7- Oxidation, reduction and redox reactions

What is meant by the oxidation number?

Charge that an atom would have if the compound consisted of separate ions, the number of electrons the atom needs to lose or gain to become neutral

Rules when calculating the oxidation number

1. An element on its own has an oxidation state of 0 (eg. Na, Cl2)

2. For simple ions (eg. K+, Na2+) the oxidation number is the charge on the ion (eg K+ =+1 , Na2+ =2+)

3.In a compound, the sum of the oxidation state must always equal 0

4. Halides > -1

5. Oxygen in peroxide (H2O2) > -1

6. H in hydride (hydrogen + metal) -1

7. Sum of oxidation state = overall charge of complex ion

What is meant by oxidation and reduction?

Oxidation:

-Gain of oxygen

-Loss of electrons

-Loss of hydrogen

Reduction:

-Loss of oxygen

-Gain of electrons

-Gain of hydrogen

What is meant by a redox reaction?

Oxidation and reduction occur simultaneously

One chemical gains electrons

One chemical looses electrons

What assumption can you make of the oxidation number of an atom?

Usually, their oxidation state is the group they are in - similar to calculating the charge of ions.

NOTE: There are some exceptions to this

What is the oxidation state of Fe in Fe2O3

O = -2

3 x -2 = -6

Fe2 must be +6 to allow a neutral charge

6/2 = +3 For Fe alone

What is the oxidation state of nitrogen in NH4+

H = +1

1x4 = 4 (cannot be 0 as there is a +1 charge overall)

-3

Why do we use half equations?

To show more clearly what is happening in a reaction, specifically the transfer of electrons.

Shows what substances are reduced or oxidised

Cu2+ +2e- ---> Cu

Oxidation or reduction, why?

Reduction as copper is gaining electrons

2Cl- -> Cl2 + 2e-

Oxidation or reduction, why?

Oxidation as Cl has lost electrons

Half equation for I- making I2

2 atoms on the right, only one of the left so we have to balance

2I --> I2 + 2e-

Steps for writing half equations

1) Write what you know as equation

2)Balance atoms (not for O or H)

3)Balance for O by adding H2O

4) Balance for H by adding H+ ions

5) Balance charge by adding e-

Balance half equation making NO2 from NO3-

1) Write what you know as equation

NO3- ---> NO2

2) Balance for atoms that aren't O or H (unnecessary as N is already balanced)

3) Balance for O by adding H2O

3O on the left, 2O on the right.

Must add H20 to right hand side as it has one less Oxygen

NO3- ---> NO2 + H2O

4) Balance for H by adding H+ ions

Because we have added H20, there is now an uneven distribution of H atoms in the equation, right hand side has 2 but the left hand side has none, must add 2H+ ions to balance

NO3- + 2H+ --- > NO2 + H2O

5) Balance charge by adding e-

NO3- + 2H+ ---> NO2 + H2O

(overall -1)(2x1=2) (Neutral, no charges)

Left hand side = +1

Right hand side = 0

Must add 1 electron to left hand side to make neutral (=0)

How to combine half equations (writing a redox reaction)

Must have the same number of electrons being lost as gained, similar to a simultaneous equation

Combine these half equations

Al3+ + 3e- ---> Al

K ---> K+ + e-

Aluminium gains 3 electrons (reduction)

Potassium looses 1 electron (oxidation)

The electrons lost must equal the electrons gained, treat it like a simultaneous equation and find a common multiple

3 is a common multiple

The equation remains as it is

Second must be multiplied by3

AL3+ +3e- ---> Al

3K ---> 3K+ + 3e-

Add everything up on each side

Al3+ +3K + 3e- ---> Al + 3K+ + 3e-

Simplify by removing anything which is present on both sides as EQUATIONS ONLY SHOW CHANGES. (Remove both 3e-)

Al3+ + 3K ---> Al + 3K+

What is the reducing agent?

(Reducing agent itself gets oxidised)

-causes something else to be reduced

-it donates (looses) electrons

What is the oxidising agent?

(Oxidising agent itself gets reduced)

-causes something else to be oxidised

-it accepts (gains) electrons

a) What does an increased oxidation state suggest?

b) What does a decreased oxidation state suggest?

a) Oxidation has occurred

b) Reduction has occurred

O oxidation value

H oxidation value

O = -2

H = +1

IO3- ions oxidize iodide ions to iodine.

IO3- + 5I- + 6H+ ---> 3I2 + 3H2O.

Deduce the half equation to show the reduction process in this reaction.

2IO3−​ + 12H+ +10e− → I2​+ 6H2​O