Redox Reactions.

Oxidation

Reduction

Oxidation and Reduction

Oxidation:•Gain of oxygen

•Loss of electrons

Reduction:•Loss of oxygen

•Gain of electrons

Increase in oxidation

number

Decrease in oxidation

number

4 Experiments:

1. Burning magnesium

2. Copper in silver nitrate solution

3. Chlorine solution and potassium iodide solution

4. Exploding hydrogen

•Word equation•Balanced symbol equation

2Mg(s) + O2(g) 2MgO(s)

Oxidised – gains oxygen

Must be a redox!

Mg Mg2+

O O2-

CHARGE OF A - = GAINED ELECTRONS

CHARGE OF A += LOST ELECTRONS

Put the e- in.

+2e-

+2e-

Oxidised – loss of e-

Reduced – gain of e-

Cu(s) + 2AgNO3(aq) Cu(NO3 )2(aq) + 2Ag(s)

Ag+ Ag

Cu Cu2+

Complete the half-equations

+e-

+2e-

Oxidised?Reduced?

Oxidised – loss of e-

Reduced – gain of e-

H2(g) + ½ O2(g) H2O(g)

Covalent!No H+ or OH-Need a new

definition.

oxidation

reduction

Reducing agent

Oxidising agent

agents

• An oxidising agent is a substance that brings about oxidation(itself reduced)

• example- hydrogen peroxide for bleaching hair

• A reducing agent is a substance that brings about reduction.(itself oxidised)aa sulphur dioxide used to bleach straw

In terms of oxidation number

Oxidation:•Gain of oxygen

•Loss of electrons

Reduction:•Loss of oxygen

•Gain of electrons

Increase in oxidation

number

Decrease in oxidation

number

Oxidation Numbers- the seven rules

Oxidation Numbers- the seven rules

• The oxidation number of an atom in an uncombined element is zero. E.g. Mg in Mg, O in O2.

• The oxidation number of an ion of an element is the same as its charge.

• O.N Br-= -1• O.N Mg in Mg 2+ = +2• ALKALI METALS= +1• ALKALINE EARTH METALS=+2• HALOGENS= -1

Oxidation NumbersOxidation Numbers

• The oxidation numbers of atoms in a compound add up to zero.

F -1

O -2

H +1

Cl -1

Oxidation state of C in CO2?

x – 4 = 0

x = +4Put the +!

Oxidation NumbersOxidation Numbers

• The oxidation numbers of atoms in a compound add up to zero.

F -1

O -2

H +1

Cl -1

Oxidation state of Mg in MgCl2?

+2

Oxidation NumbersOxidation Numbers

• The oxidation numbers of atoms in a compound add up to zero.

F -1

O -2

H +1

Cl -1

Oxidation state of N in NH3?

-3

Oxidation NumbersOxidation Numbers• The oxidation

numbers of atoms in an ion add up to the charge on the ion.

F -1

O -2

H +1

Cl -1

Oxidation state of S in SO4

2-?

x – 8 = -2

x = +6

Oxidation NumbersOxidation Numbers• The oxidation

numbers of atoms in an ion add up to the charge on the ion.

F -1

O -2

H +1

Cl -1

Oxidation state of S in S2-?

-2

Oxidation NumbersOxidation Numbers• The oxidation

numbers of atoms in an ion add up to the charge on the ion.

F -1

O -2

H +1

Cl -1

Oxidation state of N in NH4

+?

-3

OXYGEN HYDROGEN

• Oxygen has a charge Oxygen has a charge of

• –2• EXCEPT in peroxides

where the charge • is –1 • In the compound OF2

• Where it has a value of +2.

• This is because F has a greater electronegative number than oxygen

• Hyrogen has a charge of +1 except in

• Metal hydrides where it has an O.N of –1

• Metal Hydrides are ionic compounds

• NaH• (+1)(-1)

Halogens-assign charge of –1

unless bonded to more electronegative element

Cl2O(+1)2(-2)Cl= +1

when writing formulas the most electronegative is placed

second.

• Halogen

H2(g) + ½ O2(g) H2O(g)

Covalent!

No H+ or OH-

Need a new definition.

Oxidation:•Gain of oxygen

•Loss of electrons

Reduction:•Loss of oxygen

•Gain of electrons

Increase in oxidation

number

Decrease in oxidation

number

H2(g) + ½ O2(g) H2O(g)

Covalent!

No H+ or OH-Need a new definition.

+10

-20O

H

H2(g) + ½ O2(g) H2O(g)

+10

-20O

H

Oxidised?

H – increase in oxidation

number

Reduced?

O – decrease in oxidation

number

Balancing Redox Reactions

• Using Oxidation Numbers balance the following equation.

•   • Solution:• 1 assign oxidation number • 2. Note element that changes

oxidation number.• 3.show the number of electrons

lost and gained.

• 4.Work out ratio of oxidising agent to reducing agent.

• 5.Balance remaining items by inspection method.

Worked Example

MnO 4 +Fe2+ +H+ Mn2+ +Fe3+ +H2O(+7)4(-2) (+2)(+1) (+2) (+3) 2 (+1)(-2)

>MnO 4 +Fe2+ +H+ Mn2+ +Fe3+ +H2O

(+7)4(-2) (+2)(+1) (+2) (+3) 2 (+1)(-2)

GAINS 5 ELECTRONS

LOSES 1 ELECTRON

Balance remaining items

• 1MnO 4 :5Fe2+

• 1MnO 4 +5Fe 2+ +H+ 1Mn 2+ +5Fe3+ +H2O

• 1MnO 4+5Fe2+ +8H+ 1Mn2+ +5Fe3+ +4H2O

Well done!