCM of 2 and 3, which is 6
• We must arrange them in such a way that,
   ♦ Fe donates 6 electrons.
   ♦ sulphate group accepts 6 electrons
5. This can be achieved if:
   ♦ Two Fe atoms donate 3 electrons each.
   ♦ Three sulphate groups accept 2 electrons each.
• So in the final molecule, there will be two Fe atoms and three sulphate groups.
6. Thus the formula is: Fe₂(SO₄)₃

(f) Chromium(III) oxide
1. This compound contains chromium (Cr) and oxygen (O)
2. Given that, Cr has an oxidation state of +3. So it has lost three electrons.
   ♦ Those three electrons must be accepted by the O atom.
3. The Cr donates 3 electrons. But the O can accept only 2 electrons
4. So we must take the LCM of 2 and 3, which is 6
• We must arrange them in such a way that,
   ♦ Cr donates 6 electrons.
   ♦ O accepts 6 electrons.
5. This can be achieved if:
   ♦ Two Cr atoms donate 3 electrons each.
   ♦ Three O atoms accept 2 electrons each.
• So in the final molecule, there will be two Cr atoms and three O atoms.
6. The formula is: Cr₂O₃

Solved example 8.19
Suggest a list of the substances where carbon can exhibit oxidation states from
–4 to +4 and nitrogen from –3 to +5.
Solution:
• Carbon has four valence electrons. It can either lose one or more (up to four) of those electrons to obtain oxidation numbers +1, +2, +3 and +4
(Losing four electrons will give octet)
• Or it can gain one or more (up to four) electrons from other atoms to obtain oxidation numbers -1, -2, -3 and -4.
(Gaining four electrons will give octet because, four electrons are already present)
• Further, the loss and gain may be equal, to obtain zero oxidation state.
• Let us see some examples:

Example for zero oxidation state:
1. In CH₂Cl₂, the C atom is connected to four single bonds. This is shown in fig.8.16(a) below:

Fig.8.16

2. Each of the two Cl atoms will pull one electron from C. This results in a charge of +2
3. Each of the two H atoms will donate one electron to C. This results in a charge of -2.
4, Thus the oxidation state is (+2 + -2) = 0

• We can use the basic method also:
$\mathbf\small{\rm{\overset{x}{C}_2\,\overset{+1}{H}_2\,\overset{-1}{Cl}_2}}$
• We get: 2x + 2 -2 = 0 ⇒ 2x + 0 = 0 ⇒ 2x = 0 ⇒ x = 0

Example for +1 oxidation state:
1. In C₂Cl₂, each of the two C atoms is connected to a single bond and a triple bond. This is shown in fig.8.16(b) above.
2. The triple bond creates zero net charge because both are C atoms.
3. At each of the two single bonds, the Cl will pull one electron from C. This results in a charge of +1 for C.
4. Thus the oxidation state of each of the two C atoms in C₂Cl₂ is +1.

• We can use the basic method also:
$\mathbf\small{\rm{\overset{x}{C}_2\,\overset{-1}{Cl}_2}}$
• We get: 2x -2 = 0 ⇒ 2x = 2 ⇒ x = +1

Example for -1 oxidation state:
1. In C₂H₂, each of the two C atoms is connected to a single bond and a triple bond.  This is shown in fig.8.16(c) above.
2. The triple bond creates zero net charge because both are C atoms.
3. At each of the two single bonds, the H will donate one electron to C. This results in a charge of -1 for C.
4. Thus the oxidation state of each of the two C atoms in C₂H₂ is -1.

• We can use the basic method also:
$\mathbf\small{\rm{\overset{x}{C}_2\,\overset{+1}{H}_2}}$
• We get: 2x +2 = 0 ⇒ 2x = -2 ⇒ x = -1

Example for +2 oxidation state:
• Consider CHCl₃. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{C}\,\overset{+1}{H}\,\overset{-1}{Cl}_3}}$
• We get: x +1 - 3 = 0 ⇒ x-2 = 0 ⇒ x = +2
• The reader may draw the structural formula and check the above result.

Example for -2 oxidation state:
• Consider CH₃Cl. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{C}\,\overset{+1}{H}_3\,\overset{-1}{Cl}}}$
• We get: x +3 - 1 = 0 ⇒ x+2 = 0 ⇒ x = -2
• The reader may draw the structural formula and check the above result.

