Problem. Balance the redox reaction in acidic solution:
\(\mathrm{MnO_4^- + Fe^{2+} \rightarrow Mn^{2+} + Fe^{3+}}\)
Then identify the oxidizing agent and reducing agent.
Core idea: what a redox reaction means
A redox reaction (oxidation–reduction reaction) is an electron-transfer process. Oxidation is loss of electrons (oxidation number increases), and reduction is gain of electrons (oxidation number decreases).
Step 1: Assign oxidation numbers and spot what changes
Determine oxidation numbers for the elements that change:
| Species | Key oxidation number(s) | Change | Redox role |
|---|---|---|---|
| \(\mathrm{Fe^{2+} \rightarrow Fe^{3+}}\) | Fe: \(+2 \rightarrow +3\) | Increase by \(+1\) | Oxidation (electron loss) |
| \(\mathrm{MnO_4^- \rightarrow Mn^{2+}}\) | Mn: \(+7 \rightarrow +2\) | Decrease by \(5\) | Reduction (electron gain) |
Therefore, \(\mathrm{Fe^{2+}}\) is the species being oxidized (so it acts as the reducing agent), and \(\mathrm{MnO_4^-}\) is the species being reduced (so it acts as the oxidizing agent).
Step 2: Write and balance the oxidation half-reaction
Oxidation of iron(II) to iron(III) involves one electron:
\(\mathrm{Fe^{2+} \rightarrow Fe^{3+} + e^-}\)
Step 3: Write and balance the reduction half-reaction in acidic solution
Start with the skeletal reduction:
\(\mathrm{MnO_4^- \rightarrow Mn^{2+}}\)
- Balance Mn atoms: already balanced (1 Mn on each side).
-
Balance O by adding water. There are 4 O on the left, so add \(4\) waters to the right:
\(\mathrm{MnO_4^- \rightarrow Mn^{2+} + 4H_2O}\)
-
Balance H by adding \(\mathrm{H^+}\). There are \(8\) H on the right, so add \(8\ \mathrm{H^+}\) to the
left:
\(\mathrm{MnO_4^- + 8H^+ \rightarrow Mn^{2+} + 4H_2O}\)
-
Balance charge by adding electrons. Left-side charge is \((-1) + 8(+1) = +7\). Right-side charge is \(+2\).
Add \(5\ \mathrm{e^-}\) to the left to reduce the left charge by 5, giving \(+2\) on both sides:
\(\mathrm{MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O}\)
Step 4: Equalize electrons and add half-reactions
The oxidation half-reaction produces \(1\ \mathrm{e^-}\), while the reduction half-reaction consumes \(5\ \mathrm{e^-}\). Multiply the oxidation half-reaction by 5:
\(\mathrm{5Fe^{2+} \rightarrow 5Fe^{3+} + 5e^-}\)
Add the two half-reactions and cancel electrons:
\(\mathrm{MnO_4^- + 8H^+ + 5e^- + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+} + 5e^-}\)
\(\mathrm{MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}}\)
Balanced redox reaction (acidic solution).
\(\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}\)
Oxidizing agent: \(\mathrm{MnO_4^-}\) (it is reduced).
Reducing agent: \(\mathrm{Fe^{2+}}\) (it is oxidized).
Step 5: Quick verification
- Atoms: Mn \(1=1\), Fe \(5=5\), O \(4=4\), H \(8=8\).
- Charge: left \((-1) + 5(+2) + 8(+1) = -1 + 10 + 8 = +17\); right \( (+2) + 5(+3) + 0 = 2 + 15 = +17\).
Visualization: half-reactions and electron cancellation
Summary
The balanced redox reaction in acidic solution is \(\mathrm{MnO_4^- + 5Fe^{2+} + 8H^+ \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O}\). Manganese in permanganate is reduced (\(+7 \rightarrow +2\)) so \(\mathrm{MnO_4^-}\) is the oxidizing agent, while \(\mathrm{Fe^{2+}}\) is oxidized (\(+2 \rightarrow +3\)) so it is the reducing agent.