In the previous section, we saw five steps 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 Cr2O72-
• In the dissolved state, the two ions separate away from each other.
• Cr2O72- 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 SO32-
• In the dissolved state, the two ions separate away from each other.
• SO32- 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: Cr2O72- and SO32-
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 SO42- ion.
6. Steps (1) to (4) give us the reactants. Step (5) gives us the products.
• So now we can write the skeletal equation:
Cr2O72- + SO32- → Cr3+ + SO42-
• We want to balance this equation. It can be written in 8 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: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Cr decreases from +6 to +3
⇒ Each Cr atom gains (+6 - +3) = 3 electrons.
(ii) There are two Cr atoms on the reactant side.
⇒ There is a total gain of (2 No. × 3 electrons) = 6 electrons.
(iii) Oxidation number of S increases from +4 to +6
⇒ Each S atom loses (+6 - +4) = 2 electrons.
(iv) There is only one S atom on the reactant side.
⇒ There is a total loss of (1 No. × 2 electrons) = 2 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 6 and loss of 2 do not tally. We can make them tally as follows:
If there are three S atoms on the reactant side, each one losing two electrons, will result in a total loss of 6. Then they tally.
♦ So we must put '3' in front of SO32- on the reactant side. We get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+3\,\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(g)\rightarrow
\overset{+3}{Cr^{3+}}\,(aq)+\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
• Now the loss and gain of electrons tally.
Step 5: Balance all atoms other than O and H
• In our present case, to balance the number of S atoms, we must put '3' in front of SO42- on the product side also.
• Also, to balance the number of Cr atoms, we must put 2 in front of Cr3+ in the product side. We get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+3\,\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)\rightarrow
2\,\overset{+3}{Cr^{3+}}\,(aq)+3\,\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
Step 6: Check the balancing of charges.
♦ Balance them using H+ if it is acidic medium.
♦ Balance them using OH- ions if it is in basic medium.
• In our present case, we have the following list for charges:
Reactant side:
♦ (1 No. × 2-) from the Cr2O72- ion = -2
♦ (3 No. × 2- ) from the SO32- ion = -6
♦ Total = -8
Product side:
♦ (2 No. × 3+) from the Cr3+ ion = +6
♦ (3 No. × 2- ) from the SO42- ion = -6
♦ Total = 0
✰ So there is an excess of -8 charge on the reactant side.
• In the present case, since it is an acidic medium, we must balance the charges using H+ ions.
♦ We must add 8 Nos. of H+ ions on the reactant side. We get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+3\,\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)+8\,H^{+}\,(aq)+\rightarrow
2\,\overset{+3}{Cr^{3+}}\,(aq)+3\,\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)}}$
Step 7: Balance the number of H atoms by adding appropriate number of H2O molecules.
• In our present case, there is an excess of 8 No. H atoms on the reactant side. So we must add 4 No. H2O molecules on the product side. We get:
$\mathbf\small{\rm{\overset{+6}{Cr}_2\,\overset{-2}{O_7^{2-}}\,(aq)
+3\,\overset{+4}{S}\,\overset{-2}{O_3^{2-}}\,(aq)+8\,H^{+}\,(aq)+\rightarrow
2\,\overset{+3}{Cr^{3+}}\,(aq)+3\,\overset{+6}{S}\,\overset{-2}{O_4^{2-}}\,(aq)+4\,H_2O}}$
Step 8: Due to the addition of H2O molecules, new O atoms are introduced. So we must check the balancing of O atoms.
• In our present case, we have the following list of O atoms:
Reactant side:
♦ 7 + 9 = 16
Product side:
♦ 12 + 4 = 16
✰ So the O atoms are balanced
◼ The balanced equation is:
Cr2O72- (aq) + 3SO32- (aq) + 8 H+ (aq) → 2Cr3+ (aq) + 3SO42- (aq) + 4H2O (l)
Example 5:
Balance the equation in acidic medium:
MnO4- + Cl- → Mn2+ + Cl2
Solution:
We want to balance this equation. It can be written in 8 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: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Mn decreases from +7 to +2
⇒ Each Mn atom gains (+7 - +2) = 5 electrons.
(ii) There is only one Mn atom on the reactant side.
