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.

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• 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.

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Link to some more solved examples is given below:

Solved examples from 8.21 to 8.30

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


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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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Chapter 8.13 - Half Reaction Method When Medium Is Basic

In the previous section, we saw the steps required for acidic medium. In this section, we will see the steps required for basic medium. 

Example 7
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:
1. The reactants are: MnO₄^- (aq) and Br^- (aq). (See list of common polyatomic ions)
2. The products are: MnO₂ (s) and BrO₃^- (aq).
3. So the skeletal equation is:
MnO₄^- (aq) + Br^- (aq) → MnO₂ (s) + BrO₃^- (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Br is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Br^- (aq) → BrO₃^- (aq)
• The reduction half reaction is:
MnO₄^- (aq) → MnO₂ (s) (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Br atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are three excess O atoms on the product side. So we must add three H₂O molecules on the reactant side. We get:
Br^- (aq) + 3H₂O (l) → BrO₃^- (aq)
• Now there are six excess H atoms on the reactant side. So we must put six H⁺ ions on the product side. We get:
Br^- (aq) + 3H₂O (l) → BrO₃^- (aq) + 6H⁺
• To balance the six H⁺ ions, we add six OH^- ions on both sides. We get:
Br^- (aq) + 3H₂O (l) + 6OH^- (aq) → BrO₃^- (aq) + 6H⁺ + 6OH^- (aq)
• The H⁺ and OH^- ions on the product side reacts together to give six H₂O molecules. So the equation becomes:
Br^- (aq) + 3H₂O (l) + 6OH^- (aq) → BrO₃^- (aq) + 6H₂O (l)
• There are H₂O molecules on both sides. So the net equation is:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the reactant side. So we must add two H₂O molecules on the product side. We get:
MnO₄^- (aq) → MnO₂ (s) (aq) + 2H₂O
• Now there are four extra H atoms on the product side. So we must put four H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 4H⁺ → MnO₂ (s) (aq) + 2H₂O
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
MnO₄^- (aq) + 4H⁺ (aq) + 4OH^- (aq) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give four H₂O molecules. So the equation becomes:
MnO₄^- (aq) + 4H₂O (l) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
MnO₄^- (aq) + 2H₂O (l) → MnO₂ (s) (aq) + 4OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the Br^- ion) = -1
   ♦ 6 No. × 1- (from the OH^- ion) = -6
   ♦ Total = -7
Product side:
   ♦ 1 No. × 1- (from the BrO₃^- ion) = -1
• So there is an excess charge of -6 on the reactant side. Therefore we must add 6e^- on the product side. We get:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l) + 6e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
Product side:
   ♦ 4 No. × 1- (from the OH^- ion) = -4
• So there is an excess charge of -3 on the product side. Therefore we must add 3e^- on the reactant side. We get:
MnO₄^- (aq) + 2H₂O (l) +3e^- → MnO₂ (s) (aq) + 4OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 6e^-.
    ♦ The reduction half reaction has 3e^-.
• So we multiply the reduction half reaction by '3'.
Also we keep the oxidation half reaction as such. We get:
Br^- (aq) + 6OH^- (aq) → BrO₃^- (aq) + 3H₂O (l) + 6e^-
2MnO₄^- (aq) + 4H₂O (l) +6e^- → 2MnO₂ (s) (aq) + 8OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 4H₂O (l) +6e^- + Br^- (aq) + 6OH^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + BrO₃^- (aq) + 3H₂O (l) + 6e^-
• So the net equation is:
2MnO₄^- (aq) + H₂O (l) + Br^- (aq) → 2MnO₂ (s) (aq) + 2OH^- (aq) + BrO₃^- (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 9 No., Br - 1 No., H - 2 No. 
Product side:
   ♦ Mn - 2 No., O - 9 No., Br - 1 No., H - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 1 No. × 1- (from the Br^- ion) = -1
   ♦ Total = -3
Product side:
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ 1 No. × 1- (from the BrO3- ion) = -1
   ♦ Total = -3
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + H₂O (l) + Br^- (aq) → 2MnO₂ (s) (aq) + 2OH^- (aq) + BrO₃^- (aq)

