Chemistry 2.7 (AS 90306)

Describe oxidation-reduction reactions

Questions may involve any of the following:

the properties of common oxidants and reductants, and the products of their reaction.

Common oxidants are limited to O2, I2, Cl2, Fe3+, H2O2, MnO4

(aq)/H+, Cr2O72(aq)/H+

Common reductants are limited to metals, C, CO, H2, Fe2+, Br, I, SO2, (HSO3

).

Properties are limited to appearance (colour and state), oxidation number (for polyatomic ions and single ions)

Chemistry 2.7 (AS 90306)

Describe oxidation-reduction reactions

Questions may involve any of the following:

writing balanced oxidation–reduction equations

classifying balanced half-equations as oxidation or reduction

identifying the oxidant and/or reductant from a given reaction

describing the ability of halogens to act as oxidants in reactions with other elements, water or halide ions

principles of simple electrolytic cells.

Homework for the Holidays

Complete at least Q’s 1, 2 and 3 page 73 from your text book, do more if you can.

Read Unit 18 pages 71 – 74 in your year 12 Pathfinder Text

OXIDATION - REDUCTION

Oxidation was originally defined as a gain of oxygen eg Mg reacts with O2 to form magnesium oxide, MgO.

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

Simlarly reduction was the removal of oxygen e.g. CO reduces Fe2O3 and produces Fe and CO2.

Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g)

However scientists realised that not all redox reactions involved oxygen e.g. the reaction of zinc metal with copper ions.

Zn(s) + Cu2+(aq) Zn2+ (aq) + Cu(s)

Do you remember doing this?

What metal is giving up its electrons and being oxidised ?

What metal ion is accepting these electrons and being reduced ?

What did you observe ?

So the Definition of oxidation/reduction now is:

An oxidation-reduction reaction (or redox reaction) is one that involves the transfer of electrons from one species to another.

In the reaction below the Zn metal has been oxidised as it has lost electrons, and the Cu2+ has been reduced as it has gained electrons.

1. Zn Zn2+ + 2e

2. Cu2+ + 2e Cu

Remember mnemonic -

OIL RIG

oxidation is loss reduction is gain

Oxidation (loss of e’s)Reduction

(gain of e’s)

Oxidation Numbers

A useful tool in recognising redox reactions, and for determining what is being oxidised and what is reduced in a reaction, involves the use of oxidation numbers

The oxidation number (symbol ON) describes the “degree” to which an element has been oxidised or reduced.

Chemists have developed a number of rules you must learn for assigning oxidation numbersNote: Oxidation numbers are always quoted per atom.

Oxidation number rules

The oxidation number (or state) is a number that can be assigned to each atom in an element, compound or ion, using a set of 6 rules.

These rules are as follows:

Rule 1

The oxidation number of an atom in any element is zero. For example in H2 the oxidation number of H is 0.

The oxidation number of C in carbon is 0

The oxidation of Mg in a piece of Mg is 0

The oxidation number of O in O2 is 0 .. etc

Rule 2The oxidation number of an atom in a monatomic ion is

the same as the charge on the ion

e.g. in Na+ the oxidation number is +1,

in O2 the oxidation number is -2.

In an ionic compound, the ions have the same oxidation numbers as they would alone

e.g. in Na2O the oxidation numbers are still +1 for Na+ and -2 for O2.

Rule 3

In compounds each hydrogen atom usually has an oxidation number +1 (the exception is in the metal hydrides e.g.NaH where oxidation number of H= -1).

Rule 4 In compounds each oxygen atom has an oxidation

number -2 (except in peroxides e.g. H2O2 where O has ON of -1)

Rule 5

In a molecule the sum of the oxidation numbers of all the atoms is zero.

Using rules 3, 4 and 5 it is possible to calculate the oxidation numbers of all atoms.

Eg Find the oxidation number of S in H2SO4.

H2 S O4

2 x +1 + 1x ? + 4 x -2 = 0

= 2 + ? + - 8 = 0

Can you do the math to find ON of S?

Oxidation number of S = +8 - 2 = +6

Hint ask :

What’s the ON of

H in compounds? (R3)

What’s the ON of O in compounds? (R4)

Then use these with R5 to find the ON of S

Use the rules to find the oxidation numbers of the each atom in each of the following molecules.

NO2, HNO3, NO, N2, N2O,

HNO2, N2O4

+4

-2+1

+5 -2 +2

-2 +1

-20

+3+1 -2 +4 -2

Note that the N atom has a different ON depending on which atoms it is bonded to!

Rule 6

In polyatomic ions the sum of the oxidation numbers of all the atoms is equal to the charge on the ion.

In the ion Cr2O72 the oxidation number of Cr is

calculated as follows:

2 x Cr + 7 x O

2 x ? + 7 x -2 = -2

2 x ? + -14 = -2

2 x Cr = 12

Oxidation number of Cr = = +62

12

Charge on the polyatomic ion

Note: Oxidation numbers are always quoted per atom.

(+14 to BS)

Calculate the oxidation number of S in each of the following ions:

SO42

SO32

S2O32

S4O62

S + (4 x -2) = -2 ON of S = +6

S + (3 x -2) = -2 ON of S = +4

(2 x S) + (3 x -2) = -2 ON of S = +2

4 x S + (6 x -2) = -2 ON of S = +2.5

An increase in oxidation number corresponds to

oxidation

A decrease in oxidation number corresponds to

reduction.

This is an exceptionally important thing to remember!

We now can identify which species is oxidised and which is reduced in a reaction by applying this fact:

By assigning oxidation numbers show which of the following reactions is not a redox reaction.

