Chapter 8.17 - Solved Examples Related To Redox Reactions

In the previous section, we completed the discussions about redox reactions. In this section, we will see some solved examples related to this chapter in general.

Solved example 8.16
What are the oxidation number of the underlined elements in each of the following and how do you rationalize your results ?
(a) KI₃  (b) H₂S₄O₆  (c) Fe₃O₄  (d) CH₃CH₂OH  (e) CH₃COOH  
Solution:
Part(a): KI₃
This can be written in 6 steps:
1. Consider KI₃
• It is an ionic compound. In aqueous solution, it dissociates into K⁺ and I₃^- ions
• The I₃^- ions are stable and exist independently. So we will analyze it.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{I^{-1}_3}\,}}$
• Thus we get: 3x = -1 ⇒ x = - ¹⁄₃
3. This indicates that, each I has gained ¹⁄₃ electrons.
• But electrons are whole objects. They cannot be cut into fractions.
◼ Then why do we get '¹⁄₃' ?
The following steps from (4) to (8) will give the answer:
4. The structure of I₃^- is shown below:
I 一 I ^- 一 I
• There are two bonds in total.
Consider the left side bond. On both sides of that bond, the same atom I is present. So the pull will be same. Neither of those I atoms can gain or lose electrons. So the oxidation numbers for both those I atoms should be '0'.
• Similar is the case with the right side bond. On both sides of that bond, the same atom I is present. So the pull will be same. Neither of those I atoms can gain or lose electrons. So the oxidation numbers for both those I atoms should be '0'.
• But the middle atom already has an extra electron. So it's oxidation number is -1.
5. Now we can write the oxidation numbers of the I atoms:
   ♦ Left side I atom has an oxidation number of 0
   ♦ Middle I atom has an oxidation number of -1
   ♦ Right side I atom has an oxidation number of 0
6. Thus, in the structural formula of I₃^-, we see that, different I atoms have different oxidation numbers.
• But in the molecular formula, I₃^-, we write I only once. So we write the average oxidation number. The average can be calculated in 3 steps:
(i) Total oxidation number of I atoms = 0 + -1 + 0 = -1
(ii) Number of S atoms = 3
(iii) So average oxidation number = -¹⁄₃

Part (b): H₂S₄O₆
This can be written in 8 steps:
1. Consider H₂S₄O₆
• We know that, the oxidation number of H will be +1 and that of O will be -2. We want to find the oxidation number of S.
2. For that, we write: $\mathbf\small{\rm{\overset{+1}{H}_2\,\overset{x}{S}_4\,\overset{-2}{O}_6}}$
• Thus we get: 2 + 4x - 12 = 0 ⇒ 4x - 10 = 0 ⇒ x = +⁵⁄₂
3. This indicates that, each S has lost ⁵⁄₂ electrons.
• But electrons are whole objects. They cannot be cut into fractions.
◼ Then why do we get '⁵⁄₂' ?
The following steps from (4) to (8) will give the answer:
4. The structure of H₂S₄O₆ is shown in fig.8.10(a) below:

Fig.8.10

• There are 11 bonds in total. Since we are trying to find the oxidation number of S, we need to consider only those bonds linked to S. So we need not consider the two O一H bonds. Thus there are nine bonds to analyze.
• In fig.b,
   ♦ the three S一S bonds are marked in red color. They are named R₁ to R₃.
   ♦ the two S一O bonds are marked in yellow color. They are named as Y₁ and Y₂.
   ♦ the four S＝O bonds are marked in green color. They are named G₁ to G₄.
5. Consider G₁. It is a double bond. So there will be four electrons
• Originally,
   ♦ two electrons belong to S
   ♦ two electrons belong to O
• But O being more electronegative, will pull the electrons belonging to S also. So S loses two electrons.
   ♦ This happens in G₂ also.
• So there is now a loss of 4 electrons.
• Consider Y₁. It is a single bond. So there will be two electrons
• Originally,
   ♦ one electron belong to S
   ♦ one electron belong to O
• But O being more electronegative, will pull the electron belonging to S also. So S loses one electron.
• So there is now a loss of 5 electrons.
• Consider R₁. This bond is between two S atoms. The pulls will be equal.
   ♦ Since the pulls are equal, the S will not lose or gain any electrons at R₁
• So the net effect is:
The left most S, loses five electrons and gets an oxidation number of +5.
6. By symmetry, the same steps in (5) can be written about the right most S atom. We will get:
The right most S, loses five electrons and gets an oxidation number of +5.
7. Consider the two middle S atoms. They are surrounded by S atoms on either sides. So the pulls will be equal towards the sides. They neither lose nor gain electrons towards the sides.
   ♦ Thus their oxidation numbers will be 0.
8. Thus, in the structural formula of H₂S₄O₆, we see that, different S atoms have different oxidation numbers.
• But in the molecular formula, H₂S₄O₆, we write S only once. So we write the average oxidation number. The average can be calculated in 3 steps:
(i) Total oxidation number of S atoms = 5 + 0 + 0 + 5 = 10
(ii) Number of S atoms = 4
(iii) So average oxidation number = +¹⁰⁄₄ = +⁵⁄₂

Part (c): Fe₃O₄
This can be written in 6 steps:
1. Consider Fe₃O₄
• We know that, the oxidation number of O will be -2. We want to find the oxidation number of Fe.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{Fe}_3\,\overset{-2}{O}_4}}$
• Thus we get: 3x - 8 = 0 ⇒ 3x = 8 ⇒ x = +⁸⁄₃
3. This indicates that, each S has lost ⁸⁄₃ electrons.
• But electrons are whole objects. They cannot be cut into fractions.
◼ Then why do we get '⁸⁄₃' ?
The following steps from (4) to (6) will give the answer:
4. The Fe₃O₄ is a stoichiometric mixture of FeO and Fe₂O₃.
To get one mole of Fe₃O₄, we must take one mole of FeO and one mole of Fe₂O₃ .
5. Let us find the oxidation number of Fe in FeO and Fe₂O₃ individually:
◼ For FeO, we have: $\mathbf\small{\rm{\overset{x}{Fe}\,\overset{-2}{O}}}$
• Thus we get: x - 2 = 0 ⇒ x = +2
◼ For Fe₂O₃, we have: $\mathbf\small{\rm{\overset{x}{Fe}_2\,\overset{-2}{O}_3}}$
• Thus we get: 2x - 6 = 0 ⇒ 2x = 6 ⇒ x = +3
6. So in one molecule of Fe₃O₄,
   ♦ one Fe atom has an oxidation state of +2
   ♦ two Fe atoms have an oxidation state of +3
• But in the molecular formula, Fe₃O₄, we write Fe only once. So we write the average oxidation number. The average can be calculated in 3 steps:
(i) Total oxidation number of Fe atoms = 2 + 3 + 3 = 8
(ii) Number of Fe atoms = 3
(iii) So average oxidation number = ⁸⁄₃

Part (d): CH₃CH₂OH
This can be written in 5 steps:
1. Consider CH₃CH₂OH
• We know that, the oxidation number of H will be +1 and that of O will be -2. We want to find the oxidation number of the two C atoms.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{C}\,\overset{+1}{H}_3\,\overset{x}{C}\,\overset{+1}{H}_2\,\overset{-2}{O}\,\overset{+1}{H}}}$
• Thus we get: x + 3 + x + 2 - 2 + 1 = 0 ⇒ 2x + 4 = 0 ⇒ x = -2
3. This indicates that, each C has gained 2 electrons.
• Let us cross check by analyzing the structure of CH₃CH₂OH
4. The structure of CH₃CH₂OH is shown in fig.8.11(a) below:

