3.2 d: Oxidation Numbers

Oxidation Numbers

Year 11

​Syllabus dot points

• apply the definitions of oxidation and reduction in terms of electron transfer and oxidation numbers to a range of reduction and oxidation (redox) reactions

​What we will cover

• what are oxidation numbers

• how to assign them

• ​how this indicates redox reactions

• practice problems ​

​Recap

​What is an oxidation number?

​AKA oxidation state

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The charge that atom would have if the compound was composed of ions/completely ionic

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But what does that mean?

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​Helps us to identify whether a redox reaction is occurring - loss or gain of electrons

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Somewhat arbitrary - but there is a set of rules ​

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The Rules

1. Oxidation state (OS) of an atom in a free element is 0

2. OS of monatomic ion = it's charge

3. Sum of all OS in all atoms is equal to the charge of the molecule (either neutral or an ion)

4. In compounds, Group 1 metals = +1, and Group 2 metals = +2​

5. ​In compounds, nonmetals are assigned OS in a hierarchy:

   1. Fluorine = -1

   2. Hydrogen = +1

   3. Oxygen = -2

   4. ​Group 7 = -1

   5. Group 6 = -2

   6. Group 5 = -3

1. ​Cu or Cl₂

2. ​Ca²⁺ or Cl^-

3. H₂O = 0

4. NaCl = Na has OS of 1

5. CO₂ = O has an OS of -2 ​

​Example:

​Assign an oxidation state to each atom:

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Br₂

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= ​0

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Because of Rule 1 - free element, so the OS of both Br atoms is 0 ​

​Example:

K⁺

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= +1

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Because of Rule 2, K⁺ is a monatomic ion and there its OS equals its charge ​

​Example:

LiF

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= 0

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Because, the OS of Li is +1 (Rule 4, ​Group 1 metals are +1) and OS of F is -1 (group 7 metals are -1). Then, the overall charge on the atom is 0, so therefore OS must = 0

​Example:

SO₄^-2

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= -2 ​

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Because, Rule ​3 states that the overall charge equals the OS.

Rule 5 states that O = -2, and that S is expected to = -2. ​

However, if that were the case, then Rule 3 could not apply. So S must equal +6. ​

​Example:

Na₂O₂

​

= 0

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Rule 3​ states OS is equal to the overall charge.

Rule ​4 states, Na = +1

Rule 5 states, O = -2 . However, this won't work out. Therefore, O = -1

Multiple Choice

What is the oxidation state of S in SO₃ ?

• -3

• -2

• +4

• +6

Multiple Choice

What is the oxidation state of N in NO₂^- ?

• +2

• +3

• -1

• -4

Multiple Choice

What is the oxidation number of O in Li₂O ?

• -1

• -2

• +1

• +2

​How does this help us?

Can tell us what is being oxidised and what is being reduced.

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For example:

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C + 2S --> CS₂

We cannot immediately identify where oxidation and reduction is occurring ​

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But we can assign oxidation states to all atoms within the equation.

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C = 0

S = 0

-->

C = +4

S = -2 ​

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Carbon has lost electrons, therefore has been oxidised. Sulfur has gained electrons, and has been reduced. ​

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Open Ended

Determine which species has been reduced and which has been oxidised in the following reaction:

Ca (s) + 2H₂O (l) --> Ca(OH)₂ (aq) + H₂ (g)

​What's next?

Galvanic cells ​