Example 8
Balance the following equation in basic medium:
CrO₂^- (aq) + ClO^- (aq) → CrO₄^2- (aq) + Cl^- (aq)
Solution:
The given skeletal equation is:
CrO₂^- (aq) + ClO^- (aq) → CrO₄^2- (aq) + Cl^- (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{
\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow \overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)
+\overset{-1}{Cl^{-}}(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cr is oxidized and Cl is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
CrO₂^- (aq) → CrO₄^2- (aq)
• The reduction half reaction is:
ClO^- (aq) → Cl^- (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Cr atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Cl atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the product side. So we must add two H₂O molecules on the reactant side. We get:
CrO₂^- (aq) + 2H₂O (l) → CrO₄^2- (aq)
• Now there are four excess H atoms on the reactant side. So we must put four H⁺ ions on the product side. We get:
CrO₂^- (aq) + 2H₂O (l) → CrO₄^2- (aq) + 4H⁺ (aq)
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
CrO₂^- (aq) + 2H₂O (l) + 4OH^- (aq) → CrO₄^2- (aq) + 4H⁺ (aq) + 4OH^- (aq)
• The H⁺ and OH^- ions on the product side reacts together to give four H₂O molecules. So the equation becomes:
CrO₂^- (aq) + 2H₂O (l) + 4OH^- (aq) → CrO₄^2- (aq) + 4H₂O (l)
• There are H₂O molecules on both sides. So the net equation is:
CrO₂^- (aq) + 4OH^- (aq) → CrO₄^2- (aq) + 2H₂O (l)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there is one excess O atom on the reactant side. So we must add one H₂O molecule on the product side. We get:
ClO^- (aq) → Cl^- (aq) + H₂O (l)
• Now there are two extra H atoms on the product side. So we must put two H⁺ ions on the reactant side. We get:
ClO^- (aq) + 2H⁺ → Cl^- (aq) + H₂O (l)
• To balance the two H⁺ ions, we add two OH^- ions on both sides. We get:
ClO^- (aq) + 2H⁺ + 2OH^- (aq) → Cl^- (aq) + H₂O (l) + 2OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give two H₂O molecules. So the equation becomes:
ClO^- (aq) + 2H₂O (l) → Cl^- (aq) + H₂O (l) + 2OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
ClO^- (aq) + H₂O (l) → Cl^- (aq) + 2OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the CrO₂^- ion) = -1
   ♦ 4 No. × 1- (from the OH^- ion) = -4
   ♦ Total = -5
Product side:
   ♦ 1 No. × 2- (from the CrO₄^2- ion) = -2
• So there is an excess charge of -3 on the reactant side. Therefore we must add 3e^- on the product side. We get:
CrO₂^- (aq) + 4OH^- (aq) → CrO₄^2- (aq) + 2H₂O (l) + 3e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the ClO^- ion) = -1
Product side:
   ♦ 1 No. × 1- (from the Cl^- ion) = -1
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ Total = -3
• So there is an excess charge of -2 on the product side. Therefore we must add 2e^- on the reactant side. We get:
ClO^- (aq) + H₂O (l) + 2e^- → Cl^- (aq) + 2OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 3e^-.
    ♦ The reduction half reaction has 2e^-.
• So we multiply the oxidation half reaction by '2'.
Also we multiply the reduction half reaction by '3'. We get:
2CrO₂^- (aq) + 8OH^- (aq) → 2CrO₄^2- (aq) + 4H₂O (l) + 6e^-
3ClO^- (aq) + 3H₂O (l) + 6e^- → 3Cl^- (aq) + 6OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2CrO₂^- (aq) + 8OH^- (aq) + 3ClO^- (aq) + 3H₂O (l) + 6e^- → 2CrO₄^2- (aq) + 4H₂O (l) + 6e^- + 3Cl^- (aq) + 6OH^- (aq)
The net equation is:
2CrO₂^- (aq) + 2OH^- (aq) + 3ClO^- (aq) → 2CrO₄^2- (aq) + H₂O (l) + 3Cl^- (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Cr - 2 No., O - 9 No., Cl - 3 No., H - 2 No. 
Product side:
   ♦ Cr - 2 No., O - 9 No., Cl - 3 No., H - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the CrO₂^- ion) = -2
   ♦ 2 No. × 1- (from the OH^- ion) = -2
   ♦ 3 No. × 1- (from the ClO^- ion) = -3
   ♦ Total = -7
Product side:
   ♦ 2 No. × 2- (from the CrO₄^2- ion) = -4
   ♦ 3 No. × 1- (from the 3Cl^- ion) = -3
   ♦ Total = -7
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2CrO₂^- (aq) + 2OH^- (aq) + 3ClO^- (aq) → 2CrO₄^2- (aq) + H₂O (l) + 3Cl^- (aq)

