The oxidation number (O.N.) of an element represents the total number of electrons which have been removed (a positive oxidation state) or added (a negative oxidation state) to get the element into its present state. The term oxidation describes the loss of electrons by an element and an increase in oxidation state; the term reduction describes the gain of electrons and a decrease in oxidation state. Oxidation numbers for elements in compounds can be calculated using a simple set of rules, which are reproduced below in Table 5.3. Table 5.3 Rules for Assigning Oxidation Numbers 1. The oxidation number of an element in the free state is zero. 2. A monoatomic ion will have an oxidation number that is equal to its charge. 3. In most compounds, the oxidation number of hydrogen is +1 (exception: when hydrogen is bonded to a metal, the oxidation number -1.) 4. Oxygen, within a compound, will generally have an oxidation number of –2 (exception: in peroxides, the oxidation number of oxygen is -1) 5. The oxidation numbers of combined elements from groups I and II are +1 and +2, respectively. 6. The algebraic sum of oxidation numbers of all elements in a neutral compound formula equals zero. 7. The algebraic sum of the oxidation numbers of all elements in a polyatomic ion is equal to the charge of the ion In many chemical reactions, the oxidation number of elements change. Consider the synthesis reaction shown below. In the reactants, carbon and oxygen are both elements and their oxidation numbers are zero (Rule 1). In the product, oxygen will have an oxidation number of –2 (Rule 4), therefore, carbon in CO 2 must have an oxidation number of +4 in order to balance the four negative charges on the oxygens. During this reaction, the oxidation number of carbon has changed from zero in the reactants to +4 in the products and the oxidation number of oxygen has changed from zero to –2. This is an example of a redox reaction ; a chemical reaction in which the oxidation numbers of elements change on going from reactants to products. C (s) + O 2 (g) → CO 2 (g) In a redox reaction, the element that “ loses electrons ” is said to be oxidized and will have an increase in its oxidation number . In the example above, the oxidation number of carbon increases from zero to +4; it has “lost electrons” and has been oxidized . The element that “ gains electrons ” in a redox reaction is said to be reduced and will have a decrease in its oxidation number . In the reaction above, the oxidation number of oxygen has decreased from zero to –2; it has “gained electrons” and has been reduced . An examination of the rules for assigning oxidation numbers reveals that there are many elements for which there are no specific rules, such as nitrogen, sulfur, and chlorine. These elements, as well as some others, can have variable oxidation numbers depending on the other atoms to which they are covalently bonded in a molecular compound. It is useful to analyze a few molecules in order to see the strategy to follow in assigning oxidation numbers to other atoms. Oxidation numbers for the atoms in a binary ionic compound are easy to assign because they are equal to the charge of the ion (rule 2). In CuCl 2 , the oxidation number of copper is +2, while the oxidation number of chlorine is −1. This is because CuCl 2 is an ionic compound that is actually composed of these ions. Assigning oxidation numbers for molecular compounds is trickier. The key is to remember rule 6: that the sum of all the oxidation numbers for any neutral species must be zero. Make sure to account for any subscripts which appear in the formula. As an example, consider the compound nitric acid, HNO 3 . According to rule 3, the oxidation number of hydrogen is +1. According to rule 4, the oxidation number of oxygen is −2. There is no rule regarding nitrogen, but its oxidation number can be calculated as follows. 1(+1)+x+3(−2)=0, where x is the oxidation number of nitrogen. Solving for x, we obtain x=+5 The oxidation number of the nitrogen atom in HNO 3 is +5 Examples \(\PageIndex{1}\) For each of the reactions given below, calculate the oxidation number of each of the elements in the reactants and the products and determine if the reaction involves oxidation-reduction. If it is a redox reaction, identify the elements that have been oxidized and reduced. Cu 2 S → 2 Cu + S Reactants: Cu O.N.= +1 S O.N. = -2 Products: Cu O.N.= 0 S O.N.= 0 Element oxidized: sulfur Element Reduced: copper. This is a redox reaction. CaCO 3 → CaO + CO 2 Reactants: Ca O.N.= +2 C O.N.= +4 O O.N.= -2 Products: Ca O.N.= +2 C O.N.= +4 O O.N.= -2 Element oxidized: none. Element Reduced: none. This is non-redox reaction. Fe 2 O 3 + 3 H 2 → 2 Fe + 3 H 2 O Reactants: Fe O.N.=+3 O O.N.=-2 H O.N.=0 Products: Fe O.N.=0 O O.N.=-2 H O.N.=+1 Element oxidized: hydrogen. Element Reduced: iron. This is a redox reaction AgNO 3 + NaCl → AgCl (s) + NaNO 3 Reactants: Ag O.N.=+1 (cation) N O.N.=+5 (rule 6) O O.N.=-2 Na O.N.=+1 Cl O.N.=-1 Ag O.N.=+1 (cation) N O.N.=+5 (rule 6) O O.N.=-2 Na O.N.=+1 Cl O.N.=-1 Element oxidized: none. Element Reduced: none. This is a non-redox reaction Contributors Paul R. Young , Professor of Chemistry, University of Illinois at Chicago, Wiki: AskTheNerd; PRY askthenerd.com - pyoung uic.edu; ChemistryOnline.com Oxidation numbers . Contributed by Allison Soult , Senior Lecturer (Chemistry) at University of Kentucky