Write and balance redox equations in acidic and neutral solutions

Alberta Grade 12 Chemistry

Electrochemical Changes

Alberta Chemistry 30 — Outcome 30-B1.7k

Drop a piece of iron into a solution of copper(II) sulfate and watch the blue colour fade while a reddish-brown solid appears on the iron. Electrons have moved — from the iron atoms to the copper ions in solution. Reactions in which electrons transfer from one species to another are called redox reactions. To communicate precisely what happened, chemists write balanced redox equations. Balancing these equations is more involved than balancing simple synthesis or decomposition reactions, because you must account for electron transfer as well as atoms and charge. This lesson teaches a systematic method called the half-reaction method (also called the ion-electron method) for writing and balancing redox equations in acidic and neutral aqueous solutions. This method is the standard approach in Alberta Chemistry 30.

What you will learn

Assign oxidation numbers to atoms in reactants and products to identify what is oxidized and what is reduced.
Write and balance separate half-reactions for oxidation and reduction.
Combine half-reactions into a balanced overall redox equation for acidic and neutral solutions.
Verify that both atoms and net charge are conserved in the final equation.

Prerequisite Review: Oxidation Numbers and What Redox Means

Before balancing anything, you need to identify which atoms gain electrons and which lose them. To do this, assign oxidation numbers. An oxidation number is a bookkeeping charge assigned to an atom as if the compound were made entirely of ions. The Alberta Chemistry 30 Data Booklet lists the rules; the most useful ones are reviewed here.

Key rules in priority order: (1) A pure element has an oxidation number of zero. (2) A monatomic ion has an oxidation number equal to its ionic charge. (3) Oxygen is almost always −2, except in peroxides where it is −1. (4) Hydrogen bonded to a nonmetal is +1. (5) The oxidation numbers of all atoms in a neutral compound must add to zero; in a polyatomic ion they must add to the ion's charge.

Oxidation is defined as an increase in oxidation number (the atom loses electrons). Reduction is defined as a decrease in oxidation number (the atom gains electrons). A helpful memory device used in Alberta Chemistry 30: OIL RIG — Oxidation Is Loss, Reduction Is Gain (of electrons). The species that is oxidized is the reducing agent; the species that is reduced is the oxidizing agent. These definitions are fixed — learn them exactly.

\text{Oxidation number change} = \text{final oxidation number} - \text{initial oxidation number}

Oxidation number rules establish a bookkeeping charge on each atom.
Oxidation = increase in oxidation number = loss of electrons.
Reduction = decrease in oxidation number = gain of electrons.
OIL RIG: Oxidation Is Loss, Reduction Is Gain.
The reducing agent is oxidized; the oxidizing agent is reduced.

The Half-Reaction Method: Overview and Rules

A redox reaction can always be split into two halves. The oxidation half-reaction shows only the species being oxidized and the electrons it releases. The reduction half-reaction shows only the species being reduced and the electrons it consumes. Balancing each half separately — then combining them — keeps the process organized.

The half-reaction method follows a strict sequence of steps. For acidic solutions, water molecules and hydrogen ions are available to balance oxygen and hydrogen atoms. For neutral solutions, you use the same steps but you will often find that the unbalanced atoms cancel without needing to add acid, or that the reaction itself does not change the hydrogen or oxygen count significantly. In practice, the Alberta Chemistry 30 course focuses on acidic conditions and those neutral reactions where water may still appear as a product or reactant.

Here are the six steps applied to each half-reaction: (1) Write the unbalanced skeleton for that half. (2) Balance all atoms except O and H. (3) Balance O by adding water molecules to the side that needs more oxygen. (4) Balance H by adding hydrogen ions to the side that needs more hydrogen. (5) Balance charge by adding electrons to the more positive side. (6) Multiply the half-reactions by whole-number factors so that electrons lost equal electrons gained, then add the half-reactions and cancel species that appear on both sides.

\text{electrons lost by reducing agent} = \text{electrons gained by oxidizing agent}

Split the overall reaction into an oxidation half and a reduction half.
Balance atoms in this order: elements other than O and H first, then O using water, then H using hydrogen ions.
Balance charge by adding electrons to the more positive side.
Multiply half-reactions so electron loss equals electron gain before adding.
Cancel identical species that appear on both sides of the combined equation.