Example 9
Permanganate (VII) ion MnO₄^- in basic solution oxidizes iodide ion I^- to produce molecular iodine (I₂) and manganese (iv) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction.
Solution:
The skeletal equation is:
MnO₄^- (aq) + I^- (aq) → MnO₂ (s) + I₂ (aq)
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{I^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{0}{I_2}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
I is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
I^- (aq) → I₂ (aq)
• The reduction half reaction is:
MnO₄^- (aq) → MnO₂ (s) (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, we put '2' in front of I^- in the reactant side. We get:
2I^- (aq) → I₂ (aq)
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are the same on both sides. So it is already balanced.
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are two excess O atoms on the reactant side. So we must add two H₂O molecules on the product side. We get:
MnO₄^- (aq) → MnO₂ (s) (aq) + 2H₂O
• Now there are four extra H atoms on the product side. So we must put four H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 4H⁺ → MnO₂ (s) (aq) + 2H₂O
• To balance the four H⁺ ions, we add four OH^- ions on both sides. We get:
MnO₄^- (aq) + 4H⁺ (aq) + 4OH^- (aq) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• The H⁺ and OH^- ions on the reactant side reacts together to give four H₂O molecules. So the equation becomes:
MnO₄^- (aq) + 4H₂O (l) → MnO₂ (s) (aq) + 2H₂O + 4OH^- (aq)
• There are H₂O molecules on both sides. So the net equation is:
MnO₄^- (aq) + 2H₂O (l) → MnO₂ (s) (aq) + 4OH^- (aq)
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 2 No. × 1- (from the I^- ion) = -2
Product side:
   ♦ zero
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
2I^- (aq) → I₂ (aq) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
Product side:
   ♦ 4 No. × 1- (from the OH^- ion) = -4
• So there is an excess charge of -3 on the product side. Therefore we must add 3e^- on the reactant side. We get:
MnO₄^- (aq) + 2H₂O (l) +3e^- → MnO₂ (s) (aq) + 4OH^- (aq)
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 3e^-.
• So we multiply the oxidation half reaction by '3'.
• Also we multiply the reduction half reaction by '2'.
We get:
6I^- (aq) → 3I₂ (aq) + 6e^-
2MnO₄^- (aq) + 4H₂O (l) +6e^- → 2MnO₂ (s) (aq) + 8OH^- (aq)
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 4H₂O (l) +6e^- + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq) + 6e^-
So the net equation is:
2MnO₄^- (aq) + 4H₂O (l) + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following list for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 12 No., I - 6 No., H - 8 No. 
Product side:
   ♦ Mn - 2 No., O - 12 No., I - 6 No., H - 8 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 6 No. × 1- (from the I^- ion) = -6
   ♦ Total = -8
Product side:
   ♦ 8 No. × 1- (from the OH^- ion) = -8
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + 4H₂O (l) + 6I^- (aq) → 2MnO₂ (s) (aq) + 8OH^- (aq) + 3I₂ (aq)

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The following link gives three more examples:

Solved example 8.11 - Part (a), Part (b) and Part (c)

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• We have completed a discussion on half reaction method.
• In the next section, we will see redox reactions as the basis for titrations.

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Chapter 8.12 - Half Reaction Method When Medium Is Acidic

In the previous section, we saw the steps in half reaction method, related to neutral medium. We saw them while analyzing three examples. In this section, we will see the additional steps required for acidic medium.

