Balancing Redox Reactions: Oxidation Number Method

Recognising which species gets oxidised and which gets reduced is only half the job. The real challenge arrives when you try to write a balanced equation for the reaction. Simple reactions can be balanced by trial and error, but most redox processes in solution involve ions, water molecules, and sometimes $H^+$ or $OH^-$ ions as participants. Trying to balance all of these by inspection quickly becomes impractical.

Chemists have developed two systematic approaches to handle this:

• The oxidation-number method, which tracks the increase and decrease in oxidation numbers and uses them to fix coefficients.
• The half-reaction (ion-electron) method, which splits the overall reaction into a separate oxidation half and reduction half, balances each one independently, and then combines them.

Both methods are widely used, and either one will give you the same final balanced equation. This topic walks through the oxidation-number method step by step, with fully worked examples in both acidic and basic solutions.

The Five-Step Procedure

The oxidation-number method works by first locking down the coefficients that balance electron transfer, and then tidying up the remaining atoms and charges. Here are the five steps:

Step 1: Write Correct Formulas

Write the correct chemical formula for every reactant and every product. At this stage, the equation is unbalanced (a “skeletal” equation), but every formula must be accurate. If a formula is wrong, every later step will produce a wrong answer.

Step 2: Assign Oxidation Numbers and Spot the Changes

Go through the skeletal equation and assign an oxidation number to every element on both sides. Then identify which atoms change their oxidation number from left to right. These are the atoms involved in electron transfer. The atom whose oxidation number goes up is being oxidised; the one whose oxidation number goes down is being reduced.

Built-in error check: At this point, you should find at least one species being oxidised and at least one being reduced. If you see two substances being reduced but nothing oxidised (or vice versa), something is wrong. Either a formula is incorrect or the oxidation numbers have been assigned wrongly. Go back and fix the issue before continuing.

Step 3: Equalise the Total Increase and Decrease

Calculate how much the oxidation number increases per atom for the oxidised species, and how much it decreases per atom for the reduced species. If more than one atom of the same element is involved in a single formula, multiply the per-atom change by the number of those atoms.

Now adjust coefficients so that:

$\text{Total increase in oxidation number} = \text{Total decrease in oxidation number}$

This ensures that every electron lost by the reductant is accounted for by the oxidant. Place the required coefficients in front of the appropriate species and balance those particular atoms on both sides.

Step 4: Balance the Ionic Charges

If the reaction takes place in water (which is the usual case for ionic equations), the charges on both sides may not yet be equal. Fix this by adding ions that are present in the solution:

• Acidic solution: add $H^+$ ions to the side that needs more positive charge.
• Basic solution: add $OH^-$ ions to the side that needs more negative charge.

Keep adding until the total ionic charge on the left matches the total on the right.

Step 5: Balance Hydrogen and Oxygen with Water

After adding $H^+$ or $OH^-$, the hydrogen atom count on the two sides will usually be unequal. Add $H_2O$ molecules to whichever side is short of hydrogen.

Once hydrogen is balanced, count the oxygen atoms on both sides. If they match, the equation is fully balanced. If they do not, there is an error in one of the earlier steps.

Worked Example 1: Dichromate and Sulphite in Acid (Problem 7.8)

Problem: Potassium dichromate(VI), $K_2Cr_2O_7$, reacts with sodium sulphite, $Na_2SO_3$, in acidic solution. The products are chromium(III) ion and sulphate ion. Write the balanced net ionic equation.

Applying the Five Steps

Step 1: Skeletal ionic equation

Strip out the spectator ions ($K^+$, $Na^+$) and write only the ions that participate:

$Cr_2O_7^{2-}(aq) + SO_3^{2-}(aq) \rightarrow Cr^{3+}(aq) + SO_4^{2-}(aq)$

Step 2: Assign oxidation numbers

$\overset{+6}{Cr_2}O_7^{2-}(aq) + \overset{+4}{S}O_3^{2-}(aq) \rightarrow \overset{+3}{Cr^{3+}}(aq) + \overset{+6}{S}O_4^{2-}(aq)$

• Chromium goes from $+6$ to $+3$: a decrease of $3$ per atom. This is reduction, so the dichromate ion is the oxidant (oxidising agent).
• Sulphur goes from $+4$ to $+6$: an increase of $2$ per atom. This is oxidation, so the sulphite ion is the reductant (reducing agent).

Step 3: Make the total increase equal the total decrease

Per-atom changes:

• Chromium: decrease of $3$ per atom, and there are $2$ chromium atoms in $Cr_2O_7^{2-}$, so total decrease $= 2 \times 3 = 6$.
• Sulphur: increase of $2$ per atom, and there is $1$ sulphur atom per $SO_3^{2-}$.