Example for +3 oxidation state:
• Consider Cl₃C一CCl₃. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{C}_2\,\overset{-1}{Cl}_6}}$
• We get: 2x - 6 = 0 ⇒ 2x = 6 ⇒ x = +3
• The reader may draw the structural formula and check the above result.

Example for -3 oxidation state:
• Consider H₃C一CH₃. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{C}_2\,\overset{+1}{H}_6}}$
• We get: 2x + 6 = 0 ⇒ 2x = -6 ⇒ x = -3
• The reader may draw the structural formula and check the above result.

Example for +4 oxidation state:
• Consider CCl₄. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{C}\,\overset{-1}{Cl}_4}}$
• We get: x - 4 = 0 ⇒ x = +4
• The reader may draw the structural formula and check the above result.

Another example for +4 oxidation state:
• Consider CO₂. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{C}\,\overset{-2}{O}_2}}$
• We get: x - 4 = 0 ⇒ x = +4
• The reader may draw the structural formula and check the above result.

Example for -4 oxidation state:
• Consider CH₄. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{C}\,\overset{+1}{H}_4}}$
• We get: x + 4 = 0 ⇒ x = -4
• The reader may draw the structural formula and check the above result.

---

• Nitrogen has five valence electrons. It can either lose one or more (up to five) of those electrons to obtain oxidation numbers +1, +2, +3, +4 and +5.
(Losing five electrons will give octet)
• Or it can gain one or more (up to three) electrons from other atoms to obtain oxidation numbers -1, -2, and -3.
(Gaining three electrons will give octet because, five electrons are already present)
• Further, the loss and gain may be equal, to obtain zero oxidation state.
• Let us see some examples:

Example for zero oxidation state:
1. In N₂, the two N atoms are connected by a triple bond. This is shown in fig.8.17(a) below:

Fig.8.17

2. The two N atoms will exert equal pulls on the electrons. So neither N atom can gain or lose electrons.
3. Thus the oxidation state of each N atom will be zero.

Example for +1 oxidation state:
1. In N₂O, the left N is connected to the right N only. But the right N is connected to left N and the O. This is shown in fig.8.17(b) above.
2. The triple bond creates zero net charge because both are N atoms.
3. At the single bond, the O will pull one electron from N. This results in a charge of +1 for N.
4. But when we consider the resonance structures, both N will have an oxidation state of +1. We will see it in higher classes.

• We can use the basic method also:
$\mathbf\small{\rm{\overset{x}{N}_2\,\overset{-2}{O}}}$
• We get: 2x -2 = 0 ⇒ 2x = 2 ⇒ x = +1

Example for -1 oxidation state:
1. In N₂H₂, each of the two N atoms is connected to a single bond to one H atom. Also, there is a double bond between the two N atoms.  This is shown in fig.8.17(c) above.
2. The double bond creates zero net charge because both are N atoms.
3. At each of the two single bonds, the H will donate one electron to N. This results in a charge of -1 for N.
4. Thus the oxidation state of each of the two N atoms in N₂H₂ is -1.

• We can use the basic method also:
$\mathbf\small{\rm{\overset{x}{N}_2\,\overset{+1}{H}_2}}$
• We get: 2x +2 = 0 ⇒ 2x = -2 ⇒ x = -1

Example for +2 oxidation state:
• Consider NO. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{N}\,\overset{-2}{O}}}$
• We get: x - 2 = 0 ⇒ x = +2
• The reader may draw the structural formula and check the above result.

Example for -2 oxidation state:
• Consider N₂H₄. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{N}_2\,\overset{+1}{H}_4}}$
• We get: 2x + 4 = 0 ⇒ x+2 = 0 ⇒ x = -2
• The reader may draw the structural formula and check the above result.

Example for +3 oxidation state:
• Consider N₂O₃. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{N}_2\,\overset{-2}{O}_3}}$
• We get: 2x - 6 = 0 ⇒ 2x = 6 ⇒ x = +3
• The reader may draw the structural formula and check the above result.

Example for -3 oxidation state:
• Consider NH₃. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{N}\,\overset{+1}{H}_3}}$
• We get: x + 3 = 0 ⇒ x = -3
• The reader may draw the structural formula and check the above result.

Example for +4 oxidation state:
• Consider NO₂. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{N}\,\overset{-2}{O}_2}}$
• We get: x - 4 = 0 ⇒ x = +4
• The reader may draw the structural formula and check the above result.