Example 4:
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 sulfate ion.
Solution:
1. The reactants are: K₂Cr₂O₇ and Na₂SO₃ 
2. K₂Cr₂O₇ is an ionic compound
   ♦ The ions are: K⁺ and Cr₂O₇^2-
• In the dissolved state, the two ions separate away from each other.
• Cr₂O₇^2- will not dissociate further. It is a polyatomic ion. (A list of such polyatomic ions can be seen here. It is better to remember their formula and names)
3. Na₂SO₃ is an ionic compound
   ♦ The ions are: Na⁺ and SO₃^2-
• In the dissolved state, the two ions separate away from each other.
• SO₃^2- will not dissociate further. It is a polyatomic ion.
4. K⁺ and Na⁺ are spectator ions. They do not take part in the reaction. In the reaction equation, they will appear as such on both sides, and thus will cancel out.
• So the actual reactants are: Cr₂O₇^2- and SO₃^2- 
5. We are given the products: chromium(III) ion and the sulfate ion.
• Chromium(III) ion means, the chromium ion with oxidation state +3. Obviously, it is the Cr³⁺ ion.
• Sulfate ion is known to us from the list. It is the SO₄^2- ion.
6. Steps (1) to (4) give us the reactants. Step (5) gives us the products.
• So now we can write the skeletal equation:
Cr₂O₇^2- + SO₃^2- → Cr³⁺ + SO₄^2-
• We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)\rightarrow
\overset{+3}{Cr^{3+}}\,(aq)+\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
S is oxidized and Cr is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
SO₃^2- (aq) → SO₄^2- (aq)
• The reduction half reaction is:
Cr₂O₇^2- (aq) → Cr³⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of S atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, there are two Cr atoms on the reactant side, but only one on the product side. So we must put '2' in front of Cr³⁺ on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq)
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, SO₄^2- has one O more than SO₃^2-. So we must add one H₂O molecule on the reactant side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq)
• Now there are two extra H atoms on the reactant side. So we must put two H⁺ ions on the product side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq) + 2H⁺ (aq)
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, Cr₂O₇^2- has seven O more than Cr³⁺. So we must add seven H₂O molecules on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq) + 7H₂O
• Now there are fourteen extra H atoms on the product side. So we must put fourteen H⁺ ions on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ → 2Cr³⁺ (aq) + 7H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the SO₃^2- ion) = -2
Product side:
   ♦ 1 No. × 2- (from the SO₄^2- ion) = -2
   ♦ 2 No. × 1+ (from the H⁺ ion) = +2
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
SO₃^2- (aq) + H₂O (l) → SO₄^2- (aq) + 2H⁺ (aq) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ion) = +14
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
• So there is an excess charge of +6 on the reactant side. Therefore we must add 6e^- on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 2Cr³⁺ (aq) + 7H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 6e^-.
• So we multiply the oxidation half reaction by '3'. We get:
3SO₃^2- (aq) + 3H₂O (l) → 3SO₄^2- (aq) + 6H⁺ (aq) + 6e^-
Step 11: Add the two half reactions together.
• In our present case, we get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- + 3SO₃^2- (aq) + 3H₂O (l) → 2Cr³⁺ (aq) + 7H₂O  + 3SO₄^2- (aq) + 6H⁺ (aq) + 6e^- 
• The 6e^- on either sides, cancel each other.
• The 6H⁺ on the product side will take away 6 H⁺ from the reaction side. So 8H⁺ will remain on the reaction side.
• The 3H₂O on the reaction side will take away 3H₂O from the product side. So 4H₂O will remain on the product side.
• We get:
Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8H⁺  → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + 4H₂O
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Cr - 2 No., O - 16 No., S - 3 No., H - 8 No. 
Product side:
   ♦ Cr - 2 No., O - 16 No., S - 3 No., H - 8 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 3 No. × 2- (from the SO₃^2- ions) = -6
   ♦ 8 No. × 1+ (from the H⁺ ions) = +8
   ♦ Total = 0
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
   ♦ 3 No. × 2- (from the SO₄^2- ions) = -6
   ♦ Total = 0
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
Cr₂O₇^2- (aq) + 3SO₃^2- (aq) + 8H⁺  → 2Cr³⁺ (aq) + 3SO₄^2- (aq) + 4H₂O

