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 oxidation number change method

In the oxidation number change method the underlying principle is that the gain in the oxidation number (number of electrons) in one reactant must be equal to the loss in the oxidation number of the other reactant.

* 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.

[Cr(N_{2}H_{4}CO)_{6}]_{4}[Cr(CN)_{6}]_{3} + KMnO_{4} + H_{2}SO_{4}
→
K_{2}Cr_{2}O_{7} + MnSO_{4} + CO_{2} + KNO_{3} + K_{2}SO_{4} + H_{2}O

* Step 2. Separate the process 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+3(N-3_{2}H+1_{4}C+4O-2)_{6}]_{4}[Cr+2(C+2N-3)_{6}]_{3} + K+1Mn+7O-2_{4} + H+1_{2}S+6O-2_{4}
→
K+1_{2}Cr+6_{2}O-2_{7} + Mn+2S+6O-2_{4} + C+4O-2_{2} + K+1N+5O-2_{3} + K+1_{2}S+6O-2_{4} + H+1_{2}O-2

When working with organic compounds and formulas with multiples of the same atom (such as [Cr(N_{2}H_{4}CO)_{6}]_{4}[Cr(CN)_{6}]_{3}) it is easier to use their molecular formulas instead of condensed structural formulas, so we will temporarily replace all such formulas with their molecular formulas.

C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24} + K+1Mn+7O-2_{4} + H+1_{2}S+6O-2_{4}
→
K+1_{2}Cr+6_{2}O-2_{7} + Mn+2S+6O-2_{4} + C+4O-2_{2} + K+1N+5O-2_{3} + K+1_{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). Write down the transfer of electrons. Carefully, insert coefficients, if necessary, to make the numbers of oxidized and reduced atoms equal on the two sides of each redox couples.

2C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24}
→
7K+1_{2}Cr+6_{2}O-2_{7} + 48e^{-}

(Cr)

C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24}
→
42C+4O-2_{2} + 36e^{-}

(C)

C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24}
→
66K+1N+5O-2_{3} + 528e^{-}

(N)

K+1Mn+7O-2_{4} + 5e^{-}
→
Mn+2S+6O-2_{4}

(Mn)

* 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, and depending on which reaction is dominant (exchanges more electrons) they are combined with either partial reactions of oxidation or reduction. Before combining the two reactions it is neccesary to ensure that the same coeficients are in front of both such molecules.

2C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24}
→
7K+1_{2}Cr+6_{2}O-2_{7} + 84C+4O-2_{2} + 132K+1N+5O-2_{3} + 1176e^{-}

K+1Mn+7O-2_{4} + 5e^{-}
→
Mn+2S+6O-2_{4}

* 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 to balance the number of atoms. Never change any formulas.

* 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.

2C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24} + 146KMnO_{4} + 146H_{2}SO_{4} → 7K+1_{2}Cr+6_{2}O-2_{7} + 84C+4O-2_{2} + 132K+1N+5O-2_{3} + 146MnSO_{4} + 1176e^{-}

2K+1Mn+7O-2_{4} + 3H_{2}SO_{4} + 10e^{-} → 2Mn+2S+6O-2_{4} + K_{2}SO_{4}

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

2C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24} + 146KMnO_{4} + 146H_{2}SO_{4} → 7K+1_{2}Cr+6_{2}O-2_{7} + 84C+4O-2_{2} + 132K+1N+5O-2_{3} + 146MnSO_{4} + 1176e^{-} + 1176H^{+}

2K+1Mn+7O-2_{4} + 3H_{2}SO_{4} + 10e^{-} + 10H^{+} → 2Mn+2S+6O-2_{4} + K_{2}SO_{4}

* c) 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.

2C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24} + 146KMnO_{4} + 146H_{2}SO_{4} → 7K+1_{2}Cr+6_{2}O-2_{7} + 84C+4O-2_{2} + 132K+1N+5O-2_{3} + 146MnSO_{4} + 1176e^{-} + 1176H^{+} + 19H_{2}O

2K+1Mn+7O-2_{4} + 3H_{2}SO_{4} + 10e^{-} + 10H^{+} → 2Mn+2S+6O-2_{4} + K_{2}SO_{4} + 8H_{2}O

Balanced half-reactions are well tabulated in handbooks and on the web in a 'Tables of standard electrode potentials'. These tables, by convention, contain the half-cell potentials for reduction. To make the oxidation reaction, simply reverse the reduction reaction and change the sign on the E_{1/2} value.

* Step 4. 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.

2C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24} + 146KMnO_{4} + 146H_{2}SO_{4} → 7K+1_{2}Cr+6_{2}O-2_{7} + 84C+4O-2_{2} + 132K+1N+5O-2_{3} + 146MnSO_{4} + 1176e^{-} + 1176H^{+} + 19H_{2}O

| *5

2K+1Mn+7O-2_{4} + 3H_{2}SO_{4} + 10e^{-} + 10H^{+} → 2Mn+2S+6O-2_{4} + K_{2}SO_{4} + 8H_{2}O

| *588

10C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24} + 730KMnO_{4} + 730H_{2}SO_{4} → 35K+1_{2}Cr+6_{2}O-2_{7} + 420C+4O-2_{2} + 660K+1N+5O-2_{3} + 730MnSO_{4} + 5880e^{-} + 5880H^{+} + 95H_{2}O

1176K+1Mn+7O-2_{4} + 1764H_{2}SO_{4} + 5880e^{-} + 5880H^{+} → 1176Mn+2S+6O-2_{4} + 588K_{2}SO_{4} + 4704H_{2}O

* Step 5. 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.

10C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24} + 1906K+1Mn+7O-2_{4} + 2494H_{2}SO_{4} + 5880e^{-} + 5880H^{+} → 35K+1_{2}Cr+6_{2}O-2_{7} + 1906Mn+2S+6O-2_{4} + 420C+4O-2_{2} + 588K_{2}SO_{4} + 660K+1N+5O-2_{3} + 4799H_{2}O + 5880e^{-} + 5880H^{+}

* Step 6. 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.

10C+3.14_{42}H+1_{96}Cr+2.57_{7}N-3_{66}O-2_{24} + 1906K+1Mn+7O-2_{4} + 2494H_{2}SO_{4} → 35K+1_{2}Cr+6_{2}O-2_{7} + 1906Mn+2S+6O-2_{4} + 420C+4O-2_{2} + 588K_{2}SO_{4} + 660K+1N+5O-2_{3} + 4799H_{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 |
---|---|---|---|

C | 10*42 | 420*1 | 0 |

H | 10*96 + 2494*2 | 4799*2 | -3650 |

Cr | 10*7 | 35*2 | 0 |

N | 10*66 | 660*1 | 0 |

O | 10*24 + 1906*4 + 2494*4 | 35*7 + 1906*4 + 420*2 + 588*4 + 660*3 + 4799*1 | 0 |

K | 1906*1 | 35*2 + 588*2 + 660*1 | 0 |

Mn | 1906*1 | 1906*1 | 0 |

S | 2494*1 | 1906*1 + 588*1 | 0 |

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 → [Cr(N2H4CO)6]4[Cr(CN)6]3 + KMnO4 + H2SO4 = K2Cr2O7 + MnSO4 + CO2 + KNO3 + K2SO4 + H2O.

Citing this page:

Generalic, Eni. "Balancing redox reactions by oxidation number change method." *EniG. Periodic Table of the Elements*. KTF-Split, 18 Jan. 2024. Web. {Date of access}. <https://www.periodni.com/balancing_redox_equations.php>.

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