Example for +5 oxidation state:
• Consider N₂O₅. We will use the basic method:
$\mathbf\small{\rm{\overset{x}{N}_2\,\overset{-2}{O}_5}}$
• We get: 2x - 10 = 0 ⇒ 2x =10 ⇒ x = +5
• The reader may draw the structural formula and check the above result.

Solved example 8.20
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 ?
Solution:
Given that:
• Sulphur dioxide (SO₂) can act as an oxidizing agent.
   ♦ Sulphur dioxide can act as a reducing agent also.
• Hydrogen peroxide (H₂O₂) can act as an oxidizing agent.
   ♦ Hydrogen peroxide can act as a reducing agent also.
• Ozone (O₃) can act only as an oxidizing agent.
• Nitric acid (HNO₃) can act only as an oxidizing agent.
◼ We are asked to write the reason. It can be written in 4 steps:
1. The oxidation number of S in SO₂ can be calculated as:
   ♦ $\mathbf\small{\rm{\overset{x}{S}\,\overset{-2}{O}_2}}$
   ♦ x - 4 = 0 ⇒ x = +4
• So in SO₂, the S atom has an oxidation number of +4
• But in general,
   ♦ the oxidation number of S can be greater than +4 in some compounds.
         ✰ +6 in H₂SO₄ is an example.
   ♦ the oxidation number of S can be less than +4 in some other compounds.
         ✰ -2 in H₂S is an example.
• So if SO₂ is going to react with an element which is a strong electron puller (ie., an element which is highly electronegative), the S will donate electrons. Then SO₂ will be a reducing agent. The oxidation number of S will rise above +4.
• If SO₂ is going to react with an element which is a weak electron puller, the S will accept electrons. Then SO₂ will be an oxidizing agent. The oxidation number of S will fall below +4.
2. The oxidation number of O in H₂O₂ can be calculated as:
   ♦ $\mathbf\small{\rm{\overset{+1}{H}_2\,\overset{x}{O}_2}}$
   ♦ 2 + 2x = 0 ⇒ 2x = -2 ⇒ x = -1
• So in H₂O₂, the O atom has an oxidation number of -1
• But in general,
   ♦ the oxidation number of O can be greater than -1 in some compounds.
         ✰ +2 in OF₂ is an example.
   ♦ the oxidation number of O can be less than -1 in some other compounds.
         ✰ -2 in H₂O is an example.
• So if H₂O₂ is going to react with an element which is a strong electron puller (ie., an element which is highly electronegative), the O will donate electrons. Then H₂O₂ will be a reducing agent. The oxidation number of O will rise above -1.
• If H₂O₂ is going to react with an element which is a weak electron puller, the O will accept electrons. Then H₂O₂ will be an oxidizing agent. The oxidation number of O will fall below -1.
3. Ozone is an elemental form of oxygen. So the oxidation number of O in O₃ will be 0.
• In general,
   ♦ the oxidation number of O in O₃ will not increase to a value above zero.
   ♦ the oxidation number of O in O₃ can decrease to a value below zero.
• So if O₃ is going to react with an element which is a strong electron puller (ie., an element which is highly electronegative), the O will not donate electrons. That is., O₃ cannot act as a reducing agent.
• If O₃ is going to react with an element which is a weak electron puller, the O will accept electrons. Then O₃ will be an oxidizing agent. The oxidation number of O will fall below 0.
4. The oxidation number of N in HNO₃ can be calculated as:
   ♦ $\mathbf\small{\rm{\overset{+1}{H}\,\overset{x}{N}\,\overset{-2}{O}_3}}$
   ♦ 1 + x - 6 = 0 ⇒ x = +5
• So in HNO₃, the N atom has an oxidation number of +5
• In general,
   ♦ the oxidation number of N will not increase to a value above +5.
   ♦ the oxidation number of N can decrease to a value below +5.
         ✰ -3 in NH₃ is an example.
• So if HNO₃ is going to react with an element which is a strong electron puller (ie., an element which is highly electronegative), the N will not donate electrons. That is., HNO₃ cannot act as a reducing agent.
• If HNO₃ is going to react with an element which is a weak electron puller, the N will accept electrons. Then HNO₃ will be an oxidizing agent. The oxidation number of N will fall below +5.

---

Link to some more solved examples is given below:

Solved examples from 8.21 to 8.30

---

We have completed the discussions in this chapter. In the next chapter, we will see Hydrogen.


Previous

Contents

Next

Copyright©2021 Higher secondary chemistry.blogspot.com