Example 5:
Balance the equation in acidic medium:
MnO₄^- (aq) + Cl^- (aq) → Mn²⁺ (aq) + Cl₂ (g)
Solution:
We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+\overset{-1}{Cl^{-}}\,(aq)\rightarrow\overset{+2}{Mn^{2+}}\,(aq)
+\overset{0}{Cl_2}\,(g)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cl is oxidized and Mn is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Cl^- (aq) → Cl₂ (g)
• The reduction half reaction is:
MnO₄^- (aq) → Mn²⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, we put '2' in front of Cl^- on the reactant side. We get:
2Cl^- (aq) → Cl₂ (g)
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, the number of Mn atoms are same on both sides. So it is already balanced
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms to balance.
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are extra four O atoms on the reactant side. So we must add four H₂O molecules on the product side. We get:
MnO₄^- (aq) → Mn²⁺ (aq) + 4H₂O
• Now there are eight extra H atoms on the product side. So we must put eight H⁺ ions on the reactant side. We get:
MnO₄^- (aq) + 8H⁺ → Mn²⁺ (aq) + 4H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 2 No. × 1- (from the Cl^- ion) = -2
Product side:
   ♦ zero
• So there is an excess charge of -2 on the reactant side. Therefore we must add 2e^- on the product side. We get:
2Cl^- (aq) → Cl₂ (g) + 2e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 1- (from the MnO₄^- ion) = -1
   ♦ 8 No. × 1+ (from the H⁺ ion) = +8
   ♦ Total = +7
Product side:
   ♦ 1 No. × 2+ (from the Mn²⁺ ion) = +2
• So there is an excess charge of +5 on the reactant side. Therefore we must add 5e^- on the reactant side. We get:
MnO₄^- (aq) + 8H⁺ +5e^- → Mn²⁺ (aq) + 4H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 5e^-.
• So we multiply the oxidation half reaction by '5'.
• Also we multiply the reduction half reaction by '2'.
We get:
10Cl^- (aq) → 5Cl₂ (g) + 10e^-
2MnO₄^- (aq) + 16H⁺ + 10e^- → 2Mn²⁺ (aq) + 8H₂O
Step 11: Add the two half reactions together.
• In our present case, we get:
2MnO₄^- (aq) + 16H⁺ + 10e^- + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g) + 10e-
• The 10e^- on either sides, cancel each other. We get:
2MnO₄^- (aq) + 16H⁺ + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g)
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Mn - 2 No., O - 8 No., Cl - 10 No., H - 16 No. 
Product side:
   ♦ Mn - 2 No., O - 8 No., Cl - 10 No., H - 16 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2 No. × 1- (from the MnO₄^- ion) = -2
   ♦ 10 No. × 1- (from the SO₃^2- ions) = -10
   ♦ 16 No. × 1+ (from the H⁺ ions) = +16
   ♦ Total = +4
Product side:
   ♦ 2 No. × 2+ (from the Mn²⁺ ion) = +4
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
2MnO₄^- (aq) + 16H⁺ + 10Cl^- (aq) → 2Mn²⁺ (aq) + 8H₂O + 5Cl₂ (g)

