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) - Nuovo su periodni.com [1]

by ARS method

The aggregate redox species method, or the ARS method in short, is an improved oxidation number change method that successfully solves even reactions that cannot be 'cleanly' separated into partial reactions of oxidation and reduction.

* Step 1. Write the unbalanced equation* ('skeleton equation') containing all of the reactants and products of the chemical reaction. Redox equations that need to be balanced can often be written without water molecules, H

K_{4}[Fe(SCN)_{6}] + K_{2}Cr_{2}O_{7} + H_{2}SO_{4}
→
Fe_{2}(SO_{4})_{3} + Cr_{2}(SO_{4})_{3} + CO_{2} + H_{2}O + K_{2}SO_{4} + KNO_{3}

* Step 2. Identify the redox couples in the reaction*.

* a) Assign the oxidation numbers* for each atom in the equation (see: Rules for assigning oxidation numbers). The use of the oxidation numbers greatly simplifies identifying which element in a reaction is oxidized and which element is reduced.

K+1_{4}[Fe+2(S-2C+4N-3)_{6}] + K+1_{2}Cr+6_{2}O-2_{7} + H+1_{2}S+6O-2_{4}
→
Fe+3_{2}(S+6O-2_{4})_{3} + Cr+3_{2}(S+6O-2_{4})_{3} + C+4O-2_{2} + H+1_{2}O-2 + K+1_{2}S+6O-2_{4} + K+1N+5O-2_{3}

* b) Identify and write out all redox couples* of atoms that have been oxidized (their oxidation number has increased) or reduced (their oxidation number has decreased). Determine the transfer of electrons for each redox couple, but make sure that the number of atoms that have been oxidized (or reduced) is equal on both sides of the equation. If necessary, write down the stoichiometric coefficients in front of the species.

2K+1_{4}[Fe+2(S-2C+4N-3)_{6}]
→
Fe+3_{2}(S+6O-2_{4})_{3} + 2e^{-}

(Fe)

K+1_{4}[Fe+2(S-2C+4N-3)_{6}]
→
2Fe+3_{2}(S+6O-2_{4})_{3} + 48e^{-}

(S)

K+1_{4}[Fe+2(S-2C+4N-3)_{6}]
→
2Cr+3_{2}(S+6O-2_{4})_{3} + 48e^{-}

(S)

K+1_{4}[Fe+2(S-2C+4N-3)_{6}]
→
6K+1_{2}S+6O-2_{4} + 48e^{-}

(S)

K+1_{4}[Fe+2(S-2C+4N-3)_{6}]
→
6K+1N+5O-2_{3} + 48e^{-}

(N)

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

(Cr)

* c) Remove unnecessary redox couples.* Some atoms, even though they have different oxidation numbers on the left and right side of the equation, do not have to be redox atoms. It's necessary to know the chemical reaction described by the equation in order to know which redox couples should be retained and which rejected (see: Dividere la reazione redox in due semireazioni). The program has automatically marked redox couples that need to be removed (grayed out equations). Rules for deciding which redox pairs should be kept and which removed can be found on the Dividere la reazione redox in due semireazioni page. You can change the selection by clicking on the check box before the desired redox couples.

* Step 3. Aggregate the redox species* into one equation (ARS equation). Species with redox atoms are combined in such a way that all redox atoms are balanced in the ARS equation.

* a) Balance redox atoms that aren't part of a redox couple.* When more than one atom of an element that changes oxidation number is present in a formula, we must calculate the number of electrons transferred per formula unit. Because of this it is necessary to combine all redox pairs with species containing several different redox atoms into one equation.

2K+1_{4}[Fe+2(S-2C+4N-3)_{6}]
→
Fe+3_{2}(S+6O-2_{4})_{3} + 12K+1_{2}S+6O-2_{4} + 12K+1N+5O-2_{3} + 194e^{-}

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

* b) Balance non-redox atoms that are present only in redox species.* When the non-redox atom is found only in redox species it must be balanced before the electron transfer is equalized. Since the redox atoms are already balanced, non-redox atoms can only be balanced by multiplying the entire equation by some factor and then summing it with another oxidation or reduction equation.

This step can be skipped. There is no non-redox atoms present only in redox species.

* c) Combine the remaining reactions into two partial reactions.* The remaining equations are combined into two partial equations: one for oxidation and one for reduction. Take special care to ensure that the balance of non-redox atoms is not disrupted.

This step can be skipped. We already have two partial equations.

* d) Equalize the electron transfer in the two partial reactions.* Now that all redox atoms are balanced we can make the number of electrons lost in the oxidation equal to the number of electrons gained in the reduction.