To get a total increase of $6$ (matching the decrease), we need $3$ sulphite ions:

$3 \times 2 = 6$

Place the coefficients and balance chromium and sulphur on both sides:

$Cr_2O_7^{2-}(aq) + 3SO_3^{2-}(aq) \rightarrow 2Cr^{3+}(aq) + 3SO_4^{2-}(aq)$

Step 4: Balance ionic charges

Count the total charge on each side:

• Left side: $(-2) + 3(-2) = -2 - 6 = -8$
• Right side: $2(+3) + 3(-2) = +6 - 6 = 0$

The left side is at $-8$ and the right side is at $0$. Since the reaction takes place in acid, add $H^+$ ions to the left to bring the charge up from $-8$ to $0$. That requires $8H^+$:

$Cr_2O_7^{2-}(aq) + 3SO_3^{2-}(aq) + 8H^+(aq) \rightarrow 2Cr^{3+}(aq) + 3SO_4^{2-}(aq)$

Step 5: Balance hydrogen and oxygen with water

There are $8$ hydrogen atoms on the left (from $8H^+$) and none on the right. Add $4H_2O$ to the right side:

$Cr_2O_7^{2-}(aq) + 3SO_3^{2-}(aq) + 8H^+(aq) \rightarrow 2Cr^{3+}(aq) + 3SO_4^{2-}(aq) + 4H_2O(l)$

Verification (oxygen check):

• Left side oxygen: $7$ (from $Cr_2O_7^{2-}$) $+ 3 \times 3$ (from $3SO_3^{2-}$) $= 7 + 9 = 16$
• Right side oxygen: $3 \times 4$ (from $3SO_4^{2-}$) $+ 4 \times 1$ (from $4H_2O$) $= 12 + 4 = 16$

Both sides have $16$ oxygen atoms. The equation is balanced.

Worked Example 2: Permanganate and Bromide in Basic Medium (Problem 7.9)

Problem: Permanganate ion reacts with bromide ion in basic solution to produce manganese dioxide and bromate ion. Write the balanced ionic equation.

Applying the Five Steps

Step 1: Skeletal ionic equation

$MnO_4^-(aq) + Br^-(aq) \rightarrow MnO_2(s) + BrO_3^-(aq)$

Step 2: Assign oxidation numbers

$\overset{+7}{Mn}O_4^-(aq) + \overset{-1}{Br^-}(aq) \rightarrow \overset{+4}{Mn}O_2(s) + \overset{+5}{Br}O_3^-(aq)$

• Manganese goes from $+7$ to $+4$: a decrease of $3$. This is reduction, so permanganate is the oxidant.
• Bromine goes from $-1$ to $+5$: an increase of $6$. This is oxidation, so bromide is the reductant.

Step 3: Make the total increase equal the total decrease

Per-atom changes:

• Manganese: decrease of $3$ per atom.
• Bromine: increase of $6$ per atom.

To equalise, we need $2$ permanganate ions because $2 \times 3 = 6$, which matches the single bromine’s increase of $6$:

$2MnO_4^-(aq) + Br^-(aq) \rightarrow 2MnO_2(s) + BrO_3^-(aq)$

Step 4: Balance ionic charges

Count the total charge on each side:

• Left side: $2(-1) + (-1) = -3$
• Right side: $0 + (-1) = -1$

The left side is at $-3$ and the right side is at $-1$. Since the reaction is in basic medium, add $OH^-$ ions to the right to bring it from $-1$ to $-3$. That requires $2OH^-$:

$2MnO_4^-(aq) + Br^-(aq) \rightarrow 2MnO_2(s) + BrO_3^-(aq) + 2OH^-(aq)$

Step 5: Balance hydrogen and oxygen with water

There are now $2$ hydrogen atoms on the right (from $2OH^-$) and none on the left. Add $1H_2O$ to the left:

$2MnO_4^-(aq) + Br^-(aq) + H_2O(l) \rightarrow 2MnO_2(s) + BrO_3^-(aq) + 2OH^-(aq)$

Verification (oxygen check):

• Left side oxygen: $2 \times 4$ (from $2MnO_4^-$) $+ 1$ (from $H_2O$) $= 8 + 1 = 9$
• Right side oxygen: $2 \times 2$ (from $2MnO_2$) $+ 3$ (from $BrO_3^-$) $+ 2 \times 1$ (from $2OH^-$) $= 4 + 3 + 2 = 9$

Both sides have $9$ oxygen atoms. The equation is balanced.

Quick-Reference Summary of the Method

Step	What You Do	Key Question to Ask
1	Write correct formulas for all species	Are all reactant and product formulas accurate?
2	Assign oxidation numbers to every element	Which atoms change oxidation number?
3	Equalise total increase and total decrease	How many of each species are needed to balance electron transfer?
4	Add $H^+$ (acid) or $OH^-$ (base)	Are the total ionic charges equal on both sides?
5	Add $H_2O$ to balance H, then check O	Do the hydrogen and oxygen atom counts match?

Tip: After completing all five steps, always run a final check: count every type of atom and verify the total charge on both sides. If everything matches, the balanced equation is correct. If anything is off, trace back through the steps to find where the mismatch crept in.