# How to Balance Redox Equations

By Jack Brubaker; Updated April 25, 2017

Oxidation-reduction, or “redox,” reactions represent one of the major reaction classifications in chemistry. The reactions necessarily involve the transfer of electrons from one species to another. Chemists refer to the loss of electrons as oxidation and to the gain of electrons as reduction. The balancing of a chemical equation refers to the process of adjusting the numbers of each reactant and product so that the compounds on the left and right sides of the reaction arrow — the reactants and products, respectively — contain the same number of each type of atom. This process represents a consequence of the first law of thermodynamics, which states that matter can be neither created nor destroyed. Redox reactions take this process one step further by also balancing the number of electrons on each side of the arrow because, like atoms, electrons possess mass and are therefore governed by the first law of thermodynamics.

Write the unbalanced chemical equation on a piece of paper and identify the species being oxidized and reduced by examining the charges on the atoms. For example, consider the unbalanced reaction of permanganate ion, MnO4(-), where (-) represents a charge on the ion of negative one, and oxalate ion, C2O4(2-) in the presence of an acid, H(+): MnO4(-) + C2O4(2-) + H(+) → Mn(2+) + CO2 + H2O. Oxygen almost always assumes a charge of negative two in compounds. Thus, MnO4(-), if each oxygen maintains a negative two charge and the overall charge is negative one, then the manganese must exhibit a charge of positive seven. The carbon in C2O4(2-) similarly exhibits a charge of positive three. On the product side, the manganese possesses a charge of positive two and the carbon is positive four. Thus, in this reaction, the manganese is reduced because its charge decreases and the carbon is oxidized because its charge increases.

Write separate reactions — called half-reactions — for the oxidation and reduction processes and include the electrons. The Mn(+7) in MnO4(-) becomes Mn(+2) by taking on five additional electrons (7 - 2 = 5). Any oxygen in the MnO4(-), however, must become water, H2O, as a byproduct, and the water cannot form with hydrogen atoms, H(+). Therefore, protons, H(+) must be added to the left side of the equation. The balanced half-reaction now becomes MnO4(-) + 8 H(+) + 5 e → Mn(2+) + 4 H2O, where e represents an electron. The oxidation half-reaction similarly becomes C2O4(2-) - 2e → 2 CO2.

Balance the overall reaction by ensuring that the number of electrons in the oxidation and reduction half-reactions are equal. Continuing the previous example, the oxidation of the oxalate ion, C2O4(2-), only involves two electrons, whereas the reduction of manganese involves five. Consequently, the entire manganese half reaction must be multiplied by two and the entire oxalate reaction must be multiplied by five. This will bring the number of electrons in each half reaction to 10. The two half reactions now become 2 MnO4(-) + 16 H(+) + 10 e → 2 Mn(2+) + 8 H2O, and 5 C2O4(2-) - 10 e → 10 CO2.

Obtain the balanced overall equation by summing the two balanced half reactions. Note the manganese reaction includes the gain of 10 electrons, whereas the oxalate reaction involves the loss of 10 electrons. The electrons therefore cancel. In practical terms, this means that five oxalate ions transfer a total of 10 electrons to two permanganate ions. When summed, the overall balanced equation becomes 2 MnO4(-) + 16 H(+) + 5 C2O4(2-) → 2 Mn(2+) + 8 H2O + 10 CO2, which represents a balanced redox equation.