Applying the Steps in Acidic Solution

Consider the reaction between dichromate ions and iron(II) ions in acidic solution. The dichromate ion,

\text{Cr}_2\text{O}_7^{2-}

, is the oxidizing agent; it gets reduced to

\text{Cr}^{3+}

. The iron(II) ion,

\text{Fe}^{2+}

, is the reducing agent; it gets oxidized to

\text{Fe}^{3+}

. Acidic solution means that

\text{H}^+

and

\text{H}_2\text{O}

are available as balancing tools.

For the reduction half: Start with

\text{Cr}_2\text{O}_7^{2-} \rightarrow \text{Cr}^{3+}

. Balance Cr: add a second

\text{Cr}^{3+}

on the right. Balance O: add 7

\text{H}_2\text{O}

to the right. Balance H: add 14

\text{H}^+

to the left. Balance charge: left side has a net charge of

(14 \times +1) + (-2) = +12

; right side has

2 \times (+3) = +6

. Add 6 electrons to the left to make both sides equal to +6. The balanced reduction half-reaction is

\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6e^- \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l)

.

For the oxidation half:

\text{Fe}^{2+} \rightarrow \text{Fe}^{3+}

. No O or H to balance. Left charge is +2, right is +3. Add 1 electron to the right:

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

. This half-reaction already loses 1 electron. To match the 6 electrons in the reduction half, multiply the oxidation half by 6. Then add both half-reactions and cancel the 6 electrons that appear on both sides. The final balanced equation is

\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{Fe}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 6\text{Fe}^{3+}(aq) + 7\text{H}_2\text{O}(l)

.

\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{Fe}^{2+}(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 6\text{Fe}^{3+}(aq) + 7\text{H}_2\text{O}(l)

In acidic solution, use water to balance O and hydrogen ions to balance H.
Add electrons to the more positive side of each half-reaction.
Multiply half-reactions by factors so electrons cancel exactly when added.
Check: count atoms and net charge on each side of the final equation.

Balancing Redox Equations in Neutral Solution

In neutral solution, you cannot assume a large supply of free

\text{H}^+

ions. However, the same six-step half-reaction method still applies. The key difference is that any hydrogen ions added during balancing must cancel out in the final step — if they do not, the equation is not truly neutral. In many neutral redox reactions, the half-reactions involve species that do not require additional H or O for balancing, so the hydrogen-ion and water steps produce no net addition.

A common neutral-solution example is the displacement reaction between zinc metal and copper(II) ions:

\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)

. Oxidation half:

\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + 2e^-

. Reduction half:

\text{Cu}^{2+}(aq) + 2e^- \rightarrow \text{Cu}(s)

. Electrons are equal at 2; add directly. No water or hydrogen ions are needed because there are no oxygen or hydrogen atoms to balance. The overall equation is balanced as written, and the net charge is equal on both sides: left

= 0 + (+2) = +2

; right

= +2 + 0 = +2

.

Some neutral reactions do produce water as a product or use water as a reactant, for example the reaction of permanganate with a neutral organic substance. In Alberta Chemistry 30, such cases are handled by following the same six steps — and the

\text{H}^+

ions needed during balancing will cancel across the two half-reactions when you add them, leaving no net

\text{H}^+

in the final equation. Always verify charge and atom balance in the final combined equation regardless of solution type.

\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)

In neutral solution, apply the same six steps as in acidic solution.
Any hydrogen ions introduced during balancing must cancel in the final equation.
If no O or H atoms are involved in a neutral-solution half-reaction, no water or hydrogen ions are needed.
Always verify both atoms and net charge in the final combined equation.

Verifying a Balanced Redox Equation

A balanced redox equation must satisfy two independent checks. First, atom balance: every element must have the same total count on both sides. Count each element separately. Second, charge balance: the total net charge on the left must equal the total net charge on the right. Add up all ionic charges and electron charges on each side. These two checks are equally important — an equation can balance atoms but fail the charge check, making it wrong.

To carry out the charge check, list each species, multiply its charge by its coefficient, and sum all values for each side. For the dichromate–iron example: left side charge

= (-2) + 14(+1) + 6(+2) = -2 + 14 + 12 = +24

. Right side charge

= 2(+3) + 6(+3) + 7(0) = +6 + 18 + 0 = +24

. Both sides give +24, so charge is balanced. The atom count (Cr: 2 = 2; Fe: 6 = 6; O: 7 = 7; H: 14 = 14) confirms full balance.

State symbols matter in Alberta Chemistry 30. Aqueous ions carry the symbol

(aq)

, pure metals and solids carry

(s)

, liquids carry

(l)

, and gases carry

(g)

. Water in redox equations is liquid, written as

\text{H}_2\text{O}(l)

. Include state symbols in all final balanced equations. They are expected on diploma-style questions.