⇒ There is a total gain of (1 No. × 5 electrons) = 5 electrons.
(iii) Oxidation number of Cl increases from -1 to 0
⇒ Each Cl atom loses (0 - -1) = 1 electron.
(iv) There is only one Cl atom on the reactant side.
⇒ There is a total loss of (1 No. × 1 electron) = 1 electron.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 5 and loss of 1 do not tally. We can make them tally using 2 steps:
(i)
If there are 5 No. Cl- ions on the reactant side, each one losing one
electron, will result in a total loss of 5. Now they tally.
♦ So we must put '5' in front of Cl- on the reactant side. We get:
$\mathbf\small{\rm{
\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+5\,\overset{-1}{Cl^{-}}\,(aq)
\rightarrow
\overset{+2}{Mn^{2+}}\,(aq)
+\overset{0}{Cl_2}\,(g)}}$
• Now the loss and gain of electrons tally.
Step 5: Balance all atoms other than O and H
• In our present case, Mn is already balanced.
• Cl atoms are not balanced. 5 Cl atoms on the reactant side, cannot give 2 Cl atoms on the product side. So we use the least common multiple of 5 and 2, which is 10.
• 10 Cl atoms on the reactant side will give 5 Cl2 molecules on the product side. We get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+10\,\overset{-1}{Cl^{-}}\,(aq)\rightarrow\overset{+2}{Mn^{2+}}\,(aq)
+5\,\overset{0}{Cl_2}\,(g)}}$
• But now, the tally of electrons that we obtained in step 4 is lost:
♦ One Mn atom gains 5 electrons.
♦ Ten Cl- ions lose 10 electrons.
• So we must put '2' in front of MnO4- ion on the reactant side.
• To compensate, we must put '2' in front of Mn2+ ion on the product side also.
We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+10\,\overset{-1}{Cl^{-}}\,(aq)\rightarrow2\,\overset{+2}{Mn^{2+}}\,(aq)
+5\,\overset{0}{Cl_2}\,(g)}}$
Step 6: Check the balancing of charges.
♦ Balance them using H+ if it is acidic medium.
♦ Balance them using OH- ions if it is in basic medium.
• In our present case, we have the following list for charges:
Reactant side:
♦ (2 No. × 1-) from the MnO4- ion = -2
♦ (10 No. × 1- ) from the Cl- ion = -10
♦ Total = -12
Product side:
♦ (2 No. × 2+) from the Mn2+ ion = +4
♦ Total = +4
✰ So there is an excess of -8 charge on the reactant side.
• In the present case, since it is an acidic medium, we must balance the charges using H+ ions.
♦ We must add 8 Nos. of H+ ions on the reactant side. We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+10\,\overset{-1}{Cl^{-}}\,(aq)+8\,H^{+}\rightarrow2\,\overset{+2}{Mn^{2+}}\,(aq)+5\,\overset{0}{Cl_2}\,(g)}}$
Step 7: Balance the number of H atoms by adding appropriate number of H2O molecules.
• In
our present case, there is an excess of 8 No. H atoms on the reactant
side. So we must add 4 No. H2O molecules on the product side. We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O_4^{-}}\,(aq)
+10\,\overset{-1}{Cl^{-}}\,(aq)+8\,H^{+}\rightarrow2\,\overset{+2}{Mn^{2+}}\,(aq)+5\,\overset{0}{Cl_2}\,(g)+4H_2O\,(l)}}$
Step 8: Due to the addition of H2O molecules, new O atoms are introduced. So we must check the balancing of O atoms.
• In our present case, we have the following list of O atoms:
Reactant side:
♦ 8
Product side:
♦ 4
✰ So there is an excess of 4 Nos. O atoms on the reactant side.
• This can be balanced by providing a total of 8 Nos. H2O molecules on the product side.
• To compensate, the H atoms, there must be a total of 16 H+ ions on the reactant side.
• The balanced equation is:
2MnO4- + 10Cl- + 16H+ (aq) → 2Mn2+ + 5Cl2 + 8H2O (l)
The following link gives three more examples:
Solved example 8.8 - Part (a), Part (b) and Part (c)
• We
have seen the additional steps required in acidic medium.
• 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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