REDOX & OXIDATION NUMBERS - PowerPoint Presentation

Slide1

REDOX & OXIDATION NUMBERSSlide2

REDOX REACTIONS

OILRIG –

Oxidation Is Loss of electrons Reduction Is gain of electrons

This is really important

Think of January sales!Slide3

OXIDATION NUMBER RULES

Oxidation numbers which are assigned to ions (of each element) in a compound help us to write half equations and full ionic equations.

This is very important when deciding which species in an ionic equation is being oxidised and which is being reducedSlide4

RULES

All

uncombined elements have an oxidation umber of zero e.g. Zn, Cl2, O2, Ar all have an oxidation number of zero

The oxidation numbers of the elements in a compound add up to zero

e.g. In

NaCl Na = +1 Cl = -1 Sum = +1-1 = 0

The oxidation number of a monoatomic ion is equal to the charge on the ion

e.g. Zn2+

= +2 Cl- = -1

In a polyatomic ion (CO

3

2-

) the sum of the individual oxidation numbers of the elements adds up to the charge on the ion

e.g. in CO

3

2-

C = +4 and O = -2

Sum = +4 (3 x -2) = -2Slide5

Several elements have invariable oxidation numbers in their common compoundsSlide6

Some examples – Give the oxidation number of:

N in NO

2 S in SO42-Fe in FeCl3N in NO3-F in F2S in SO3Na in NaOHSlide7

QUESTIONSSlide8

Consider the following redox reaction:

The half equation to show what is happening to Br2 is:

Br2 + 2e-  2Br

-

0 -2Here Br

2 has gained electrons since its gone from 0 to -2So Br2

has been reduced

The half equation to show what is happening to the I- ion is:

2I

-

 I

2

+ 2e-

-2 0

Here I

-

has lost electrons since it has gone from -2 to 0

So I- has been oxidised to form I

2Slide9

NOTE//

An oxidising agent (or oxidant) is the species that causes another element to oxidise. It is itself reduced in the reaction

A reducing agent (or reductant) is the species that causes another element reduce. It is itself oxidised in the reactionNOTE ALSO//When naming oxidising and reducing agents always refer to full name of substance and not just name of elementSlide10

WRITING HALF EQUATIONS

RULES

Work out oxidation numbers for element being oxidised/ reduced i.e.Zn  Zn2+ Zn changes from 0 to +2

2. Add electrons equal to the change in oxidation number For reduction

add e’s to reactants For oxidation add e’s to products

Check to see that the sum of the charges on the reactant side equals

the sum of the charges on the product side Slide11

QUESTIONS

Write half equations for the following reactions and state whether the half equation is showing a reduction or oxidation:

Fe2+/Fe3+F2/F-O2

/O2-

Al

3+/AlSn

2+/Sn4+

Cu2+/CuSlide12

Say you needed to show a half equation for the following:

MnO

4-/Mn2+Then we would have:MnO4-

 Mn

2+

This equation is impossible to balance as we do not have oxygen on the RHSIn these situations we MUST add H

2O on the RHS and H+

on the LHS of the equation i.e.MnO

4

-

+ H

+

 Mn

2+

+ H

2

O

Now we have all the atoms of each element on each side to enable us to balance this half equation.Slide13

Example:

Write the half equation for the change MnO

4-  Mn2+ MnO4- + H+  Mn

2+

+ H2

O

Add H2

O and H+

on appropriate sides if needed

2.

Then balance the number of atoms of each element on both sides:

MnO

4

-

+ 8H

+

 Mn

2+

+ 4H

2

O

3. Then balance the change in oxidation state with electrons:

MnO

4

-

+ 8H

+

 Mn

2+

+ 4H

2

O

-1 +8  +2 0

+7  +2Slide14

5e- must be added to the RHS to balance the charges

i.e. MnO

4- + 8H+ + 5e-  Mn2+ + 4H2O

 Slide15

MnO

4

- + 8H+ + 5e-  Mn2+ + 4H2O

The oxidation state of Mn

here is +7

The oxidation state of

Mn

here is +2

So this half equation is showing that

Mn

is being reduced from +7 to +2

NOTE//

The oxidation state of oxygen and hydrogen does not change

Oxygen remains as -2

Hydrogen remains as +1 Slide16

Example:

Write the half equation for the change SO

42-  SO2Again it is impossible to balance this equation despite the fact that we have the same atoms of each element on both sides, so again we must turn to using H2O and H+ on appropriate sides to balance this equation

