higher chemistry: unit 2 chemical changes and structure

Higher Chemistry St. Ninian’s High School

CIS: Redox Revision of Redox Reactions

Higher Chemistry: Unit 2 – Chemical Changes and Structure

Revision of Redox Equations and Titrations

The lesson will help you revise the following topics

1. Oxidising and Reducing Agents

2. Complex Redox Equations

3. Redox Titrations

You will have been successful in this lesson if you:

1. Read the summary given (do not copy these notes – you already have them)

2. Watch the Recorded Lesson – Link provided

3. Complete questions provided

4. EXTENSION: There is a further reading section to help you gain more depth of

understanding for this section. There are also suggested questions for you to try from the

blue book of revision questions.

If you have any questions about the content of this lesson, you should ask your class teacher

either through your class MS team or via email. The teams will be monitored through the

week and someone will get back to you as soon as they can.

You may wish to revise the following to help you understand this lesson:

Higher chemistry – Unit 3 – Redox Reactions

Learning Outcomes

Success Criteria

Links to Prior Knowledge

You may wish to have a copy of the data booklet handy for this lesson. Download or

print a copy of the Higher Chemistry Data Booklet from MS Teams or the SQA website -

https://www.sqa.org.uk/sqa/files_ccc/ChemistryDataBooklet_NewH_AH-Sep2016.pdf

https://www.sqa.org.uk/sqa/files_ccc/ChemistryDataBooklet_NewH_AH-Sep2016.pdf



Redox Equations and Titrations


Redox Revision from Nat 5

OILRIG: Oxidation Is Loss, Reduction Is Gain (of electrons)

Oxidation and Reduction reactions always occur together.

These types of reactions are called redox reactions.

Ion-electron equations for the oxidation and reduction in a typical redox reaction are (from

data booklet page 12):

Cu → Cu2+ + 2e- Oxidation

Ag+ + e- → Ag Reduction

Cu + 2Ag+ → 2Ag + Cu2+ Balanced Redox Equation

Points to Note:

- Electrons and spectator ions are not included in balanced redox equations.

- For the reaction above, the silver equation had to be multiplied by 2 to balance the

electrons in the redox equation.

Do not copy the notes below – these are a summary of the class notes you

already have for this unit.

WATCH: Click on the link for a recorded lesson for this topic:

RECORDED LESSON from Ms Hastie

https://glowscotland-my.sharepoint.com/:v:/g/personal/gw08hastiepaula_glow_sch_uk/EfyYsNKoxKBDqF2QaTHm2SsBQBpjvfwvcFpCsyqiJf8xwA?e=QJa09i



Oxidising Agents

An oxidising agent is a substance which accepts electrons.

In the above reaction, the silver ion, Ag+ is the oxidising agent: Ag+ + e- → Ag

The oxidising agent causes the oxidation of another reactant and in the process is reduced.

Elements with a high electronegativity tend to form ions by gaining electrons (reduction)

and so often act as oxidising agents. Strong oxidising agents are found at the BOTTOM LEFT

in the Electrochemical Series (ECS), page 12 of your Data Booklet.

Group 7 elements, especially fluorine and chlorine are strong oxidising agents. Common

oxidising agents also include:

Acidified dichromate ions (Cr2O72-)

Acidified permanganate ions (MnO4-)

Hydrogen peroxide (H2O2)

Reducing Agents

A reducing agent is a substance which donates electrons.

In the above reaction, copper is the reducing agent: Cu → Cu2+ + 2e-

In donating electrons, a reducing agent causes the reduction of another reactant, and in the

process, is oxidised.

Elements with low electronegativities tend to form ions by losing electrons (oxidation) and

so often act as reducing agents. Strong reducing agents are found at the TOP RIGHT in the

Electrochemical Series (ECS), page 12 of your Data Booklet.

Group 1 metals are strong reducing agents. Another example of a reducing agent is carbon

monoxide (CO).



Identifying Oxidising and Reducing Agents

In the reaction: Cu + 2AgNO3 → 2Ag + Cu (NO3 )2

The oxidising agent is the one being reduced (or the one that is causing the oxidation). This

is the silver ion, Ag+

The reducing agent is the one being oxidized. This is the copper atom, Cu

Nitrate, NO3- , is the spectator ion

Everyday Examples of Oxidising Agents

Oxidising agents are widely used in cleaning products because of the

effectiveness with which they can kill fungi and bacteria, and can

inactivate viruses.

The oxidation process is also an effective means of breaking down

coloured compounds making oxidising agents ideal for use as

‘bleach’ for clothes and hair.


Example 1. Write an ion–electron equation for the conversion of bromine to bromate, BrO3¯

Step 1. Balance the main element

(2 x BrO3¯ because Br2 on left hand side, LHS ) