Chemical Equation Balancer

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balance ionic equation acidic basic medium

Ionic equations and half-reactions: balancing in acidic and basic medium

In ionic form the electric charge has to close as well as the atoms. Type the species with their charge (MnO4^-, Fe^2+) and the tool returns the equation and the half-reactions, in acidic or basic medium.

The ion-electron method

Two half-reactions are separated, one where the element is oxidized and one where it is reduced. Each is balanced in five passes: atoms other than O and H; oxygen, with H2O; hydrogen, with H+; charge, with electrons; and, in basic medium, replacing each H+ by H2O by adding OH- to both sides. For permanganate in acid, MnO4- → Mn2+ takes 4 H2O, 8 H+ and 5 electrons: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. The charge checks: -1 + 8 - 5 = +2, equal to that of Mn2+. The iron oxidation half is Fe2+ → Fe3+ + e-.
Then the electrons are equalized by multiplying each half-reaction (here, the iron one by 5) and added, cancelling the electrons, H+ and H2O that appear on both sides: MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O. The total charge is -1 + 10 + 8 = +17 in the reactants and +2 + 15 = +17 in the products. Dichromate follows the same path, with 14 H+ because it has 7 oxygens and two chromiums: Cr2O7 2- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O, with charge +24 on both sides. In basic medium the balance uses OH-, as in 2MnO4- + 6I- + 4H2O → 2MnO2 + 3I2 + 8OH-, where the charge is -8 on both sides.
The tool solves the same problem another way, treating charge as one more row of the system and the electron as one more species, and shows the half-reactions in the Redox tab with the medium you pick (automatic, acid or basic). It also accepts the equation without spectators, which is what half-reactions require: with whole KMnO4 in place of MnO4-, the loose K+ and Cl- block the separation, and the tool explains that instead of forcing a result. Chlorine disproportionation in base, 3Cl2 + 6OH- → 5Cl- + ClO3- + 3H2O, is a case where the same reactant appears in both half-reactions.

Balanced ionic equations

Permanganate and iron(II) in acid

Input: MnO4^- + Fe^2+ + H^+ -> Mn^2+ + Fe^3+ + H2O
Expected output: MnO4^- + 5Fe^2+ + 8H^+ -> Mn^2+ + 5Fe^3+ + 4H2O

Charge +17 on both sides; 5 electrons exchanged.

Dichromate and iron(II) in acid

Input: Cr2O7^2- + Fe^2+ + H^+ -> Cr^3+ + Fe^3+ + H2O
Expected output: Cr2O7^2- + 6Fe^2+ + 14H^+ -> 2Cr^3+ + 6Fe^3+ + 7H2O

Charge +24 on both sides; 6 electrons exchanged.

Permanganate and iodide in base

Input: MnO4^- + I^- + H2O -> MnO2 + I2 + OH^-
Expected output: 2MnO4^- + 6I^- + 4H2O -> 2MnO2 + 3I2 + 8OH^-

Charge -8 on both sides; 6 electrons exchanged.

Chlorine disproportionation in base

Input: Cl2 + OH^- -> Cl^- + ClO3^- + H2O
Expected output: 3Cl2 + 6OH^- -> 5Cl^- + ClO3^- + 3H2O

Cl2 is oxidized (0 → +5) and reduced (0 → -1) at the same time.

Zinc and copper(II) ion

Input: Zn + Cu^2+ -> Zn^2+ + Cu
Expected output: Zn + Cu^2+ -> Zn^2+ + Cu

Already balanced: charge +2 on both sides, every coefficient 1.

Iron complex ion with permanganate

Input: [Fe(CN)6]^4- + MnO4^- + H^+ -> [Fe(CN)6]^3- + Mn^2+ + H2O
Expected output: 5[Fe(CN)6]^4- + MnO4^- + 8H^+ -> 5[Fe(CN)6]^3- + Mn^2+ + 4H2O

The hexacyanoferrate keeps the CN group intact; only Fe changes from +2 to +3.

Full tool FAQ

How do you balance a chemical equation?

Count the atoms of each element on both sides and adjust the coefficients (never the subscripts) until the counts match. By hand this is done by trial; the tool solves the exact linear system. For Fe + O2 → Fe2O3 the answer is 4Fe + 3O2 → 2Fe2O3, with 4 Fe atoms and 6 O atoms on each side.

Why does an equation need to be balanced?

How does balancing by trial and error work?

How do you balance equations with ions?

How do you balance a redox reaction?

What does "more than one independent solution" mean?

Why does the tool say a substance is on the wrong side or takes no part?

Can I type the equation with coefficients?

How do you find the limiting reagent and the yield?

Does the tool predict the products of a reaction?

Frequently asked questions

How do I write an ion's charge in the tool?

After a caret: Fe^3+, SO4^2-, Cl^-. Forms like Fe3+ and SO42- also work, and the electron is e^- or e-. For bracketed species such as hexacyanoferrate, write [Fe(CN)6]^4-.

When do I use OH- and when H+?

H+ in acidic medium and OH- in basic medium, as the problem states. The final result changes: the same MnO4- becomes Mn2+ in acid and MnO2 in neutral or basic medium, so the product depends on the medium too. The tool balances the species you type.

