Balancing Redox Reactions: Half-Reaction (Ion-Electron) Method

The oxidation-number method is powerful, but there is a second equally popular approach that many chemists find even more intuitive. Instead of tracking oxidation numbers across the whole equation, what if you simply split the reaction in two: one piece that shows only the oxidation, and another that shows only the reduction? Balance each piece on its own, then stitch them back together. That is exactly what the half-reaction method (also called the ion-electron method) does.

This approach is especially handy for reactions in aqueous solution, where water molecules, $H^+$ ions, or $OH^-$ ions participate as reactants or products. By dealing with each half separately, the bookkeeping stays clean and manageable, even for reactions that look intimidating at first glance.

The Seven-Step Procedure

Here is the complete procedure, broken into clear steps. The first six steps build the balanced equation, and the seventh is a verification check.

Step 1: Write the Unbalanced Ionic Equation

Start by writing the reaction in its ionic form. Drop spectator ions (ions that appear unchanged on both sides) and keep only the species that actually take part in the redox process. At this point, the equation will not be balanced, and that is fine.

Step 2: Split into Two Half-Reactions

Identify which species is being oxidised and which is being reduced. Write each process as a separate half-reaction:

• Oxidation half: the species losing electrons (its oxidation number goes up)
• Reduction half: the species gaining electrons (its oxidation number goes down)

Step 3: Balance Atoms Other Than O and H

In each half-reaction separately, balance every element except oxygen and hydrogen. If the same element appears on both sides but in different numbers, adjust the coefficient to make them equal.

Why leave O and H for later? Because these two elements are handled by adding $H_2O$ and $H^+$ (or $OH^-$) in the next step, so touching them now would only create extra work.

Step 4: Balance O and H Using Water and Hydrogen Ions

Now handle oxygen and hydrogen in each half-reaction:

• For reactions in acidic medium:
  • Balance oxygen atoms by adding $H_2O$ molecules to whichever side is short on oxygen.
  • Balance hydrogen atoms by adding $H^+$ ions to whichever side is short on hydrogen.

This works because $H_2O$ and $H^+$ are both readily available in an acidic aqueous solution.

Step 5: Balance Charges by Adding Electrons

At this stage, atoms are balanced in each half-reaction but the charges may not be. Add electrons ($e^-$) to one side of each half-reaction so that the total charge on the left equals the total charge on the right.

• In an oxidation half-reaction, electrons appear on the product side (electrons are released).
• In a reduction half-reaction, electrons appear on the reactant side (electrons are consumed).

If the two half-reactions do not involve the same number of electrons, multiply one or both by suitable integers so that both halves have the same electron count. This is essential because every electron released by the oxidation half must be absorbed by the reduction half.

Step 6: Add the Two Half-Reactions

Combine the two balanced half-reactions by adding them together. The electrons on opposite sides cancel out completely. Simplify by cancelling any species (like $H_2O$) that appears on both sides. The result is the net ionic equation.

Step 7: Verify the Final Equation

Count every type of atom on the left and on the right. They must match. Then count the total ionic charge on each side. They must also match. If both checks pass, the equation is fully balanced.

Adapting for Basic Medium

Everything described above works perfectly for acidic solutions. For basic solutions, there is one extra adjustment:

1. First, balance the half-reaction exactly as you would in acidic medium (using $H_2O$ and $H^+$).
2. Then, for every $H^+$ ion in the equation, add an equal number of $OH^-$ ions to both sides.
3. Wherever $H^+$ and $OH^-$ appear on the same side, combine them into $H_2O$.
4. Cancel any $H_2O$ molecules that appear on both sides.

This procedure removes all free $H^+$ ions (which would not actually be present in a basic solution) and replaces them with $OH^-$ and $H_2O$, which are the species that genuinely exist in basic conditions.

Worked Example 1: Iron(II) and Dichromate in Acid

Problem: $Fe^{2+}$ ions are oxidised to $Fe^{3+}$ by dichromate ions ($Cr_2O_7^{2-}$) in acidic solution, with $Cr_2O_7^{2-}$ being reduced to $Cr^{3+}$. Write the balanced ionic equation.

Applying the Seven Steps

Step 1: Unbalanced ionic equation

$Fe^{2+}(aq) + Cr_2O_7^{2-}(aq) \rightarrow Fe^{3+}(aq) + Cr^{3+}(aq) \qquad \text{(7.50)}$

Step 2: Separate into half-reactions

• Oxidation half:

$\overset{+2}{Fe^{2+}}(aq) \rightarrow \overset{+3}{Fe^{3+}}(aq) \qquad \text{(7.51)}$

• Reduction half:

$\overset{+6}{Cr_2}\overset{-2}{O_7^{2-}}(aq) \rightarrow \overset{+3}{Cr^{3+}}(aq) \qquad \text{(7.52)}$

Iron is losing electrons (oxidation number rises from $+2$ to $+3$), and chromium is gaining electrons (oxidation number drops from $+6$ to $+3$).

Step 3: Balance atoms other than O and H

The oxidation half already has one iron atom on each side, so it is balanced for iron.

For the reduction half, there are two chromium atoms in $Cr_2O_7^{2-}$ but only one $Cr^{3+}$ on the right. Place a coefficient of $2$ in front of $Cr^{3+}$:

$Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) \qquad \text{(7.53)}$

Step 4: Balance O and H (acidic medium)

The left side has $7$ oxygen atoms (from $Cr_2O_7^{2-}$) and the right side has none. Add $7H_2O$ to the right to supply those $7$ oxygens:

$Cr_2O_7^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$

Now the right side has $7 \times 2 = 14$ hydrogen atoms and the left has none. Add $14H^+$ to the left:

$Cr_2O_7^{2-}(aq) + 14H^+(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l) \qquad \text{(7.54)}$

Atoms are now balanced in this half-reaction: $2$ Cr, $7$ O, $14$ H on each side.

Step 5: Balance charges with electrons

Check the charges on each side of the oxidation half:

$Fe^{2+}(aq) \rightarrow Fe^{3+}(aq)$

• Left: $+2$
• Right: $+3$

The right side is more positive by $1$, so add $1$ electron to the right to bring it down to $+2$:

$Fe^{2+}(aq) \rightarrow Fe^{3+}(aq) + e^- \qquad \text{(7.55)}$

Now check the charges on the reduction half:

$Cr_2O_7^{2-}(aq) + 14H^+(aq) \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$

• Left: $(-2) + 14(+1) = +12$
• Right: $2(+3) + 0 = +6$

The left side is more positive by $12 - 6 = 6$ units. Add $6$ electrons to the left to reduce its charge from $+12$ to $+6$:

$Cr_2O_7^{2-}(aq) + 14H^+(aq) + 6e^- \rightarrow 2Cr^{3+}(aq) + 7H_2O(l) \qquad \text{(7.56)}$

Equalise electrons: The oxidation half produces $1$ electron; the reduction half consumes $6$. Multiply the entire oxidation half by $6$:

$6Fe^{2+}(aq) \rightarrow 6Fe^{3+}(aq) + 6e^- \qquad \text{(7.57)}$

Both halves now involve $6$ electrons.

Step 6: Add the half-reactions and cancel electrons

$6Fe^{2+}(aq) \rightarrow 6Fe^{3+}(aq) + \cancel{6e^-}$

$Cr_2O_7^{2-}(aq) + 14H^+(aq) + \cancel{6e^-} \rightarrow 2Cr^{3+}(aq) + 7H_2O(l)$

Adding these together:

$6Fe^{2+}(aq) + Cr_2O_7^{2-}(aq) + 14H^+(aq) \rightarrow 6Fe^{3+}(aq) + 2Cr^{3+}(aq) + 7H_2O(l) \qquad \text{(7.58)}$

Step 7: Verification

Element/Charge	Left Side	Right Side
Fe	$6$	$6$
Cr	$2$	$2$
O	$7$	$7$
H	$14$	$14$
Total charge	$6(+2) + (-2) + 14(+1) = +24$	$6(+3) + 2(+3) + 0 = +24$

All atoms and charges match. The equation is fully balanced.

Worked Example 2: Permanganate and Iodide in Basic Solution (Problem 7.10)

Problem: Permanganate ion ($MnO_4^-$) in basic solution oxidises iodide ion ($I^-$) to molecular iodine ($I_2$), while permanganate itself is reduced to manganese(IV) oxide ($MnO_2$). Write a balanced ionic equation.

Applying the Seven Steps

Step 1: Unbalanced ionic equation

$MnO_4^-(aq) + I^-(aq) \rightarrow MnO_2(s) + I_2(s)$

Step 2: Separate into half-reactions

• Oxidation half:

$\overset{-1}{I^-}(aq) \rightarrow \overset{0}{I_2}(s)$

Iodine’s oxidation number rises from $-1$ to $0$, so this is oxidation.

• Reduction half:

$\overset{+7}{Mn}O_4^-(aq) \rightarrow \overset{+4}{Mn}O_2(s)$

Manganese’s oxidation number drops from $+7$ to $+4$, so this is reduction.

Step 3: Balance atoms other than O and H

In the oxidation half, $I_2$ contains two iodine atoms but there is only one $I^-$ on the left. Place a $2$ in front of $I^-$:

$2I^-(aq) \rightarrow I_2(s)$

The reduction half already has one manganese on each side, so it is balanced for Mn.