2K+1_{4}[Fe+2(S-2C+4N-3)_{6}]
→
Fe+3_{2}(S+6O-2_{4})_{3} + 12K+1_{2}S+6O-2_{4} + 12K+1N+5O-2_{3} + 194e^{-}

| *3

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

| *97

6K+1_{4}[Fe+2(S-2C+4N-3)_{6}]
→
3Fe+3_{2}(S+6O-2_{4})_{3} + 36K+1_{2}S+6O-2_{4} + 36K+1N+5O-2_{3} + 582e^{-}

97K+1_{2}Cr+6_{2}O-2_{7} + 582e^{-}
→
97Cr+3_{2}(S+6O-2_{4})_{3}

* e) Sum up the partial equations and determine the 'free' species.* Both partial equations are summed into the ARS equation. The electrons are cancelled. In the ARS equation, all redox and non-redox atoms that appear only in redox species are balanced. All stoichiometric coefficients of species in the ARS equation are 'frozen', except those containing redox atoms with the same oxidation number on both sides of the equation (marked red). Its coefficients can be changed independently of other species in the ARS equation.

6K+1_{4}[Fe+2(S-2C+4N-3)_{6}] + 97K+1_{2}Cr+6_{2}O-2_{7}
→
3Fe+3_{2}(S+6O-2_{4})_{3} + 36K+1_{2}S+6O-2_{4} + 36K+1N+5O-2_{3} + 97Cr+3_{2}(S+6O-2_{4})_{3}

* Step 4. Balance the charges and other atoms.* The ARS equation is viewed as one whole and all mathematical operations that change one species must be applied to all other species in the ARS equation. If you wish, the ARS equation can be divided with the largest common divisor in order to minimize the coefficients.

* a) Add non-redox species and balance all atoms besides hydrogen and oxygen.* In the 'frozen' ARS equation add species that don't contain redox atoms (marked blue). Balance all other atoms besides hydrogen and oxygen. The resulting equation will often be simple enough that it can be easily balanced by inspection. In this step the program uses the Gauss method of elimination.

6K_{4}[Fe(SCN)_{6}] + 97K_{2}Cr_{2}O_{7} + 355H_{2}SO_{4}
→
3Fe_{2}(SO_{4})_{3} + 91K_{2}SO_{4} + 36KNO_{3} + 97Cr_{2}(SO_{4})_{3} + 36CO_{2}

* b) Balance the charges.* For reactions in an acidic solution, balance the charge so that both sides have the same total charge by adding an H

* c) Balance the oxygen atoms.* Check if the number of oxygen atoms on the left side of the equation is equal to their number on the right side. If they aren't equal, equilibrate the atoms by adding water molecules to the side lacking oxygen atoms.

6K_{4}[Fe(SCN)_{6}] + 97K_{2}Cr_{2}O_{7} + 355H_{2}SO_{4}
→
3Fe_{2}(SO_{4})_{3} + 91K_{2}SO_{4} + 36KNO_{3} + 97Cr_{2}(SO_{4})_{3} + 36CO_{2} + 355H_{2}O

* Step 5. Simplify the equation.* The same species on opposite sides of the arrow can be canceled. If necessary, the whole equation can be divided with the largest common divisor in order to make the coefficients as small as possible.

* Step 6. Check the balance of charges and elements.* Like any chemical reaction, a redox reaction must be balanced by mass and charge. Check if the sum of each type of atom on one side of the equation is equal to the sum of the same atoms on the other side. Check if the sum of electrical charges on the left side of the equation is equal to those on the right side. It doesn't matter what the sum of the charges is as long as it's equal on both sides.

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

K | 6*4 + 97*2 | 91*2 + 36*1 | 0 |

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

S | 6*6 + 355*1 | 3*3 + 91*1 + 97*3 | 0 |

C | 6*6 | 36*1 | 0 |

N | 6*6 | 36*1 | 0 |

Cr | 97*2 | 97*2 | 0 |

O | 97*7 + 355*4 | 3*12 + 91*4 + 36*3 + 97*12 + 36*2 + 355*1 | 0 |

H | 355*2 | 355*2 | 0 |

Charge | 0 | 0 | 0 |

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.

6K_{4}[Fe(SCN)_{6}] + 97K_{2}Cr_{2}O_{7} + 355H_{2}SO_{4} → 3Fe_{2}(SO_{4})_{3} + 97Cr_{2}(SO_{4})_{3} + 36CO_{2} + 355H_{2}O + 91K_{2}SO_{4} + 36KNO_{3}

- E. Generalić, N. Vladislavić, Aggregate Redox Species Method - An Improved Oxidation Number Change Method for Balancing Redox Equations,
*Chemistry Journal*, Vol.**4**, No. 3, 43-49 (2018)

Citazione della questa pagina:

Generalic, Eni. "Balancing redox reactions by aggregate redox species method." *EniG. Tavola periodica degli elementi*. KTF-Split, 18 Jan. 2024. Web. {Data di accesso}. <https://www.periodni.com/it/ars_metodo.php>.

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