Example 6:
Balance the equation in acidic medium:
Fe²⁺ (aq) + Cr₂O₇^2- → Fe³⁺ (aq) + Cr³⁺ (aq)
Solution:
We want to balance this equation. It can be written in 12 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+2}{Fe^{2+}}\,(aq)+\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)\rightarrow\overset{+3}{Fe^{3+}}\,(aq)
+\overset{+3}{Cr^{3+}}\,(aq)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Fe is oxidized and Cr is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Fe²⁺ (aq) → Fe³⁺ (aq)
• The reduction half reaction is:
Cr₂O₇^2- → Cr³⁺ (aq)
Step 4: Balance the atoms other than O and H in the oxidation half reaction.
• In our present case, the number of Fe atoms are the same on both sides. So it is already balanced.
Step 5: Balance the atoms other than O and H in the reduction half reaction.
• In our present case, there are two Cr atoms on the reactant side, but only one on the product side. So we must put '2' in front of Cr³⁺ on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq)
Step 6: Balance O atoms in the oxidation half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, there are no O and H atoms to balance.
Step 7: Balance O atoms in the reduction half reaction by using H₂O molecules. Also balance the H atoms by using H⁺ ions.
• In our present case, Cr₂O₇^2- has seven O more than Cr³⁺. So we must add seven H₂O molecules on the product side. We get:
Cr₂O₇^2- (aq) → 2Cr³⁺ (aq) + 7H₂O
• Now there are fourteen extra H atoms on the product side. So we must put fourteen H⁺ ions on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ → 2Cr³⁺ (aq) + 7H₂O
Step 8: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2+ (from the Fe²⁺ ion) = +2
Product side:
   ♦ 1 No. × 3+ (from the Fe³⁺ ion) = +3
• So there is an excess charge of +1 on the product side. Therefore we must add 1e^- on the product side. We get:
Fe²⁺ (aq) → Fe³⁺ (aq) + 1e^-
Step 9: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the list of charges are as follows:
Reactant side:
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ion) = +14
Product side:
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
• So there is an excess charge of +6 on the reactant side. Therefore we must add 6e^- on the reactant side. We get:
Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 2Cr³⁺ (aq) + 7H₂O
Step 10: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 1e^-.
    ♦ The reduction half reaction has 6e^-.
• So we multiply the oxidation half reaction by '6'. We get:
6Fe²⁺ (aq) → 6Fe³⁺ (aq) + 6e^-
Step 11: Add the two half reactions together.
• In our present case, we get:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ + 6e^- → 6Fe³⁺ (aq) + 6e^- + 2Cr³⁺ (aq) + 7H₂O
• The 6e^- on either sides, cancel each other.
• We get:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ → 6Fe³⁺ (aq) + 2Cr³⁺ (aq) + 7H₂O
Step 12: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Fe - 6 No., Cr - 2 No., O - 7 No., H - 14 No. 
Product side:
   ♦ Fe - 6 No., Cr - 2 No., O - 7 No., H - 14 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 6 No. × 2+ (from the Fe²⁺ ions) = +12
   ♦ 1 No. × 2- (from the Cr₂O₇^2- ion) = -2
   ♦ 14 No. × 1+ (from the H⁺ ions) = +14
   ♦ Total = 24
Product side:
   ♦ 6 No. × 3+ (from the Fe³⁺ ions) = +18
   ♦ 2 No. × 3+ (from the Cr³⁺ ion) = +6
   ♦ Total = 24
         ✰ All charges are balanced.
◼  So the balanced equation is same as that obtained in step 11:
6Fe²⁺ (aq) + Cr₂O₇^2- (aq) + 14H⁺ → 6Fe³⁺ (aq) + 2Cr³⁺ (aq) + 7H₂O

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The following link gives three more examples:

Solved example 8.10 - Part (a), Part (b) and Part (c)

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• We have seen the additional steps required in acidic medium when half reaction method is used.
• In the next section, we will see some examples, which take place in basic medium.
   ♦ There we will see the steps required for basic medium.


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Chapter 8.11 - Half Reaction Method When The medium Is Neutral

In the previous section, we completed the discussion on oxidation number method. In this section, we will see half reaction method.

