Do you have a redox equation you don't know how to balance? Besides simply balancing the equation in question, these programs will also give you a detailed overview of the entire balancing process with your chosen method.

- Ion-electron method (also called the half-reaction method)
- Oxidation number change method
- Aggregate redox species method (or ARS method) - New on periodni.com [1]

by the ion-electron method

In the ion-electron method (also called the half-reaction method), the redox equation is separated into two half-equations - one for oxidation and one for reduction. Each of these half-reactions is balanced separately and then combined to give the balanced redox equation.

* Step 1. Write down the unbalanced equation* ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.

CrI_{3} + Cl_{2}
→
CrO_{4}^{2-} + IO_{4}^{-} + Cl^{-}

* Step 2. Separate the redox reaction into half-reactions.* A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

* a) Assign oxidation numbers* for each atom in the equation. Oxidation number (also called oxidation state) is a measure of the degree of oxidation of an atom in a substance (see: Rules for assigning oxidation numbers).

Cr+3I-1_{3} + Cl0_{2}
→
Cr+6O-2_{4}^{2-} + I+7O-2_{4}^{-} + Cl-1^{-}

* b) Identify and write out all redox couples in reaction.* Identify which reactants are being oxidized (the oxidation number increases when it reacts) and which are being reduced (the oxidation number goes down).

Cr+3I-1_{3}
→
Cr+6O-2_{4}^{2-}

(Cr)

Cr+3I-1_{3}
→
I+7O-2_{4}^{-}

(I)

Cl0_{2}
→
Cl-1^{-}

(Cl)

* c) Combine these redox couples into two half-reactions:* one for the oxidation, and one for the reduction (see: Divide the redox reaction into two half-reactions). It is necessary to combine all redox pairs with species containing several different redox atoms into one equation. Before combining the two reactions it is neccesary to ensure that the same coeficients are in front of both such molecules.

Cr+3I-1_{3}
→
Cr+6O-2_{4}^{2-} + I+7O-2_{4}^{-}

Cl0_{2}
→
Cl-1^{-}

* Step 3. Balance the atoms in each half reaction.* A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas. Never change a formula when balancing an equation. Balance each half reaction separately.

* a) Balance all other atoms except hydrogen and oxygen.* We can use any of the species that appear in the skeleton equations for this purpose. Keep in mind that reactants should be added only to the left side of the equation and products to the right.

CrI_{3} → CrO_{4}^{2-} + 3IO_{4}^{-}

Cl_{2} → 2Cl^{-}

* b) Balance the oxygen atoms.* Check if there are the same numbers of oxygen atoms on the left and right side, if they aren't equilibrate these atoms by adding water molecules.

CrI_{3} + 16H_{2}O → CrO_{4}^{2-} + 3IO_{4}^{-}

Cl_{2} → 2Cl^{-}

* c) Balance the hydrogen atoms.* Check if there are the same numbers of hydrogen atoms on the left and right side, if they aren't equilibrate these atoms by adding protons (H

CrI_{3} + 16H_{2}O → CrO_{4}^{2-} + 3IO_{4}^{-} + 32H^{+}

Cl_{2} → 2Cl^{-}

* d) For reactions in a basic medium*, add one OH

CrI_{3} + 16H_{2}O + 32OH^{-} → CrO_{4}^{2-} + 3IO_{4}^{-} + 32H_{2}O

Cl_{2} → 2Cl^{-}

* Step 4. Balance the charge.* To balance the charge, add electrons (e

CrI_{3} + 16H_{2}O + 32OH^{-} → CrO_{4}^{2-} + 3IO_{4}^{-} + 32H_{2}O + 27e^{-}

Cl_{2} + 2e^{-} → 2Cl^{-}

* Step 5. Make electron gain equivalent to electron lost.* The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

CrI_{3} + 16H_{2}O + 32OH^{-} → CrO_{4}^{2-} + 3IO_{4}^{-} + 32H_{2}O + 27e^{-}

| *2

Cl_{2} + 2e^{-} → 2Cl^{-}

| *27

2CrI_{3} + 32H_{2}O + 64OH^{-} → 2CrO_{4}^{2-} + 6IO_{4}^{-} + 64H_{2}O + 54e^{-}

27Cl_{2} + 54e^{-} → 54Cl^{-}

* Step 6. Add the half-reactions together.* The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

2CrI_{3} + 27Cl_{2} + 32H_{2}O + 54e^{-} + 64OH^{-} → 2CrO_{4}^{2-} + 54Cl^{-} + 6IO_{4}^{-} + 64H_{2}O + 54e^{-}

* Step 7. Simplify the equation.* The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

2CrI_{3} + 27Cl_{2} + 64OH^{-} → 2CrO_{4}^{2-} + 54Cl^{-} + 6IO_{4}^{-} + 32H_{2}O

* Finally, always check to see that the equation is balanced.* First, verify that the equation contains the same type and number of atoms on both sides of the equation.

ELEMENT | LEFT | RIGHT | DIFFERENCE |
---|---|---|---|

Cr | 2*1 | 2*1 | 0 |

I | 2*3 | 6*1 | 0 |

Cl | 27*2 | 54*1 | 0 |

O | 64*1 | 2*4 + 6*4 + 32*1 | 0 |

H | 64*1 | 32*2 | 0 |

Second, verify that the sum of the charges on one side of the equation is equal to the sum of the charges on the other side. It doesn't matter what the charge is as long as it is the same on both sides.

2*0 + 27*0 + 64*-1 = 2*-2 + 54*-1 + 6*-1 + 32*0

-64 = -64

-64 = -64

Since the sum of individual atoms on the left side of the equation matches the sum of the same atoms on the right side, and since the charges on both sides are equal we can write a balanced equation.

2CrI_{3} + 27Cl_{2} + 64OH^{-} → 2CrO_{4}^{2-} + 6IO_{4}^{-} + 54Cl^{-} + 32H_{2}O

Citing this page:

Generalic, Eni. "Balancing redox reactions by the ion-electron method." *EniG. Periodic Table of the Elements*. KTF-Split, 22 June 2019. Web. {Date of access}. <https://www.periodni.com/half-reaction_method.php>.

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