Balance the following oxidation-reduction reaction in acid solution:
SO32- + MnO4- --> SO42- + Mn2+ + H2O
S changes from +4 in SO3^-2 to +6 in SO4^-2, a loss of 2 electrons.
Mn changes from +7 in MnO4^- to +2 in Mn^+2, a gain of 5 electrons.
Multiply the Mn half reaction x 2.
Multiply the SO3^-2 half reaction x 5.
Count up oxygen atoms and balance H2O, then add H^+ to the left side to balance H atoms. Check that the charge balances.
What is the pH ( 3 significant digits) of a 0.01 M solution of sulfuric acid? Hint is that the first ionization constant is "large", and the second is finite, and the [H+] donated by the second equilibrium depends upon the amount of [H+] already donated by the first ionization. What is the answer please.
To balance the given oxidation-reduction reaction in acid solution, follow these steps:
Step 1: Break the reaction into half-reactions.
The oxidation half-reaction involves the reaction of SO32- to form SO42-:
SO32- → SO42-
The reduction half-reaction involves the reaction of MnO4- to form Mn2+:
MnO4- → Mn2+
Step 2: Balance the atoms other than hydrogen and oxygen in each half-reaction.
For the oxidation half-reaction, both sides already have the same number of sulfur atoms.
For the reduction half-reaction, both sides already have the same number of manganese atoms.
Step 3: Balance the oxygen atoms.
In the oxidation half-reaction, there are 3 oxygen atoms on the reactant side (SO32-) and 4 oxygen atoms on the product side (SO42-). To balance the number of oxygen atoms, add water (H2O) to the reactant side:
SO32- + H2O → SO42-
Step 4: Balance the hydrogen atoms.
In the oxidation half-reaction, there are no hydrogen atoms on either side.
In the reduction half-reaction, there are no hydrogen atoms on either side.
Step 5: Balance the charge in each half-reaction.
In the oxidation half-reaction, the charge is balanced.
In the reduction half-reaction, the charge is not balanced. To balance the charge, add hydrogen ions (H+) to the product side (acidic condition):
MnO4- + 8H+ → Mn2+
Step 6: Multiply the half-reactions by integers to equalize the total number of electrons transferred.
To balance the electrons, we multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:
5(SO32- + H2O → SO42-)
+
2(MnO4- + 8H+ → Mn2+)
This gives us:
5SO32- + 5H2O → 5SO42-
+
2MnO4- + 16H+ → 2Mn2+
Step 7: Combine the half-reactions.
Add the balanced half-reactions together:
5SO32- + 5H2O + 2MnO4- + 16H+ → 5SO42- + 2Mn2+
Step 8: Check the balanced equation for mass and charge balance.
The equation is now balanced in terms of mass and charge.
The balanced oxidation-reduction reaction in acid solution is:
5SO32- + 5H2O + 2MnO4- + 16H+ → 5SO42- + 2Mn2+
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