Balance the following oxidation-reduction reactions, which occur in acidic solution, using the half-reaction method. (Use the lowest possible coefficients. Include states-of-matter under the given conditions in your answer.)
Mn2+(aq) + NaBiO3(s) → Na+(aq) + Bi3+(aq) + MnO4‾(aq)
See your posts above.
To balance the given oxidation-reduction reaction in acidic solution using the half-reaction method, follow these steps:
Step 1: Write the unbalanced half-reactions
First, identify the oxidation and reduction half-reactions in the given reaction.
Oxidation half-reaction: Mn2+(aq) → MnO4‾(aq)
Reduction half-reaction: NaBiO3(s) → Na+(aq) + Bi3+(aq)
Step 2: Balance the atoms in each half-reaction
Balance the atoms in each half-reaction, excluding hydrogen and oxygen.
Oxidation half-reaction: Mn2+(aq) → MnO4‾(aq)
Reduction half-reaction: NaBiO3(s) → Na+(aq) + Bi3+(aq)
The atoms appear balanced in both half-reactions.
Step 3: Balance the oxygen atoms by adding water (H2O)
In the reduction half-reaction, there are no oxygen atoms. In the oxidation half-reaction, there are four oxygen atoms in MnO4‾(aq).
Since you are balancing the reaction under acidic conditions, you can add water (H2O) to balance the oxygen atoms on the side with fewer oxygen atoms.
Oxidation half-reaction: Mn2+(aq) + 4H2O(l) → MnO4‾(aq)
Step 4: Balance the hydrogen atoms by adding H+
Next, balance the hydrogen atoms by adding H+ ions to the side with fewer hydrogen atoms. Since you are balancing the reaction in acidic solution, you can add H+ ions.
Oxidation half-reaction: Mn2+(aq) + 4H2O(l) → MnO4‾(aq) + 8H+(aq)
Step 5: Balance the charges by adding electrons (e-)
Now, balance the charges by adding electrons (e-) to the side with the higher charge. In this case, the reduction half-reaction has a higher overall charge, so we add electrons to balance the charge.
Oxidation half-reaction: Mn2+(aq) + 4H2O(l) → MnO4‾(aq) + 8H+(aq) + 5e-
Reduction half-reaction: NaBiO3(s) + 3H2O(l) → Na+(aq) + Bi3+(aq)
Note: The reduction half-reaction can be balanced without the need for electrons since the overall charge is already balanced.
Step 6: Multiply the half-reactions to equalize the number of electrons
To balance the number of electrons in both half-reactions, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 1.
5(Mn2+(aq) + 4H2O(l) → MnO4‾(aq) + 8H+(aq) + 5e-)
NaBiO3(s) + 3H2O(l) → Na+(aq) + Bi3+(aq)
Now, both half-reactions have the same number of electrons.
Step 7: Combine the half-reactions and cancel out common species
Combine the balanced half-reactions, canceling out common species on both sides of the equation.
5(Mn2+(aq) + 4H2O(l) → MnO4‾(aq) + 8H+(aq) + 5e-)
+ NaBiO3(s) + 3H2O(l) → Na+(aq) + Bi3+(aq)
--------------------------------------------------
5Mn2+(aq) + 8H2O(l) + NaBiO3(s) → MnO4‾(aq) + Na+(aq) + Bi3+(aq)
Step 8: Verify the balance and add state-of-matter
Finally, check that all atoms are balanced and add the state-of-matter for each component.
5Mn2+(aq) + 8H2O(l) + NaBiO3(s) → MnO4‾(aq) + Na+(aq) + Bi3+(aq)
The balanced equation for the given oxidation-reduction reaction in acidic solution is:
5Mn2+(aq) + 8H2O(l) + NaBiO3(s) → MnO4‾(aq) + Na+(aq) + Bi3+(aq)
Note: The states-of-matter are indicated by (aq) for aqueous (dissolved in water) and (s) for solid.