Why does the tool not show half-reactions for whole KMnO4?

Because K+ and the other spectator ions are part of the formula and get in the way of the separation. Rewrite only the ions that react (MnO4-, Fe2+) and the half-reactions appear. Balancing the molecular equation still works as usual.

Next steps

Atom count, before and after the coefficients
Element	Before (all 1)	After	Check
C	3	1	3	3	✓
H	8	2	8	8	✓
O	2	3	10	10	✓

Atom count, before and after the coefficients

Element

Before (all 1)

After

Check

React.

Prod.

React.

Prod.

✓

The ion-electron method

Two half-reactions are separated, one where the element is oxidized and one where it is reduced. Each is balanced in five passes: atoms other than O and H; oxygen, with H2O; hydrogen, with H+; charge, with electrons; and, in basic medium, replacing each H+ by H2O by adding OH- to both sides. For permanganate in acid, MnO4- → Mn2+ takes 4 H2O, 8 H+ and 5 electrons: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O. The charge checks: -1 + 8 - 5 = +2, equal to that of Mn2+. The iron oxidation half is Fe2+ → Fe3+ + e-.

Then the electrons are equalized by multiplying each half-reaction (here, the iron one by 5) and added, cancelling the electrons, H+ and H2O that appear on both sides: MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O. The total charge is -1 + 10 + 8 = +17 in the reactants and +2 + 15 = +17 in the products. Dichromate follows the same path, with 14 H+ because it has 7 oxygens and two chromiums: Cr2O7 2- + 6Fe2+ + 14H+ → 2Cr3+ + 6Fe3+ + 7H2O, with charge +24 on both sides. In basic medium the balance uses OH-, as in 2MnO4- + 6I- + 4H2O → 2MnO2 + 3I2 + 8OH-, where the charge is -8 on both sides.

The tool solves the same problem another way, treating charge as one more row of the system and the electron as one more species, and shows the half-reactions in the Redox tab with the medium you pick (automatic, acid or basic). It also accepts the equation without spectators, which is what half-reactions require: with whole KMnO4 in place of MnO4-, the loose K+ and Cl- block the separation, and the tool explains that instead of forcing a result. Chlorine disproportionation in base, 3Cl2 + 6OH- → 5Cl- + ClO3- + 3H2O, is a case where the same reactant appears in both half-reactions.

Balanced ionic equations

Permanganate and iron(II) in acid

Input: MnO4^- + Fe^2+ + H^+ -> Mn^2+ + Fe^3+ + H2O
Expected output: MnO4^- + 5Fe^2+ + 8H^+ -> Mn^2+ + 5Fe^3+ + 4H2O

Charge +17 on both sides; 5 electrons exchanged.

Dichromate and iron(II) in acid

Input: Cr2O7^2- + Fe^2+ + H^+ -> Cr^3+ + Fe^3+ + H2O
Expected output: Cr2O7^2- + 6Fe^2+ + 14H^+ -> 2Cr^3+ + 6Fe^3+ + 7H2O

Charge +24 on both sides; 6 electrons exchanged.

Permanganate and iodide in base

Input: MnO4^- + I^- + H2O -> MnO2 + I2 + OH^-
Expected output: 2MnO4^- + 6I^- + 4H2O -> 2MnO2 + 3I2 + 8OH^-

Charge -8 on both sides; 6 electrons exchanged.

Chlorine disproportionation in base

Input: Cl2 + OH^- -> Cl^- + ClO3^- + H2O
Expected output: 3Cl2 + 6OH^- -> 5Cl^- + ClO3^- + 3H2O

Cl2 is oxidized (0 → +5) and reduced (0 → -1) at the same time.

Zinc and copper(II) ion

Input: Zn + Cu^2+ -> Zn^2+ + Cu
Expected output: Zn + Cu^2+ -> Zn^2+ + Cu

Already balanced: charge +2 on both sides, every coefficient 1.

Iron complex ion with permanganate

Input: [Fe(CN)6]^4- + MnO4^- + H^+ -> [Fe(CN)6]^3- + Mn^2+ + H2O
Expected output: 5[Fe(CN)6]^4- + MnO4^- + 8H^+ -> 5[Fe(CN)6]^3- + Mn^2+ + 4H2O

The hexacyanoferrate keeps the CN group intact; only Fe changes from +2 to +3.

Full tool FAQ

How do you balance a chemical equation?

Why does an equation need to be balanced?

How does balancing by trial and error work?

How do you balance equations with ions?

How do you balance a redox reaction?

What does "more than one independent solution" mean?

Why does the tool say a substance is on the wrong side or takes no part?

Can I type the equation with coefficients?

How do you find the limiting reagent and the yield?

Does the tool predict the products of a reaction?

Frequently asked questions

How do I write an ion's charge in the tool?

After a caret: Fe^3+, SO4^2-, Cl^-. Forms like Fe3+ and SO42- also work, and the electron is e^- or e-. For bracketed species such as hexacyanoferrate, write [Fe(CN)6]^4-.