Redox

The oxidation state of an ion refers to the charge that the ion possess. An ion’s oxidation state can be either negative or positive. Many elements can only assume one or two oxidations states. Others, like the transition metal elements, are able to adopt a wide range of oxidation states. The oxidation state of an ion can be changed using two processes:

1. Oxidation: The loss of electrons, leading to an increase in charge and an increase oxidation state.

\text{X}\rarr\text{X}^{+}+\text{e}^{-}

 

2. Reduction: The gain of electrons, leading to a reduction in charge a decrease in oxidation state.

\text{X}+\text{e}^{-}\rarr\text{X}^{-}

 

Elements existing as lone atoms, covalent molecules, and Nobel gasses, all have an oxidation state of 0 (i.e. they have no charge).  Elements that are existing as a monatomic ion (an lone ion outside of a compound) will have an oxidation state equal to their charge. For example a single ion of \text{Fe(III)} will have an oxidation state of +3.  

For  molecules with no charge (and hence and oxidation state of 0) the sum of the oxidation states of the constituent elements equals 0. For molecules that do have an overall charge, the sum oxidation states of the constituent ions should equals the overall charge of the molecules.

It is important to note that in both cases we are considering the oxidation state per atom of an element, e.g \text{Cl} in \text{FeCl}_3 has an oxidation state of -1 not -3.

There are set rules for the oxidation states of particular elements. These rules are essential as they help to determine the oxidation states of elements that have varying oxidation states.

Element	Oxidation State	Exceptions
Group 1 Metals	+1
Group 2 Metals	+2
Aluminum (\text{Al})	+3	-1 in metal hydrides
Hydrogen (\text{H})	+1
Fluorine (\text{F})	-1
Chlorine (\text{Cl}), Bromine (\text{Br}), Iodine (\text{I})	-1	Can change if in a compound with either oxygen or fluorine
Oxygen (\text{O})	-2	-1 in peroxides and in compounds with fluorine.
Transition Metals	Variable

A full redox equation is formed by combining two half equations, one showing oxidation and the other showing reduction. Take for example the redox equation between bromine and magnesium:

Half Equation for Oxidation: \text{Mg}\rarr\text{Mg}^{2+}+2\text{e}^-
Half Equation for Reduction: \text{Br}_2+2\text{e}^-\rarr2\text{Br}^-
Redox Equation: 2\text{Mg}+\text{Br}_2\rarr2\text{MgBr}

In the above reactions, Magnesium (\text{Mg}) is being oxidised while Bromine (\text{Br}) is being reduced. Since \text{Mg} is donating electrons to reduce \text{Br}, we would call \text{Mg} a reducing agent. Similarly, since \text{Br} is accepting electrons to oxidise \text{Mg}, we call it an oxidising agent.

 

An oxidising agent – A species that oxidises another element while it is itself reduced by accepting electrons
A reducing agent – A species that reduces another element while it is itself oxidised by losing electrons

Forming Half Equations
1. Write down the reactants and the products. Make sure to write down the oxidation state or charges (not necessary if it is an element).

2. Balance the number of atoms if needed.

3. Balance the charges on both sides of the equation using electrons

Oxygen ions can oxidise each other to form oxygen molecules in the following reaction:

2\text{O}^{2-}\rarr\text{O}_2

Step 1: Write an initial redox half equation.

\text{O}^{2-}\rarr\text{O}_2

An oxygen ion is oxidised to an oxygen molecule. However the equation is not balanced. There are two oxygens on the right and only one on the left.

Step 2: Balance the elements.

2\text{O}^{2-}\rarr\text{O}_2

We multiply the oxide ions by two. We now have balance in the elements, but there is an imbalance in charge. The overall charge of the left is 4+ but on the right it is o

Step 3: Balance the charge.

2\text{O}^{2-}\rarr\text{O}_2+4\text{e}^-

The total charge on the left side was -4 and, on the right, it was 0 so four electrons were needed on the right side to make the charges equal.