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 6
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 8 steps:
Step 1: Write the oxidation numbers of all elements.
• For our present case, we get:
$\mathbf\small{\rm{\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow \overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
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: 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 +4
⇒ Each Mn atom gains (+7 - +4) = 3 electrons.
(ii) There is only one Mn atom on the reactant side.
⇒ There is a total gain of (1 No. × 3 electrons) = 3 electrons.
(iii) Oxidation number of Br increases from -1 to +5
⇒ Each Br atom loses (+5 - -1) = 6 electrons.
(iv) There is only one Br atom on the reactant side.
⇒ There is a total loss of (1 No. × 6 electrons) = 6 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 3 and loss of 6 do not tally. We can make them tally as follows:
If
there are two Mn atoms on the reactant side, each one gaining three
electrons, will result in a total gain of 6. Then they tally.
♦ So we must put '2' in front of MnO4- on the reactant side. We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow
\overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
• 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 Mn atoms, we must put '2' in front of MnO4 on the product side also. We get:
$\mathbf\small{\rm{2\,\overset{+7}{Mn}\,\overset{-2}{O^{-}_4}\,(g)+\overset{-1}{Br^{-}}\,(aq)\rightarrow
2\,\overset{+4}{Mn}\,\overset{-2}{O_2}\,(s)+\overset{+5}{Br}\,\overset{-2}{O_3^{-}}\,(aq)}}$
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:
♦ (2 No. × 1-) from the MnO4- ion = -2
♦ (1 No. × 1- ) from the Br- ion = -1
♦ Total = -3
Product side:
♦ (1 No. × 1-) from the BrO3- ion = -1
♦ Total = -1
✰ So there is an excess of -2 charge on the reactant side.
• In the present case, since it is a basic medium, we must balance the charges using OH- ions.
♦ We must add 2 Nos. of OH- ions on the product side. We get:
2MnO4- (aq) + Br- (aq) → 2MnO2 (s) + BrO3- (aq) + 2OH- (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 2 No. H atoms on the product side. So we must add 1 No. H2O molecules on the reactant side. We get:
2MnO4- (aq) + Br- (aq) + H2O (l) → 2MnO2 (s) + BrO3- (aq) + 2OH- (aq)
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 + 1 = 9
Product side:
♦ 4 + 3 + 2 = 9
✰ So the O atoms are balanced
◼ The balanced equation is:
2MnO4- (aq) + Br- (aq) + H2O (l) → 2MnO2 (s) + BrO3- (aq) + 2OH- (aq)
Example 7
Balance the following equation in basic medium:
CrO2- (aq) + ClO- (aq) → CrO42- (aq) + Cl- (aq)
Solution:
The given skeletal equation is:
CrO2- (aq) + ClO- (aq) → CrO42- (aq) + Cl- (aq)
• 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:
$\mathbf\small{\rm{
\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow \overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)
+\overset{-1}{Cl^{-}}(aq)}}$
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: Find the number of electrons which are transferred
• In our present case, we can find it in 4 steps:
(i) Oxidation number of Cl decreases from +1 to -1
⇒ Each Cl atom gains (+1 - -1) = 2 electrons.
(ii) There is only one Cl atom on the reactant side.
⇒ There is a total gain of (1 No. × 2 electrons) = 2 electrons.
(iii) Oxidation number of Cr increases from +3 to +6
⇒ Each Cr atom loses (+6 - +3) = 3 electrons.
(iv) There is only one Cr atom on the reactant side.
⇒ There is a total loss of (1 No. × 3 electrons) = 3 electrons.
Step 4: Number of electrons gained must be equal to the number of electrons lost.
• In our present case, gain of 2 and loss of 3 do not tally. We can make them tally as follows:
If
there are two Cr atoms on the reactant side, each one losing 3
electrons, will result in a total loss of 6.
If
there are three Cl atoms on the reactant side, each one gaining 2
electrons, will result in a total gain of 6.
Then they tally.
♦ So we must put '2' in front of CrO2- on the reactant side.
♦ Also, we must put '3' in front of ClO- on the reactant side.We get:
$\mathbf\small{\rm{2\,\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+3\,\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow \overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)+\overset{-1}{Cl^{-}}(aq)}}$
• 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 Cr atoms, we must put '2' in front of CrO42- on the product side.
Also, to balance the number of Cl atoms, we must put '3' in front of Cl- on the product side. We get:
$\mathbf\small{\rm{2\,\overset{+3}{Cr}\,\overset{-2}{O_2^{-}}\,(aq)+3\,\overset{+1}{Cl}\,\overset{-2}{O^{-}}(aq)\rightarrow 2\,\overset{+6}{Cr}\,\overset{-2}{O_4^{2-}}\,(aq)+3\,\overset{-1}{Cl^{-}}(aq)}}$
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:
♦ (2 No. × 1-) from the CrO2- ion = -2
♦ (3 No. × 1- ) from the ClO- ion = -3
♦ Total = -5
Product side:
♦ (2 No. × 2-) from the CrO42- ion = -4
♦ (3 No. × 1-) from the Cl- ion = -3
♦ Total = -7
✰ So there is an excess of -2 charge on the product side.
• In the present case, since it is a basic medium, we must balance the charges using OH- ions.
♦ We must add 2 Nos. of OH- ions on the reactant side. We get:
2CrO2- (aq) + 3ClO- (aq) + 2OH- → 2CrO42- (aq) + 3Cl- (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 2 No. H atoms on the reactant side. So we must add 1 No. H2O molecules on the product side. We get:
2CrO2- (aq) + 3ClO- (aq) + 2OH- → 2CrO42- (aq) + 3Cl- (aq) + H2O
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:
♦ 4 + 3+ 2 = 9
Product side:
♦ 8 + 1 = 9
✰ So the O atoms are balanced
◼ The balanced equation is:
2CrO2- (aq) + 3ClO- (aq) + 2OH- → 2CrO42- (aq) + 3Cl- (aq) + H2O
Now we will see some solved examples. The link is given below:
Solved example 8.9 Parts (a), (b), (c) and (d)
• We
have completed the discussion on oxidation number method. In the next section, we will see the half reaction method.
Previous
Contents
Next
Copyright©2021 Higher secondary chemistry.blogspot.com
No comments:
Post a Comment