Fig.8.11

• There are 8 bonds in total. Since we are trying to find the oxidation number of C, we need to consider only those bonds linked to C. So we need not consider the O一H bond. Thus there are 7 bonds to analyze.
In fig.b,
• Consider the left side C atom
   ♦ From each of the three green bonds, this C will gain -1 charge.
         ✰ This is because, C is more electronegative than H.
   ♦ From the red bond, this C has no gain or loss of electrons.
         ✰ This is because, C and C are equally powerful.
   ♦ So in total, this C has a charge of -3
• Consider the right side C atom
   ♦ From each of the two green bonds, this C will gain -1 charge.
         ✰ This is because, C is more electronegative than H.
   ♦ From the red bond, this C has no gain or loss of electrons.
         ✰ This is because, C and C are equally powerful.
   ♦ From the yellow bond, this C has a loss of one electron.
         ✰ This is because, O is more electronegative than C.
   ♦ So in total, this C has a charge of (-2 + 1) = -1
5. Thus, in the structural formula of CH₃CH₂OH, we see that, different C atoms have different oxidation numbers.
   ♦ The left C atom has an oxidation number of -3
   ♦ The right C atom has an oxidation number of -1
• If we write the formula as C₂H₆O, we will have to take the average.
• The average can be calculated in 3 steps:
(i) Total oxidation number of C atoms = -3 + -1 = -4
(ii) Number of C atoms = 2
(iii) So average oxidation number = ^-4⁄₂ = -2

Part (e): CH₃COOH
This can be written in 8 steps:
1. Consider CH₃COOH 
• We know that, the oxidation number of H will be +1 and that of O will be -2. We want to find the oxidation number of the two C atoms.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{C}\,\overset{+1}{H}_3\,\overset{x}{C}\,\overset{-2}{O}\,\overset{-2}{O}\,\overset{+1}{H}}}$
• Thus we get: x + 3 + x - 2 - 2 + 1 = 0 ⇒ 2x + 4 - 4 = 0 ⇒ 2x = 0 ⇒ x = 0
3. This indicates that, each C has neither gained nor lost any electrons.
• Let us cross check by analyzing the structure of CH₃COOH 
4. The structure of CH₃COOH is shown in fig.8.12(a) below:

Fig.8.12

• There are 7 bonds in total. Since we are trying to find the oxidation number of C, we need to consider only those bonds linked to C. So we need not consider the O一H bond. Thus there are 6 bonds to analyze.
In fig.b,
• Consider the left side C atom
   ♦ From each of the three green bonds, this C will gain -1 charge.
         ✰ This is because, C is more electronegative than H.
   ♦ From the red bond, this C has no gain or loss of electrons.
         ✰ This is because, C and C are equally powerful.
   ♦ So in total, this C has a charge of -3
• Consider the right side C atom
   ♦ From the red bond, this C has no gain or loss of electrons.
         ✰ This is because, C and C are equally powerful.
   ♦ From the yellow bond, this C has a loss of one electron.
         ✰ This is because, O is more electronegative than C.
   ♦ From the magenta double bond, this C has a loss of two electrons.
         ✰ This is because, O is more electronegative than C.
   ♦ So in total, this C has a charge of (+1 + 2) = +3
5. Thus, in the structural formula of CH₃COOH we see that, different C atoms have different oxidation numbers.
   ♦ The left C atom has an oxidation number of -3
   ♦ The right C atom has an oxidation number of +3
• If we write the formula as C₂H₄O₂, we will have to take the average.
• The average can be calculated in 3 steps:
(i) Total oxidation number of C atoms = -3 + +3 = 0
(ii) Number of C atoms = 2
(iii) So average oxidation number = ⁰⁄₂ = 0

Solved example 8.17
Fluorine reacts with ice and results in the change:
H₂O (s) + F₂ (g) gives HF (g) + HOF (g)
Justify that this reaction is a redox reaction 
Solution:
1. Writing the oxidation numbers in reactants and products, we get:
$\mathbf\small{\rm{\overset{+1}{H_2}\,\overset{-2}{O}\,(s)+\overset{0}{F_2}\,(g)\rightarrow \overset{+1}{H}\,\overset{-1}{F}\,(g)+\overset{+1}{H}\,\overset{-2}{O}\,\overset{+1}{F}\,(g)}}$
2. We see that,
    ♦ Oxidation number of F increases from 0 to +1.
    ♦ Oxidation number of F decreases from 0 to -1.
3. So F undergoes both oxidation and reduction.
    ♦ Thus it is a disproportionation redox reaction.

