# Oxidation Numbers and Redox Reactions

Submitted by ChemPRIME Staff on Thu, 12/16/2010 - 14:26

Redox reactions may involve protonThe positively charged particle in an atomic nucleus; its mass is similar to the mass of a hydrogen atom. transfers and other bond-breaking and bond-making processes, as well as electronA negatively charged, sub-atomic particle with charge of 1.602 x 10-19 coulombs and mass of9.109 x 1023 kilograms; electrons have both wave and particle properties; electrons occupy most of the volume of an atom but represent only a tiny fraction of an atom's mass. transfers, and therefore the equations involved are much more difficult to deal with than those describing acid-base reactions. In order to be able to recognize redox reactions, we need a method for keeping a careful account of all the electrons. This is done by assigning oxidation numbers to each atom before and after the reaction.

For example, in NO3 the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of –2. This arbitrary assignment corresponds to the nitrogen’s having lost its original five valence electrons to the electronegative oxygens. In NO2, on the other hand, the nitrogen has an oxidation number of + 4 and may be thought of as having one valence electron for itself, that is, one more electron than it had in NO3.

This arbitrarily assigned gain of one electron corresponds to reductionThat part of a chemical reaction in which a reactant gains electrons; simultaneous oxidation of a reactant must occur. of the nitrogen atom on going from NO3 to NO2. As a general rule, reduction corresponds to a lowering of the oxidation number of some atom. Oxidation corresponds to increasing the oxidation number of some atom. Applying the oxidation number rules to the following equation, we have

$\overset{\text{0}}{\mathop{\text{Cu}}}\,\text{ + }\overset{\text{+5 }-\text{2}}{\mathop{\text{2NO}_{\text{3}}^{-}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{4H}_{\text{3}}\text{O}^{\text{+}}}}\,\text{ }\to \text{ }\overset{\text{+1}}{\mathop{\text{Cu}^{\text{2+}}}}\,\text{ + }\overset{\text{+5 }-\text{2}}{\mathop{\text{2NO}_{\text{2}}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{6 H}_{\text{2}}\text{O}}}\,$

Since the oxidation number of copper increased from 0 to +2, we say that copper was oxidized and lost two negatively charged electrons. The oxidation number of nitrogen went down from 5 to 4, and so the nitrogen (or nitrate ion) was reduced. Each nitrogen gained one electron, so 2e were needed for the 2 NO3. The nitrogen was reduced by electrons donated by copper, and so copper was the reducing agentA chemical species that donates electrons in order to reduce another species. In the process the reducing agent is itself oxidized.. Copper was oxidized because its electrons were accepted by an oxidizing agentA chemical species that accepts electrons in order to oxidize another species. In the process the oxidizing agent is itself reduced., nitrogen (or nitrate ion).

Although they are useful and necessary for recognizing redox reactions, oxidation numbers are a highly artificial device. The nitrogen atom in NO3 does not really have a +5 charge which can be reduced to +4 in NO2. Instead, there are covalent bonds and electron-pair sharing between nitrogen and oxygen in both species, and nitrogen has certainly not lost its valence electrons entirely to oxygen. Even though this may (and indeed should) make you suspicious of the validity of oxidation numbers, they are undoubtedly a useful tool for spotting electron-transfer processes. So long as they are used for that purpose only, and not taken to mean that atomsThe smallest particle of an element that can be involved in chemical combination with another element; an atom consists of protons and neutrons in a tiny, very dense nucleus, surrounded by electrons, which occupy most of its volume. in covalent species actually have the large charges oxidation numbers often imply, their use is quite valid.

EXAMPLE 1 Identify the redox reactions and the reducing and oxidizing agents from the following:

a) 2MnO4 + 5SO2 + 6H2O → 5SO42– + 2Mn2+ + 4H3O+

b) NH4+ + PO43– → NH3 + PO42–

c) HClO + H2S → H3O+ + Cl + S

Solution

a) The appropriate oxidation numbers are

$\overset{\text{+7 }-\text{2}}{\mathop{\text{2MnO}_{\text{4}}^{-}}}\,\text{ + }\overset{\text{+4 }-\text{2}}{\mathop{\text{5SO}_{\text{2}}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{6H}_{\text{2}}\text{O}}}\,\text{ }\to \text{ }\overset{\text{+6 }-\text{2}}{\mathop{\text{5SO}_{\text{4}}^{\text{2}-}}}\,\text{ + }\overset{\text{+2}}{\mathop{\text{2Mn}^{\text{2+}}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{4H}_{\text{3}}\text{O}^{\text{+}}}}\,$

The only atoms which change are Mn, from +7 to +2, a reduction, and S, from +4 to +6, an oxidation. The reaction is a redox process. SO2 has been oxidized by MnO4, and so MnO4is the oxidizing agent. MnO4 has been reduced by SO2, and so SO2 is the reducing agent.

b) The oxidation numbers

$\overset{-\text{3 +1}}{\mathop{\text{NH}_{\text{4}}^{\text{+}}}}\,\text{ + }\overset{\text{+5 }-\text{2}}{\mathop{\text{PO}_{\text{4}}^{\text{3}-}}}\,\text{ }\to \text{ }\overset{-\text{3 +1}}{\mathop{\text{NH}_{\text{3}}}}\,\text{ + }\overset{\text{+1 +5 }-\text{2}}{\mathop{\text{HPO}_{\text{4}}^{\text{2}-}}}\,$

show that no redox has occurred. This is an acid-base reaction because a proton, but no electrons, has been transferred.

c)    $\overset{\text{+1 +1 }-\text{2}}{\mathop{\text{HClO}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{H}_{\text{2}}\text{S}}}\,\text{ }\to \text{ }\overset{\text{+1 }-\text{2}}{\mathop{\text{H}_{\text{3}}\text{O}^{\text{+}}}}\,\text{ + }\overset{-\text{1}}{\mathop{\text{Cl}^{-}}}\,\text{ + }\overset{\text{0}}{\mathop{\text{S}}}\,$

H2S has been oxidized, losing two electrons to form elemental S. Since H2S donates electrons, it is the reducing agent. HClO accepts these electrons and is reduced to Cl. Since it accepts electrons, HClO is the oxidizing agent.