(i) CuCO3 CuO + CO2

(ii) Cu + 2AgNO3 Cu(NO3)2 + 2Ag

(iii) Cr2O72 + 6Fe2+ 6Fe3+ + 2Cr3+ + 7H2O

+2 +4 -2 +2 -2+4-2

0 +1 -2+5 -2+5+2 0

Not a redox reaction no change in ON’s

Ag reduced

Cu oxidised

+6 +3 +3 +2 Cr2O72reduced

Fe 2+ oxidised

Looks like you’ve mastered the whole assigning ON thing!

Now it’s time to apply them in balancing more complex redox reactions

Balancing Redox Equations.

The following method for balancing more complex redox equations is commonly called the ion-electron half-equation method.

*These are 6 more steps we must memorise

This is all you need to memorize – honest!

Step 1 Identify the species undergoing oxidation and the species undergoing reduction

Remember to use ON for this:

The species that decreases in ON is reduced

The species that increases in ON is oxidised

e.g. when a solution of potassium dichromate reacts with iron II nitrate the species oxidised is Fe2+ and the species reduced is Cr2O7

2.

The reactions are:Fe2+ Fe3+ and Cr2O72 Cr3+

+2 +3

oxidation

+6 +3

reduction

Step 2 Balance all atoms undergoing a change in oxidation number in each half equation.

Fe2+ Fe3+

Step 3 Balance the number of O atoms by adding the appropriate number of water molecules.

Fe2+ Fe3+ and Cr2O72 2Cr3+ + 7H2O

Step 4 Balance the H atoms by adding H+ ions.

Fe2+ Fe3+ and Cr2O72 + 14H + 2Cr3+ + 7H2O

and Cr2O72 2Cr3+

Step 5 Balance the charge by adding electrons, e-. This gives 2 balanced half-equations.

Fe2+ Fe3+ + e

and

Cr2O72 + 14H+ + 6e 2Cr3+ + 7H2O

Step 6 To obtain an overall balanced equation the 2 half equations must be added together.

Before doing this the equations may have to be multiplied so that the number of electrons in each half-equation is the same. In this way, the electrons will be eliminated in the final equation.

Fe2+ Fe3+ + e

is now 6Fe2+ 6Fe3+ + 6e

Cr2O72 + 14H+ + 6e 2Cr3+ + 7H2O

Now combine both equations to give the final balanced redox equation – cancelling any electrons, H2O or H+

Finally check that the equation is balanced, particularly for charge!!

(6 x 2) + (-2) + (14) = +24

(2 x 3) + (6 x 3) = +24

6Fe2+ + Cr2O72 + 14H+ 2Cr3+ + 7H2O + 6Fe3+

(x6)

Question

A solution of Fe2+ is added to a solution of purple potassium permanganate (KMnO4) which went colourless showing that the Mn2+ ion had formed.

Using the steps for balancing redox equations write the balanced redox equation.

Hint – The Fe 2+ is oxidised to Fe 3+

Now turn to page 230 in your lab book

Common oxidants and reductants

Then colours of species – issue sheet to be coloured

Turn to page 236 in your lab books

“more oxidants”

Read the experiment carefully

Oxidant (aka oxidisng agent)

An oxidant is the substance that a______ electrons and is _________

Oxidants and Reductants revisited

Reductant (aka reducing agent)

A reductant is the substance that d______ electrons and is _______

cceptsreduced

onatesoxidised

Look at the Demo then Complete the following

KMnO4 is an oxidising agent and reacts with H2O2 .

1.What is the colour of KMnO4?

2.Write the reaction for the reduction of the KMnO4

3.Write the reaction for the oxidation of H2O2

4.Write the full redox reaction by combining both half equations

5.What would you observe in the this reaction?

Reactions of Halogens (has appeared in exams)

Halogens e.g. chlorine, Cl2 (a yellow-green gas), bromine, Br2 (an orange/brown liquid), and iodine, I2 (a shiny black solid) are all reduced to their respective colourless halide ions, Cl, Br, I.

The order of oxidising strength is Cl2 > Br2 > I2 In other words Cl2 will oxidise Br- ions to form Br2 the reaction is:

Cl2 + 2Br- 2Cl- + Br2 (brown bromine appears)

But !! 2Cl- + Br2 no reaction Why?

Any halogen is able to oxidise the halide ion from a weaker halogen e.g. Cl2 can successfully oxidise I to I2. But I2 cannot oxidise Cl- because it is weaker oxidising agent 

How could you introduce Cl- ions into a solution?

How could you introduce Br- ions into a solution?

What are the colours of

I-

Br-

Cl-

What are the colours ofI2

Br2

Cl2

QUESTION SEVEN 2.7 2005 examGroup 17 elements, the halogens (F, Cl, Br, I) act as oxidants in reactions.Aqueous chlorine, Cl2(aq), can react with a solution containing iodide ions, I–(aq).

Write balanced half-equations for the oxidation and reduction reactions that occur below.Then use these to write a balanced equation for the overalloxidation-reduction reaction that occurs.

oxidation: reduction: overall equation: Use the balanced equation to predict expected observations for thisreaction, and justify these observations by referring to the species involved.

2I- I2 + 2e-

Cl2 + 2e- 2Cl-

2I- + Cl2 I2 + 2Cl-

Colourless Cl2 solution oxidises colourless I- ions to form an orange/brown solution of iodine (I2) (E)

Look at the Demo then Complete the following

K2Cr2O7 (Cr2O72-)is an oxidising agent and reacts with

NaHSO3 (HSO3-).

1.What is the colour of Cr2O72- ion?

2.Write the reaction for the reduction of the Cr2O72-

3.Write the reaction for the oxidation of HSO3-

4.Write the full redox reaction by combining both half equations

5.What would you observe in the this reaction?

What’s occurring in this Demo

Cu + HNO3

K2Cr2O7/SO2

SO2 gas is bubbled through some acidified K2Cr2O7 solution