Oxidation Numbers (VCE SSCE Chemistry): Revision Notes

Oxidation Numbers

Introduction

Oxidation numbers (also called oxidation states) are a useful tool in chemistry for identifying which substances act as reducing agents and which act as oxidising agents in redox reactions. They help us extend the concept of redox reactions beyond simple ionic compounds to include reactions where electron transfer is less obvious.

It's important to understand that oxidation numbers have no physical meaning. They don't necessarily indicate a formal charge or describe the physical or chemical properties of a substance. Instead, they serve as a systematic way to track electron movement in chemical reactions.

Oxidation numbers vs ionic charges

As an example, consider the magnesium ion $\text{Mg}^{2+}$. Its oxidation number is $+2$ while its charge is written as $2+$. Similarly, the oxide ion $\text{O}^{2-}$ has an oxidation number of $-2$ and a charge of $2-$.

While the numerical values may be the same in these cases, oxidation numbers don't always represent actual ionic species that carry a charge.

Determining oxidation numbers

To assign oxidation numbers to elements in a reaction, chemists follow a specific set of rules. When applying these rules, we treat all compounds and polyatomic ions as if they were composed of individual ions.

Oxidation number rules

There are six main rules for determining oxidation numbers:

Rule 1: Free elements have an oxidation number of zero

Any element in its uncombined state has an oxidation number of $0$. This applies to monatomic elements like sodium ($\text{Na}$) and chlorine ($\text{Cl}$), as well as diatomic molecules like $\text{Cl}_2$ and polyatomic molecules like $\text{P}_4$.

Rule 2: Simple ions have oxidation numbers equal to their charge

For a simple ion (a single atom with a charge), the oxidation number equals the charge on that ion. For example:

• $\text{Na}^+$ has an oxidation number of $+1$
• $\text{Cl}^-$ has an oxidation number of $-1$
• $\text{Mg}^{2+}$ has an oxidation number of $+2$
• $\text{O}^{2-}$ has an oxidation number of $-2$
• $\text{Al}^{3+}$ has an oxidation number of $+3$
• $\text{N}^{3-}$ has an oxidation number of $-3$

Rule 3: Certain elements have fixed oxidation numbers in compounds

Some elements consistently have the same oxidation number in their compounds, with only a few exceptions:

(a) Main group metals have oxidation numbers equal to the charge on their ions. For example, in ionic compounds like $\text{KCl}$ and $\text{MgSO}_4$, potassium has an oxidation number of $+1$ and magnesium has $+2$.

(b) Hydrogen usually has an oxidation number of $+1$ when combined with non-metals. For instance, in $\text{H}_2\text{O}$, hydrogen has an oxidation number of $+1$.

(c) Oxygen typically has an oxidation number of $-2$ in most compounds.

Rule 4: Oxidation numbers in neutral compounds sum to zero

In any neutral compound, the sum of all oxidation numbers must equal zero. For example, in $\text{CO}_2$, carbon has an oxidation number of $+4$ and each oxygen has $-2$. The calculation is: $(+4) + 2(-2) = 0$.

Rule 5: Oxidation numbers in polyatomic ions sum to the ion's charge

For a polyatomic ion, the sum of oxidation numbers equals the overall charge on the ion. For instance, in the sulfate ion $\text{SO}_4^{2-}$, the oxidation numbers must sum to $-2$. In the ammonium ion $\text{NH}_4^+$, they sum to $+1$.

Rule 6: The most electronegative element gets the negative oxidation number

When two non-metals combine, the more electronegative element is assigned the negative oxidation number. For example, in $\text{OF}_2$, fluorine (the most electronegative element) has an oxidation number of $-1$, while oxygen has $+2$.

Common oxidation states

The periodic table below shows the most common oxidation states for the first 36 elements in their compounds. The asterisk (*) indicates elements that can have a range of oxidation states, particularly transition metals and some non-metals.

Calculating oxidation numbers using algebra

When a compound contains several elements, you can use algebra combined with the oxidation number rules to calculate unknown oxidation numbers.

Colourful oxidation states

A characteristic property of transition elements is that they form brightly coloured compounds. Different oxidation states of the same transition metal can result in different colours.

Using oxidation numbers to name chemicals

Since transition elements can form ions with several different charges, many have variable oxidation numbers. This can create confusion when naming compounds.

For example, iron forms two different chloride compounds:

• $\text{FeCl}_2$
• $\text{FeCl}_3$

Using the oxidation number rules, we know chloride has an oxidation number of $-1$. Therefore:

• In $\text{FeCl}_2$, iron has an oxidation number of $+2$
• In $\text{FeCl}_3$, iron has an oxidation number of $+3$

Using oxidation numbers to identify oxidation and reduction

The concept of oxidation numbers allows us to extend our definition of oxidation and reduction beyond simple electron transfer.

In this expanded definition:

• Oxidation involves an increase in oxidation number
• Reduction involves a decrease in oxidation number

This alternative definition is particularly useful for non-ionic compounds where it's difficult to determine whether electrons have actually been transferred.

Identifying redox reactions

Let's examine the combustion of carbon in excess oxygen:

$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)}$

At first glance, this may not appear to be a redox reaction because none of the species are ionic compounds. However, by assigning oxidation numbers, we can identify both oxidation and reduction processes.

Step-by-step method for identifying oxidation and reduction

The systematic approach involves:

1. Determining the oxidation number of each element on both sides of the equation
2. Identifying which elements have changed oxidation number
3. Determining whether each change represents an increase (oxidation) or decrease (reduction)

Using oxidation numbers to identify conjugate redox pairs

When we write a half-equation for an oxidation reaction, the reactant acts as a reducing agent and loses electrons. The product formed is an oxidising agent. The reactant and the product it forms together make up a conjugate redox pair.

Oxidation half-equations

Consider the oxidation of zinc:

$\text{Zn(s)} \rightarrow \text{Zn}^{2+}\text{(aq)} + 2e^-$

In this half-equation:

• Zinc metal ($\text{Zn}$) is the reducing agent
• It forms $\text{Zn}^{2+}\text{(aq)}$, which is an oxidising agent
• Together, $\text{Zn(s)}$ and $\text{Zn}^{2+}\text{(aq)}$ form a conjugate redox pair, written as $\text{Zn(s)}/\text{Zn}^{2+}\text{(aq)}$

Looking at oxidation numbers in this pair, zinc's oxidation number increases from $0$ to $+2$, confirming this is an oxidation half-reaction.

Reduction half-equations

In a reduction half-equation, the reactant is an oxidising agent that gains electrons, forming a reducing agent as the product. This creates another conjugate redox pair.

For example, consider the reduction of silver ions:

$\text{Ag}^+\text{(aq)} + e^- \rightarrow \text{Ag(s)}$

In this half-equation:

• $\text{Ag}^+\text{(aq)}$ is the oxidising agent
• It forms $\text{Ag(s)}$, which is a reducing agent
• $\text{Ag}^+\text{(aq)}$ and $\text{Ag(s)}$ form a conjugate redox pair

The oxidation number of silver decreases from $+1$ to $0$, confirming this is a reduction half-reaction.

Understanding conjugate pairs in complete reactions

For every redox reaction, there are two conjugate redox pairs - one associated with oxidation and one with reduction.

Remember!