1. So,

SO42- + H

+  SO2 + H

2O

2. Now balance the number of atoms on each side: SO

4

2-

+ 4H

+

 SO

2

+ 2H

2

OSlide17

3. Now balance the charges on each side using electrons;

SO

42- + 4H+  SO2 + 2H2O -2 +4  0 0 +2 0So, 2e- must be added to the LHS SO42- + 4H+ + 2e-  SO

2 + 2H

2OSlide18

SO

4

2- + 4H+ + 2e-  SO2 + 2H2OThe oxidation state of S in SO

4

2- is +6

The oxidation state of S in SO2

is +4

So here S is being reduced from +6 to +4

Again the oxidation states of both oxygen and hydrogen do not changeSlide19

QUESTIONS

Write half equations for the following reactions and using oxidation numbers explain which element is being oxidised/reduced:

 PbO2  Pb

2+

Cl

-

 Cl

2

S

2

O

3

2-



S

4

O

6

2-

I

2



I

-

IO

3

-



I

2

ClO

-



ClO

3

-

,

ClO

-



Cl

-

H

2

SO

4



SO

2

Br

-



Br

2

H

2

SO

4



S

H

2

SO

4



H

2

S

PbO

2



Pb

2+

SO

3

2-



SO

4

2-

 Slide20

COMBINING HALF EQUATIONS TO MAKE FULL IONIC EQUATIONS

Write both half equations first (one which will show an oxidation and one which will show a reduction)

To combine two half equations there must be equal numbers of electrons in the two half equations so that the electrons cancel outWrite two half equations to show:MnO4-/Mn2+C2O4

2-/CO2

ANSWER

MnO4- + 8H+ + 5e  Mn

2+ + 4H2

OC2

O

4

2-

 2CO

2

+ 2e-Slide21

Now combine both half equations by eliminating the electrons:

(REDUCTION)

MnO4- + 8H+ + 5e-  Mn2+ + 4H2O (X2)(OXIDATION) C2O42-  2CO

2 + 2e-

(X5)

2MnO4

- + 16H+

+ 5C2O

4

2-

→ 2Mn

2+

+ 10 CO

2

+ 8H

2

O

2MnO

4

-

+ 16H

+

+ 5C

2O42- → 2Mn2+ + 10CO

2 + 8H2O

The oxidation state of

Mn

here is +7

The oxidation state of

Mn

here is +2

The oxidation state of C here is +3

The oxidation state of C here is +4

So this is a redox reaction as:

Mn

has been reduced from +7 to +2

And

C has been oxidised from +3 to +4

So

MnO

4

-

is acting as an

OXIDISING AGENT

C

2

O

4

2-

is acting as a

REDUCING AGENTSlide22

Write two half equations to show:

SO

42-/H2SI-/I2ANSWER

REDUCTION

SO

4

2- + 10H+

+ 8e-  H

2

S+ 4H

2

O

OXIDATION

2I

-

→ I

2

+ 2e-

Now combine both half equations by eliminating the electrons:

SO

4

2-

+ 10H

+

+ 8e-



H

2

S+ 4H

2

O (X1)

2I

-

→ I

2

+ 2e- (X4)

8I

-

+ SO

4

2-

+ 10H

+



H

2

S + 4I

2

+ 4H

2

OSlide23

8I

- + SO

42- + 10H+  H2S + 4I2 + 4H2OHere oxidation state of I is -1 Here the oxidation state of S is +6 Here the oxidation state if S is -2

Here the oxidation state of I is zero

So in this redox reaction:

I is being oxidised from -1 to 0

S is being reduced from +6 to -2

I

-

is acting as a REDUCING AGENT

SO

4

2-

is acting as an OXIDISING AGENTSlide24

Write two half equations for each question and then combine the half equations.

State which species has been oxidised and which has been reduced.

Give the name of the oxidising agent and reducing agent in each full ionic equation. a) PbO2 

Pb2+

, Cl

-

 Cl

2

b) S

2

O

3

2-



S

4

O

6

2-

, I

2



2I

-

c) IO

3

-



I

2

, I

-



I

2

d) ClO

-



ClO

3

-

, ClO

-



Cl

-

e) H

2

SO

4



SO

2

, Br

-



Br

2

f) H

2

SO

4



S, I

-



I

2

g) H

2

SO

4



H

2

S, I

-



I

2

h)

ClO

-

 Cl-, I-  I2 i) PbO2  Pb2+, SO32-  SO42-

QUESTIONS