• First we will see redox reactions taking place in neutral medium.
Example 1: We want to balance the equation:
Fe₂O₃ (s) + CO (g) → Fe (s) + CO₂ (g)
It can be written in 10 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+3}{Fe}_2\,\overset{-2}{O}_3\,(s)
+\overset{+2}{C}\,\overset{-2}{O}\,(g)\rightarrow\overset{0}{Fe}\,(s)
+\overset{+4}{C}\,\overset{-2}{O}_2\,(g)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Fe is reduced and C is oxidized.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
CO (g) + O₂ (g) → CO₂ (g)
• The reduction half reaction is:
Fe₂O₃ (s) → Fe (s) + O₂ (g)
Step 4: Balance the oxidation half reaction.
• In our present case, we get:
2CO (g) + O₂ (g) → 2CO₂ (g)
Step 5: Balance the reduction half reaction.
• In our present case, we get:
2Fe₂O₃ (s) → 4Fe (s) + 3O₂ (g)
Step 6: Make the number of O₂ molecules the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has one O₂ molecule.
    ♦ The reduction half reaction has three O₂ molecules.
• So we multiply the oxidation half reaction by '3'. We get:
6CO (g) + 3O₂ (g) → 6CO₂ (g)
Step 7: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the oxidation half reaction do not have any charges. So we need not consider this step.
Step 8: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, the reduction half reaction do not have any charges. So we need not consider this step.
Step 9: Add the two reactions together.
• In our present case, we get:
2Fe₂O₃ (s) + 6CO (g) + 3O₂ (g) → 4Fe (s) + 3O₂ (g) + 6CO₂ (g)
• The three O₂ molecules on either sides, cancel each other. We get:
2Fe₂O₃ (s) + 6CO (g) → 4Fe (s) + 6CO₂ (g)
• We see that, all the coefficients have a common factor '2'. So we can divide all the coefficients by '2'. We get:
Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g)
Step 10: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Fe - 2 Nos, O - 6 Nos, C - 3 Nos. 
Product side:
   ♦ Fe - 2 Nos, O - 6 Nos, C - 3 Nos.
         ✰ So all atoms are balanced.
• In our present case, there are no ions. So we do not have to check charges.
◼  We see that, all atoms are balanced. So we can write the balanced equation as:
Fe₂O₃ (s) + 3CO (g) → 2Fe (s) + 3CO₂ (g)

Example 2:
Consider the reaction that we saw at the beginning of this chapter. (Fig.1 of section 8.1)
• The skeletal equation can be written as:
Zn (s) + CuSO₄ (aq)  → ZnSO₄ (aq) + Cu (s)
• Since CuSO₄ and ZnSO₄ are in aqueous solution, they will be dissociated into ions. So we can write the dissociated form:
Zn (s) + Cu²⁺ + SO₄^2- (aq)  → Zn²⁺ +SO₄^2- (aq) + Cu (s)
• We see that, SO₄^2- appears on both sides. It does not take part in the reaction. It is a spectator ion. So it can be cancelled.
(Remember that, polyatomic ions like SO₄^2-, NO₃^- etc., do not split into individual atoms when dissolved in water. Some common polyatomic ions can be seen here. It is useful to remember their formula and names)
• So the net reaction is:
Zn (s) + Cu²⁺  → Zn²⁺ + Cu (s)
• We want to balance this equation. It can be written in 5 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{0}{Zn}\,(s)+\overset{+2}{Cu^{2+}}\,(aq)
\rightarrow \overset{+2}{Zn^{2+}}\,(aq)+\overset{0}{Cu}\,(s)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Zn is oxidized and Cu is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Zn (s) → Zn²⁺ (aq)
• The reduction half reaction is:
Cu²⁺ (aq) → Cu (s)
Step 4: Balance the oxidation half reaction.
• In our present case, the number of Zn atoms are the same on both sides. So it is already balanced.
Step 5: Balance the reduction half reaction.
• In our present case, the number of Cu atoms are the same on both sides. So it is already balanced.
Step 6: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, Zn²⁺ creates an excess charge of 2+ on the product side. So we must add 2e^- on the product side. We get:
Zn (s) → Zn²⁺ (aq) + 2e^-
Step 7: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, Cu²⁺ creates an excess charge of 2+ on the reactant side. So we must add 2e^- on the reactant side. We get:
Cu²⁺ (aq) + 2e^- → Cu (s)
Step 8: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 2e^-.
• So they are already equal.
Step 9: Add the two half reactions together.
• In our present case, we get:
Zn (s) + Cu²⁺ (aq) + 2e^- → Zn²⁺ (aq) + 2e^- + Cu (s)
• The 2e^- on either sides, cancel each other. We get:
Zn (s) + Cu²⁺ (aq) → Zn²⁺ (aq) + Cu (s)
Step 10: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Zn - 1 No, Cu - 1 No. 
Product side:
   ♦ Zn - 1 No, Cu - 1 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2+ (from the Cu²⁺ ion)
Product side:
   ♦ 2+ (from the Zn²⁺ ion)
         ✰ So all charges are balanced.
◼  Now we can write the balanced equation as:
Zn (s) + Cu²⁺  → Zn²⁺ + Cu (s)