\sum (charge × coefficient)_{left} = \sum (charge × coefficient)_{right}

Check atom balance: every element must match on both sides.
Check charge balance: net charge on left must equal net charge on right.
Both checks must pass; passing only one is not sufficient.
Include state symbols in all final equations.

Half-Reaction Method: Six Steps at a Glance

Step	What to Do	Tool Used
1	Write the skeleton half-reaction (just the redox-active species)	Oxidation numbers
2	Balance all atoms except O and H	Coefficients
3	Balance O atoms	Add $\text{H}_2\text{O}(l)$
4	Balance H atoms	Add $\text{H}^+(aq)$ (acidic/neutral)
5	Balance net charge	Add $e^-$ to the more positive side
6	Equalise electrons, add half-reactions, cancel common species	Lowest common multiple of electrons

Worked example

Balancing a Redox Equation in Acidic Solution: Permanganate and Hydrogen Peroxide

Potassium permanganate reacts with hydrogen peroxide in acidic solution. The permanganate ion is reduced to manganese(II) ion. The hydrogen peroxide is oxidized to oxygen gas. Write the complete, balanced ionic equation including state symbols.

Step 1 — Assign oxidation numbers and identify the change
In the permanganate ion, oxygen is −2 (four oxygens = −8) and the ion has a charge of −1, so Mn must be +7. In the product, the manganese(II) ion has Mn at +2. Mn decreases from +7 to +2: that is a decrease of 5, so Mn is reduced. In hydrogen peroxide, oxygen is −1 (each O). In the product oxygen gas, O is 0. O increases from −1 to 0: that is an increase of 1 per O atom, so H₂O₂ is oxidized.
Mn: $+7 \rightarrow +2 (gain of 5 electrons, reduced); \quad O in H_2O_2: -1 \rightarrow 0 (loss of 1 electron per O, oxidized)$
Step 2 — Write the skeleton half-reactions
Write one skeleton for reduction (permanganate to Mn²⁺) and one for oxidation (hydrogen peroxide to oxygen gas). Do not add balancing species yet.
$\text{Reduction: } \text{MnO}_4^-(aq) \rightarrow \text{Mn}^{2+}(aq) \\ \text{Oxidation: } \text{H}_2\text{O}_2(aq) \rightarrow \text{O}_2(g)$
Step 3 — Balance atoms other than O and H
In the reduction half, there is one Mn on each side — already balanced. In the oxidation half, there are two O atoms on each side — already balanced. No other non-O, non-H atoms are present.
$\text{Reduction: } \text{MnO}_4^-(aq) \rightarrow \text{Mn}^{2+}(aq) \quad (\text{Mn balanced: 1 = 1}) \\ \text{Oxidation: } \text{H}_2\text{O}_2(aq) \rightarrow \text{O}_2(g) \quad (\text{O balanced: 2 = 2})$
Step 4 — Balance O with water, then H with hydrogen ions
Reduction half needs 4 oxygen on the right (from MnO₄⁻). Add 4 H₂O to the right. Then balance H: 4 H₂O on the right gives 8 H, so add 8 H⁺ to the left. Oxidation half: O is already balanced (2 = 2). Balance H: H₂O₂ has 2 H on the left and O₂ has 0 H on the right. Add 2 H⁺ to the right.
$\text{Reduction: } \text{MnO}_4^-(aq) + 8\text{H}^+(aq) \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \\ \text{Oxidation: } \text{H}_2\text{O}_2(aq) \rightarrow \text{O}_2(g) + 2\text{H}^+(aq)$
Step 5 — Balance charge with electrons
Reduction half — left charge: (−1) + 8(+1) = +7. Right charge: (+2) + 0 = +2. The left is more positive by 5. Add 5 electrons to the left to make both sides +2. Oxidation half — left charge: 0. Right charge: 0 + 2(+1) = +2. The right is more positive by 2. Add 2 electrons to the right to make both sides 0.
$\text{Reduction: } \text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l) \\ \text{Oxidation: } \text{H}_2\text{O}_2(aq) \rightarrow \text{O}_2(g) + 2\text{H}^+(aq) + 2e^-$
Step 6 — Equalise electrons and combine
The reduction half involves 5 electrons; the oxidation half involves 2 electrons. The least common multiple of 5 and 2 is 10. Multiply the reduction half by 2 and the oxidation half by 5. Then add the two half-reactions. Cancel species that appear identically on both sides: 10 electrons cancel. Also cancel H⁺: 16 H⁺ on the left minus 10 H⁺ on the right leaves a net of 6 H⁺ on the left.
$2\text{MnO}_4^-(aq) + 16\text{H}^+(aq) + 10e^- + 5\text{H}_2\text{O}_2(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 8\text{H}_2\text{O}(l) + 5\text{O}_2(g) + 10\text{H}^+(aq) + 10e^-$
Step 7 — Write and verify the final equation
After cancelling 10 electrons from both sides and 10 H⁺ from both sides (16 − 10 = 6 H⁺ net on the left), the final equation is shown below. Atom check — Mn: 2 = 2; O: 8 + 10 = 18 on left (8 from MnO₄⁻, 10 from H₂O₂), right: 8 + 10 = 18 (8 in H₂O, 10 in O₂). H: 6×2 + 5×2 = 12 + 10 = 22 wait — recount: left H = 6(+1)×1 H each + H₂O₂ 5×2 H = 6 + 10 = 16 H. Right H = 8×2 = 16 H. ✓ Charge check — left: 2(−1) + 6(+1) = −2 + 6 = +4. Right: 2(+2) + 0 + 0 = +4. ✓
$2\text{MnO}_4^-(aq) + 6\text{H}^+(aq) + 5\text{H}_2\text{O}_2(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 8\text{H}_2\text{O}(l) + 5\text{O}_2(g)$