Step 4: Balance O and H (start with acidic-medium procedure, then adapt for basic)

Focus on the reduction half first. The left side has $4$ oxygen atoms (from $MnO_4^-$) and the right has $2$ (from $MnO_2$). The right side is $2$ short, so add $2H_2O$ there:

$MnO_4^-(aq) \rightarrow MnO_2(s) + 2H_2O(l)$

Now the right has $4$ hydrogen atoms and the left has none. Add $4H^+$ to the left:

$MnO_4^-(aq) + 4H^+(aq) \rightarrow MnO_2(s) + 2H_2O(l)$

Convert to basic medium: The reaction occurs in basic solution, so free $H^+$ ions should not appear. For the $4H^+$ on the left, add $4OH^-$ to both sides:

$MnO_4^-(aq) + 4H^+(aq) + 4OH^-(aq) \rightarrow MnO_2(s) + 2H_2O(l) + 4OH^-(aq)$

On the left side, $4H^+$ and $4OH^-$ combine to form $4H_2O$:

$MnO_4^-(aq) + 4H_2O(l) \rightarrow MnO_2(s) + 2H_2O(l) + 4OH^-(aq)$

Cancel $2H_2O$ that appears on both sides ($4H_2O$ on the left minus $2H_2O$ on the right leaves $2H_2O$ on the left):

$MnO_4^-(aq) + 2H_2O(l) \rightarrow MnO_2(s) + 4OH^-(aq)$

The oxidation half ($2I^- \rightarrow I_2$) has no oxygen or hydrogen to worry about, so it needs no adjustment.

Step 5: Balance charges with electrons

For the oxidation half:

$2I^-(aq) \rightarrow I_2(s)$

• Left: $2(-1) = -2$
• Right: $0$

Add $2$ electrons to the right to bring it from $0$ to $-2$:

$2I^-(aq) \rightarrow I_2(s) + 2e^-$

For the reduction half:

$MnO_4^-(aq) + 2H_2O(l) \rightarrow MnO_2(s) + 4OH^-(aq)$

• Left: $(-1) + 0 = -1$
• Right: $0 + 4(-1) = -4$

The right side is more negative by $4 - 1 = 3$ units. Add $3$ electrons to the left to bring it from $-1$ to $-4$:

$MnO_4^-(aq) + 2H_2O(l) + 3e^- \rightarrow MnO_2(s) + 4OH^-(aq)$

Equalise electrons: The oxidation half produces $2$ electrons and the reduction half consumes $3$. The least common multiple of $2$ and $3$ is $6$. Multiply the oxidation half by $3$ and the reduction half by $2$:

$6I^-(aq) \rightarrow 3I_2(s) + 6e^-$

$2MnO_4^-(aq) + 4H_2O(l) + 6e^- \rightarrow 2MnO_2(s) + 8OH^-(aq)$

Step 6: Add the half-reactions and cancel electrons

$6I^-(aq) \rightarrow 3I_2(s) + \cancel{6e^-}$

$2MnO_4^-(aq) + 4H_2O(l) + \cancel{6e^-} \rightarrow 2MnO_2(s) + 8OH^-(aq)$

Combined:

$6I^-(aq) + 2MnO_4^-(aq) + 4H_2O(l) \rightarrow 3I_2(s) + 2MnO_2(s) + 8OH^-(aq)$

Step 7: Verification

Element/Charge	Left Side	Right Side
I	$6$	$3 \times 2 = 6$
Mn	$2$	$2$
O	$2 \times 4 + 4 \times 1 = 12$	$2 \times 2 + 8 \times 1 = 12$
H	$4 \times 2 = 8$	$8 \times 1 = 8$
Total charge	$6(-1) + 2(-1) = -8$	$0 + 0 + 8(-1) = -8$

Every atom count and the total charge match on both sides. The equation is fully balanced.

Comparing the Two Methods at a Glance

Feature	Oxidation-Number Method	Half-Reaction Method
Starting point	Track oxidation-number changes across the whole equation	Split into separate oxidation and reduction halves
How electrons are handled	Implied through oxidation-number changes	Explicitly written as $e^-$ in each half-reaction
Balancing O and H	Done after equalising oxidation numbers	Done within each half-reaction before adding electrons
Basic medium adaptation	Add $OH^-$ to balance charges directly	Balance as acid first, then neutralise $H^+$ with $OH^-$
Best suited for	Molecular equations and quick balancing	Ionic equations in solution, especially electrochemistry

Both methods always give the same final balanced equation. Choose whichever one feels more natural for the problem at hand.