To balance the redox reaction:
MnO4^- + NO2 -> Mn^2+ + NO3^-
We need to assign oxidation numbers to each element in the reaction. Let's start by assigning oxidation numbers:
MnO4^-: Mn has an oxidation number of +7, and each oxygen atom (O) has an oxidation number of -2. Since there are four oxygen atoms, the total oxidation number contributed by oxygen is -8. So, to balance the charge, the oxidation number of Mn must be +7.
NO2: The oxidation number of N is +4, and each oxygen atom (O) has an oxidation number of -2. Therefore, the total oxidation number contributed by oxygen is -4. So, to balance the charge, the oxidation number of N must be +4.
Mn^2+: Mn has an oxidation number of +2.
NO3^-: The oxidation number of N is +5, and each oxygen atom (O) has an oxidation number of -2. Since there are three oxygen atoms, the total oxidation number contributed by oxygen is -6. So, to balance the charge, the oxidation number of N must be +5.
Now, we can observe the changes in oxidation numbers and identify the species being oxidized and reduced:
The Mn atom changes from +7 to +2, which means it is reduced (gaining electrons). The N atom changes from +4 to +5, which means it is oxidized (losing electrons).
To balance the equation, we need the same number of electrons lost in oxidation to be gained in reduction. In this case, we need 5 electrons:
MnO4^- + 5NO2 -> Mn^2+ + 5NO3^-
Now, let's balance the oxygen atoms by adding water (H2O):
MnO4^- + 5NO2 -> Mn^2+ + 5NO3^- + 2H2O
Next, let's balance the hydrogen atoms by adding hydrogen ions (H+):
MnO4^- + 5NO2 + 8H+ -> Mn^2+ + 5NO3^- + 2H2O
Finally, let's balance the charges by adding electrons to the appropriate side. Since there are 8H+ ions on the left side, we need 8 electrons on the right side:
MnO4^- + 5NO2 + 8H+ + 8e^- -> Mn^2+ + 5NO3^- + 2H2O
The balanced redox equation for the reaction is:
MnO4^- + 5NO2 + 8H+ + 8e^- -> Mn^2+ + 5NO3^- + 2H2O