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Redox Revision


Redox is a term used to describe reactions in which both oxidation and reduction are taking place simultaneously. The term redox comes from the combination of the two processes involved. 

Oxidation States

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.



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



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   
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Redox Equations

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:


Step 1: Write an initial redox half equation.


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.


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.


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.

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Example 1: Forming more Complex Redox Reactions

Sometimes, the oxygen (\text{O}) in substances needs to be balanced. We do this by adding \text{H}^{+} ions (\text{OH}^{-} in alkaline conditions) and \text{H}_2\text{O}.

  1. Write down the reactants and the products. Make sure to write down the oxidation states or charges (not necessary if it is an element).
  2. Balance the number of atoms if needed. For molecules with excess oxygen, add an equal amount of water molecules to balance out the oxygen molecules.
  3. Balance excess \text{H}^{+} ions in the equation by adding equal amount of \text{H}^{+} ions need to the opposite side.
  4. Finally, add electrons to balance the total charges on each side of the equation.

[4 marks]

Step 1: A nitrate ion is reduced to nitrogen dioxide.


Step 2: There are 3 oxygen atoms on the left side and 2 on the right side so 1 water molecule is needed on the right side for the oxygen molecules to be balanced.


Step 3: To balance the hydrogens on the right side, 2\text{ H}^{+} ions are added to the left side.


Step 4: The total charge on the left side was +1 and the total charge on the right was 0. To balance this, an electron is added to the left side.


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Example 2: Forming Full Redox Equations

Using two of the half equations we created above, one oxidation and one reduction, we can form a full redox equation. To form a full redox equation, the number of electrons in each of the half equations must be equal.

[4 marks]

Equation 1: \text{NO}_3^-+2\text{H}^++\text{e}^-\rarr\text{NO}_2+\text{H}_2\text{O}
Equation 2: 2\text{O}^{2-}\rarr\text{O}_2+4\text{e}^-

We can multiply equation 1 by 4 to balance the electrons in the equation with the electrons in equation 2.

4\text{NO}_3^-+2\text{H}^++\text{e}^-\rarr4\text{NO}_2+4\text{H}_2\text{O    }\times4


Now that the electrons are equal, they can be cancelled out and the equations can be combined

Redox Equation: 2\text{O}^{2-}+4\text{NO}_3^-+8\text{H}^+\rarr4\text{NO}_2+4\text{H}_2\text{O}+\text{O}

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Redox Example Questions

The loss of electrons

\text{Charge of HNO}_3=0





2\text{Cl}^-\rarr\text{Cl}_2+ 2\text{e}^- 


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