Answer:

2\text{MnO}_4^-(aq) + 6\text{H}^+(aq) + 5\text{H}_2\text{O}_2(aq) \rightarrow 2\text{Mn}^{2+}(aq) + 8\text{H}_2\text{O}(l) + 5\text{O}_2(g)

Check: Atom balance: Mn 2 = 2; O: (8 + 10) = 18 left, (8 + 10) = 18 right ✓; H: (6 + 10) = 16 left, (16) right ✓. Charge balance: left = 2(−1) + 6(+1) = +4; right = 2(+2) = +4 ✓. Both checks pass.

Common mistakes and how to avoid them

Adding electrons to the wrong side of a half-reaction, for example putting electrons on the right side of a reduction half-reaction.

Correction: Always add electrons to the more positive side. For reduction, the reactant side is more positive (the species gains electrons from the left), so electrons go on the left. For oxidation, electrons are released as products, so electrons go on the right.

Forgetting to multiply the coefficients of all species — including water and hydrogen ions — when scaling a half-reaction by a factor.

Correction: When you multiply a half-reaction by a factor, every coefficient in that half-reaction is multiplied, including the water molecules, hydrogen ions, and electrons. Treat the entire equation as a unit.

Checking only atom balance and skipping the charge balance check.

Correction: Both checks are required. An equation can have correct atom counts but unequal net charges, making it wrong. Always calculate total net charge on each side as a separate step.

Cancelling electrons that are not equal on both sides before equalising them.

Correction: Electrons can only be cancelled after you have multiplied the half-reactions so that the number of electrons is identical on both sides. Find the least common multiple first, then cancel.

Omitting state symbols in the final balanced equation.

Correction: Alberta Chemistry 30 and diploma-style questions expect state symbols: (aq) for dissolved species, (s) for solids, (l) for pure liquids, and (g) for gases. Include them in every final answer.

Lesson summary

A redox reaction involves electron transfer: the reducing agent is oxidized (loses electrons, oxidation number increases) and the oxidizing agent is reduced (gains electrons, oxidation number decreases).
Use oxidation numbers to identify which species is oxidized and which is reduced before writing any half-reaction.
The half-reaction method balances each half separately in this order: atoms other than O and H → O using water → H using hydrogen ions → charge using electrons.
Multiply half-reactions by whole-number factors so that electrons lost equal electrons gained, then add and cancel identical species on both sides.
In neutral solution, apply the same six steps; any hydrogen ions added during balancing must cancel in the final combined equation.
Always verify the final equation with two independent checks: atom balance and net charge balance on both sides.

Check your understanding

Question 1

In the half-reaction

\text{MnO}_4^-(aq) + 8\text{H}^+(aq) + 5e^- \rightarrow \text{Mn}^{2+}(aq) + 4\text{H}_2\text{O}(l)

, what is the net charge on the left side before the electrons are added (i.e., with only

\text{MnO}_4^-

and

8\text{H}^+

)?

+5
+7
+6
+8

Show answer and explanation

+7

The permanganate ion has a charge of −1 and the eight hydrogen ions each contribute +1, giving a total of −1 + 8(+1) = +7. Adding 5 electrons (each −1) brings the left side to +7 − 5 = +2, which matches the right side (

\text{Mn}^{2+}

= +2, water = 0).