Solved example 8.18
Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇^2- and NO₃^-. Suggest structure of these compounds. Count for fallacy.
Solution:
H₂SO₅
1. We know that, the oxidation number of H will be +1 and that of O will be -2. We want to find the oxidation number of S.
2. For that, we write: $\mathbf\small{\rm{\overset{+1}{H}_2\,\overset{x}{S}\,\overset{-2}{O}_5}}$
• Thus we get: 2 + x - 10 = 0 ⇒ x - 8 = 0 ⇒ x = 8
3. This indicates that, the S has lost 8 electrons.
• The subshell electronic configuration of S is: 1s²2s²2p⁶3s²3p⁴. So the outermost shell has only 6 electrons. That means, S cannot lose more than 6 electrons.
◼ Then why do we get '8' ?
The following step (4) will give the answer:
4. The structure of H₂S₄O₆ is shown in fig.8.13(a) below:

Fig.8.13

• There are 7 bonds in total. Since we are trying to find the oxidation number of S, we need to consider only those bonds linked to S. So we need not consider the O一H bonds and the O一O bond. Thus there are 4 bonds to analyze.
In fig.b,
• Consider the S atom
   ♦ From each of the two yellow single bonds, this S will gain +1 charge.
         ✰ This is because, O is more electronegative than S.
   ♦ From each of the two green double bonds, this S will gain +2 charge.
         ✰ This is because, O is more electronegative than S.
   ♦ So in total, this S has a charge of (2  × +1) + (2  × +2) = +6
• The value obtained in (2) is wrong.

Cr₂O₇^2-
1. We know that, the oxidation number of O will be -2. We want to find the oxidation number of Cr.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{Cr}_2\,\overset{-2}{O_7^{-1}}}}$
• Thus we get: 2x - 14 = -2 ⇒ x = +6
3. This indicates that, each Cr has lost 6 electrons. Cr is a transition metal. Sometimes the electrons in the d subshell also takes part in bonding. So Cr can lose 6 electrons. Thus there is no fallacy.
4. Let us check the structure of Cr₂O₇^2-. It is shown in fig.8.14(a) below:

Fig.8.14

• There are 7 bonds in total. We need to consider them all because, all are related to Cr.
In fig.b,
• Consider the left side Cr atom
   ♦ From each of the two red single bonds, this Cr will gain +1 charge.
         ✰ This is because, O is more electronegative than Cr.
   ♦ From each of the two yellow double bonds, this Cr will gain +2 charge.
         ✰ This is because, O is more electronegative than Cr.
   ♦ So in total, this Cr has a charge of (2  × +1) + (2  × +2) = +6
• By symmetry, we can write, the right side Cr also will have a charge of +6.

NO₃^-
1. We know that, the oxidation number of O will be -2. We want to find the oxidation number of N.
2. For that, we write: $\mathbf\small{\rm{\overset{x}{N}\,\overset{-2}{O^{-1}_3}}}$
• Thus we get: x - 6 = -1 ⇒ x - 5 = 0 ⇒ x = 5
3. This indicates that, the N has lost 5 electrons.
• The subshell electronic configuration of N is: 1s²2s²2p³. So the outermost shell has 5 electrons. That means, N can lose 5 electrons. Thus there is no fallacy.
4. Let us check the structure of NO₃^- It is shown in fig.8.15 below:

Fig.8.15

• Out of the five outer electrons,
    ♦ one is used to form the single bond.
    ♦ two are used to form the double bond.
    ♦ The remaining two are used to form the coordinate bond (a bond in which both electrons are donated by a single atom).
5. Let us see the effects caused by O atom:
    ♦ At the single bond, the O atom pulls the single electron of N.
    ♦ At the double bond, the O atom pulls the two electrons of N.
    ♦ At the coordinate bond, the O atom pulls both the electrons of N
• Thus in total, N attains a charge of +5.

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We have completed the discussions in this chapter. In the next section, we will see some solved examples related to this chapter in general.


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