To balance the given oxidation-reduction reaction in acid solution, we need to follow a few steps:
Step 1: Identify the atoms that are oxidized and reduced in the reaction.
In this case, we can observe that sulfur in SO32- is being oxidized from +4 to +6, and manganese in MnO4- is being reduced from +7 to +2.
Step 2: Write the half-reactions for both oxidation and reduction processes.
Oxidation half-reaction: SO32- --> SO42-
Reduction half-reaction: MnO4- --> Mn2+
Step 3: Balance the atoms in each half-reaction, excluding hydrogen and oxygen.
The oxidation half-reaction is already balanced in terms of sulfur atoms since there is only one sulfur atom on both sides.
For the reduction half-reaction, we have:
MnO4- --> Mn2+
There is one Mn atom on both sides, so no further balancing is required.
Step 4: Balance the number of oxygen atoms by adding water molecules (H2O).
In the oxidation half-reaction, we need to balance the oxygen atoms by adding water molecules. Since we have 3 oxygen atoms on the reactant side and 4 on the product side, we need to add one water molecule to the reactant side:
SO32- + H2O --> SO42-
Step 5: Balance the number of hydrogen atoms by adding hydrogen ions (H+).
In the oxidation half-reaction, there are 3 hydrogen atoms on the reactant side and 8 on the product side. To balance the hydrogen atoms, we need to add 5 hydrogen ions (H+) to the reactant side:
SO32- + H2O + 5H+ --> SO42-
Step 6: Balance the charges by adding electrons (e-) to the half-reactions.
In the oxidation half-reaction, the overall charge is -2 on both sides. So, no electrons need to be added.
In the reduction half-reaction, the MnO4- ion has a charge of -1, and the Mn2+ ion has a charge of +2. To balance the charges, we need to add 5 electrons (e-) to the reactant side:
MnO4- + 5e- --> Mn2+
Step 7: Balance the electrons between the two half-reactions.
Since the reduction half-reaction requires 5 electrons and the oxidation half-reaction does not require any electrons, we need to multiply the oxidation half-reaction by 5 to balance the electrons:
5(SO32- + H2O + 5H+ --> SO42-) + MnO4- + 5e- --> Mn2+
Step 8: Combine the balanced half-reactions and simplify, if necessary.
Finally, we can add the two balanced half-reactions together:
5SO32- + 8H+ + MnO4- --> 5SO42- + Mn2+ + 4H2O
Therefore, the balanced oxidation-reduction reaction in acid solution is:
5SO32- + 8H+ + MnO4- --> 5SO42- + Mn2+ + 4H2O