Question 2

When balancing a redox equation using the half-reaction method in acidic solution, which species is added to balance oxygen atoms?

$\text{H}^+(aq)$
$\text{OH}^-(aq)$
$\text{H}_2\text{O}(l)$
$\text{O}_2(g)$

Show answer and explanation

\text{H}_2\text{O}(l)

Oxygen atoms are balanced by adding water molecules to the side that is short on oxygen. Hydrogen ions are then used in the next step to balance the hydrogen atoms introduced by those water molecules.

Question 3

A student writes the following two half-reactions: Oxidation:

\text{Sn}^{2+}(aq) \rightarrow \text{Sn}^{4+}(aq) + 2e^-

and Reduction:

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

. What is the correct overall balanced ionic equation?

$\text{Sn}^{2+}(aq) + \text{Fe}^{3+}(aq) \rightarrow \text{Sn}^{4+}(aq) + \text{Fe}^{2+}(aq)$
$\text{Sn}^{2+}(aq) + 2\text{Fe}^{3+}(aq) \rightarrow \text{Sn}^{4+}(aq) + 2\text{Fe}^{2+}(aq)$
$2\text{Sn}^{2+}(aq) + \text{Fe}^{3+}(aq) \rightarrow 2\text{Sn}^{4+}(aq) + \text{Fe}^{2+}(aq)$
$\text{Sn}^{2+}(aq) + 2\text{Fe}^{3+}(aq) \rightarrow \text{Sn}^{4+}(aq) + \text{Fe}^{3+}(aq)$

Show answer and explanation

\text{Sn}^{2+}(aq) + 2\text{Fe}^{3+}(aq) \rightarrow \text{Sn}^{4+}(aq) + 2\text{Fe}^{2+}(aq)

The oxidation half releases 2 electrons and the reduction half consumes 1 electron. To equalise, multiply the reduction half by 2:

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

. Adding both halves and cancelling the 2 electrons gives

\text{Sn}^{2+}(aq) + 2\text{Fe}^{3+}(aq) \rightarrow \text{Sn}^{4+}(aq) + 2\text{Fe}^{2+}(aq)

. Charge check: left = +2 + 2(+3) = +8; right = +4 + 2(+2) = +8. ✓

Question 4

Which statement correctly describes the charge balance requirement for a fully balanced redox equation?

The sum of all coefficients on the left must equal the sum on the right.
The net ionic charge on the left side must equal the net ionic charge on the right side.
Each individual ion must have the same charge on both sides of the equation.
The total number of electrons on the left must equal the total number of ions on the right.

Show answer and explanation

The net ionic charge on the left side must equal the net ionic charge on the right side.

Charge balance means the total net charge — calculated by multiplying each species' charge by its coefficient and summing — must be equal on both sides. Individual ions do not need to appear on both sides, and the number of electrons is not compared to the number of ions.

Key terms

Redox reaction: A chemical reaction in which electrons are transferred from one species to another; oxidation and reduction always occur together.
Oxidation number: A bookkeeping number assigned to an atom that represents its apparent charge, as if all bonds were ionic; used to track electron transfer.
Oxidation: A process in which an atom's oxidation number increases, meaning that atom has lost electrons.
Reduction: A process in which an atom's oxidation number decreases, meaning that atom has gained electrons.
Half-reaction: An equation that shows either only the oxidation or only the reduction part of a redox reaction, including the electrons transferred.
Reducing agent: The species that is oxidized in a redox reaction; it donates electrons to the other species.
Oxidizing agent: The species that is reduced in a redox reaction; it accepts electrons from the other species.
Charge balance: The requirement that the total net ionic charge on the left side of a balanced equation equals the total net ionic charge on the right side.

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30-B1.7k · Write and balance redox equations in acidic and neutral solutions

Alberta Grade 12 Chemistry

Alberta Chemistry 30 — Outcome 30-B1.7k

What you will learn

Prerequisite Review: Oxidation Numbers and What Redox Means

The Half-Reaction Method: Overview and Rules

Applying the Steps in Acidic Solution

Balancing Redox Equations in Neutral Solution

Verifying a Balanced Redox Equation

Half-Reaction Method: Six Steps at a Glance

Balancing a Redox Equation in Acidic Solution: Permanganate and Hydrogen Peroxide

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Question 4

Key terms

Continue through CHEM30

About this lesson and its review