Example 3:
Consider the second reaction that we saw at the beginning of this chapter. (Fig.2 of section 8.1)
• The skeletal equation can be written as:
Cu (s) + AgNO₃ (aq)  → Cu(NO₃)₂ (aq) + Ag (s)
• Since AgNO₃ and Cu(NO₃)₂ are in aqueous solution, they will be dissociated into ions. So we can write the dissociated form:
Cu (s) + Ag⁺ + NO₃^- (aq)  → Cu²⁺ + NO₃^- (aq) + Ag (s)
• We see that, NO₃^- appears on both sides. It does not take part in the reaction. It is a spectator ion. So it can be cancelled.
• So the net reaction is:
Cu (s) + Ag⁺  → Cu²⁺ + Ag (s)
• We want to balance this equation. It can be written in 5 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{0}{Cu}\,(s)+\overset{+1}{Ag^{+}}\,(aq)
\rightarrow \overset{+2}{Cu^{2+}}\,(aq)+\overset{0}{Ag}\,(s)}}$
Step 2: Identify the atoms that are oxidized. Also identify the atoms that are reduced.
• For our present case, we see that:
Cu is oxidized and Ag is reduced.
Step 3: Write the oxidation half reaction and reduction half reaction.
In our present case,
• The oxidation half reaction is:
Cu (s) → Cu²⁺ (aq)
• The reduction half reaction is:
Ag⁺ (aq) → Ag (s)
Step 4: Balance the oxidation half reaction.
• In our present case, the number of Cu atoms are the same on both sides. So it is already balanced.
Step 5: Balance the reduction half reaction.
• In our present case, the number of Ag atoms are the same on both sides. So it is already balanced.
Step 6: Balance the charges in the oxidation half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, Cu²⁺ creates an excess charge of 2+ on the product side. So we must add 2e^- on the product side. We get:
Cu (s) → Cu²⁺ (aq) + 2e^-
Step 7: Balance the charges in the reduction half reaction, by adding appropriate number of electrons on the appropriate side.
• In our present case, Ag⁺ creates an excess charge of 1+ on the reactant side. So we must add e^- on the reactant side. We get:
Ag⁺ (aq) + e^- → Ag (s)
Step 8: Make the number of electrons the same, in both the half reactions.
• In our present case,
    ♦ The oxidation half reaction has 2e^-.
    ♦ The reduction half reaction has 1e^-.
• So we multiply the reduction half reaction by '2'. We get:
2Ag⁺ (aq) + 2e^- → 2Ag (s)
Step 9: Add the two half reactions together.
• In our present case, we get:
Cu (s) + 2Ag⁺ (aq) + 2e^- → Cu²⁺ (aq) + 2Ag (s) + 2e^-
• The 2e^- on either sides, cancel each other. We get:
Cu (s) + 2Ag⁺ (aq) → Cu²⁺ (aq) + 2Ag (s)
Step 10: Check the balancing of all atoms. Also check the balancing of charges.
• In our present case, we have the following table for atoms:
Reactant side:
   ♦ Cu - 1 No., Ag - 2 No. 
Product side:
   ♦ Cu - 1 No., Ag - 2 No. 
         ✰ So all atoms are balanced.
• We have the following table for charges:
Reactant side:
   ♦ 2+ (from two Ag⁺ ions)
Product side:
   ♦ 2+ (from one Cu²⁺ ion)
         ✰ So all charges are balanced.
◼  Now we can write the balanced equation as:
Cu (s) + 2Ag⁺  → Cu²⁺ + 2Ag (s)

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• We have seen the steps for balancing a reaction using half reaction method.
   ♦ The examples that we saw, take place in neutral medium.
• In the next section, we will see some examples, which take place in acidic medium.
   ♦ There will be a few more steps in addition to those that we saw above.

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