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.

K_{2}Cr_{2}O_{7} + (NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + H_{2}SO_{4}
→
Cr_{2}(SO_{4})_{3} + Fe_{2}(SO_{4})_{3} + K_{2}SO_{4} + (NH_{4})_{2}SO_{4} + H_{2}O

* 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).

K+1_{2}Cr+6_{2}O-2_{7} + (N-3H+1_{4})_{2}Fe+2(S+6O-2_{4})_{2}·6H+1_{2}O-2 + H+1_{2}S+6O-2_{4}
→
Cr+3_{2}(S+6O-2_{4})_{3} + Fe+3_{2}(S+6O-2_{4})_{3} + K+1_{2}S+6O-2_{4} + (N-3H+1_{4})_{2}S+6O-2_{4} + H+1_{2}O-2

* 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).

(N-3H+1_{4})_{2}Fe+2(S+6O-2_{4})_{2}·6H+1_{2}O-2
→
Fe+3_{2}(S+6O-2_{4})_{3}

(Fe)

K+1_{2}Cr+6_{2}O-2_{7}
→
Cr+3_{2}(S+6O-2_{4})_{3}

(Cr)

* 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).

(N-3H+1_{4})_{2}Fe+2(S+6O-2_{4})_{2}·6H+1_{2}O-2
→
Fe+3_{2}(S+6O-2_{4})_{3}

K+1_{2}Cr+6_{2}O-2_{7}
→
Cr+3_{2}(S+6O-2_{4})_{3}

* 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.

2(NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + 5H_{2}SO_{4} + K_{2}Cr_{2}O_{7} → Fe_{2}(SO_{4})_{3} + 2(NH_{4})_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4}

K_{2}Cr_{2}O_{7} + 4H_{2}SO_{4} → Cr_{2}(SO_{4})_{3} + K_{2}SO_{4}

* 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.

2(NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + 5H_{2}SO_{4} + K_{2}Cr_{2}O_{7} → Fe_{2}(SO_{4})_{3} + 2(NH_{4})_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 19H_{2}O

K_{2}Cr_{2}O_{7} + 4H_{2}SO_{4} → Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 7H_{2}O

* 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

2(NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + 5H_{2}SO_{4} + K_{2}Cr_{2}O_{7} + 4H^{+} → Fe_{2}(SO_{4})_{3} + 2(NH_{4})_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 19H_{2}O

K_{2}Cr_{2}O_{7} + 4H_{2}SO_{4} + 6H^{+} → Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 7H_{2}O

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

2(NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + 5H_{2}SO_{4} + K_{2}Cr_{2}O_{7} + 4H^{+} + 4e^{-} → Fe_{2}(SO_{4})_{3} + 2(NH_{4})_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 19H_{2}O

K_{2}Cr_{2}O_{7} + 4H_{2}SO_{4} + 6H^{+} + 6e^{-} → Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 7H_{2}O

* 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.

2(NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + 5H_{2}SO_{4} + K_{2}Cr_{2}O_{7} + 4H^{+} + 4e^{-} → Fe_{2}(SO_{4})_{3} + 2(NH_{4})_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 19H_{2}O

| *3

K_{2}Cr_{2}O_{7} + 4H_{2}SO_{4} + 6H^{+} + 6e^{-} → Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 7H_{2}O

| *2

6(NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + 15H_{2}SO_{4} + 3K_{2}Cr_{2}O_{7} + 12H^{+} + 12e^{-} → 3Fe_{2}(SO_{4})_{3} + 6(NH_{4})_{2}SO_{4} + 3Cr_{2}(SO_{4})_{3} + 3K_{2}SO_{4} + 57H_{2}O

2K_{2}Cr_{2}O_{7} + 8H_{2}SO_{4} + 12H^{+} + 12e^{-} → 2Cr_{2}(SO_{4})_{3} + 2K_{2}SO_{4} + 14H_{2}O

* 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.

6(NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + 5K_{2}Cr_{2}O_{7} + 23H_{2}SO_{4} + 24H^{+} + 24e^{-} → 3Fe_{2}(SO_{4})_{3} + 5Cr_{2}(SO_{4})_{3} + 6(NH_{4})_{2}SO_{4} + 5K_{2}SO_{4} + 71H_{2}O

* 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.

6(NH_{4})_{2}Fe(SO_{4})_{2}·6H_{2}O + 5K_{2}Cr_{2}O_{7} + 23H_{2}SO_{4} + 24H^{+} + 24e^{-} → 3Fe_{2}(SO_{4})_{3} + 5Cr_{2}(SO_{4})_{3} + 6(NH_{4})_{2}SO_{4} + 5K_{2}SO_{4} + 71H_{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 |
---|---|---|---|

N | 6*2 | 6*2 | 0 |

H | 6*8 + 6*12 + 23*2 + 24*1 | 6*8 + 71*2 | 0 |

Fe | 6*1 | 3*2 | 0 |

S | 6*2 + 23*1 | 3*3 + 5*3 + 6*1 + 5*1 | 0 |

O | 6*8 + 6*6 + 5*7 + 23*4 | 3*12 + 5*12 + 6*4 + 5*4 + 71*1 | 0 |

K | 5*2 | 5*2 | 0 |

Cr | 5*2 | 5*2 | 0 |

e | 24*1 | 0 | 24 |

The difference between atoms on the left and right side of the equation is not equal to zero.

The error most likely occurred while balancing atoms in the partial equations of oxidation and reduction (Step 3.a). You can resolve this by writing the equation in ionic form. For example, the program won't correctly balance a equation such as Na2S2O3 + I2 = Na2S4O6 + NaI while the same equation with only NaI written in ionic form (Na2S2O3 + I2 = Na2S4O6 + Na+ + I-) will be balanced with ease. Carefully review the procedure. The wise man learns from the mistakes of others.

Balance the equation with the ARS method → K2Cr2O7 + (NH4)2Fe(SO4)2*6H2O + H2SO4 = Cr2(SO4)3 + Fe2(SO4)3 + K2SO4 + (NH4)2SO4 + H2O.

Citing this page:

Generalic, Eni. "Balancing redox reactions by the ion-electron method." *EniG. Periodic Table of the Elements*. KTF-Split, 27 Oct. 2022. Web. {Date of access}. <https://www.periodni.com/half-reaction_method.php>.

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