Question AD.1: Balancing a Redox Equation Balance the following equation us...
Balancing a Redox Equation
Balance the following equation using H^{+}, but not OH^{−}:
\underset{+ 2}{Fe^{2+}} + \underset{\begin{matrix} + 7 \\ Permanganate \end{matrix} }{MnO_{4}^{−}} \rightleftharpoons \underset{+ 3}{Fe^{3+}} + \underset{+ 2}{Mn^{2+}}
- Assign oxidation numbers. They are assigned for Fe and Mn in each
species in the above reaction. - Break the reaction into two half-reactions.
Oxidation half-reaction: \underset{+ 2}{Fe^{2+}} \rightleftharpoons \underset{+ 3}{Fe^{3+}}
Reduction half-reaction: \underset{+ 7}{MnO_{4}^{−}} \rightleftharpoons \underset{+ 2}{Mn^{2+}}
3. Balance the atoms that are oxidized or reduced. Because there is
only one Fe or Mn in each species on each side of the equation, the
atoms of Fe and Mn are already balanced.
4. Balance electrons. Electrons are added to account for the change in each oxidation state.
Fe^{2+} \rightleftharpoons Fe^{3+} + e^{−}
MnO_{4}^{−} + 5e^{−} \rightleftharpoons Mn^{2+}
In the second case, we need 5e^{−} on the left side to take Mn from +7 to +2.
5. Balance oxygen atoms. There are no oxygen atoms in the Fe
half-reaction. There are four oxygen atoms on the left side of the
Mn reaction, so we add four molecules of H_{2}O to the right side:
MnO_{4}^{−} + 5e^{−} \rightleftharpoons Mn^{2+} + 4H_{2}O
6. Balance hydrogen atoms. The Fe equation is already balanced. The Mn equation needs 8H^{+} on the left.
MnO_{4}^{−} + 5e^{−} + 8H^{+} → Mn^{2+} + 4H_{2}O
At this point, each half-reaction must be completely balanced (the
same number of atoms and charge on each side) or you have made
a mistake.
7. Multiply and add the reactions. We multiply the Fe equation by 5 and the Mn equation by 1 and add:
\begin{matrix} \quad \quad \quad \quad \quad \quad 5Fe^{2+} \rightleftharpoons 5Fe^{3+} + \cancel{5e}^{−} \\ MnO_{4}^{−} + \cancel{5e}^{−} + 8H^{+} \rightleftharpoons Mn^{2+} + 4H_{2}O \\ \hline \quad \quad \quad 5Fe^{2+} + MnO_{4}^{−} + 8H^{+} \rightleftharpoons 5Fe^{3+} + Mn^{2+} + 4H_{2}O \end{matrix}
The total charge on each side is +17, and we find the same number of atoms of each element on each side. The equation is balanced.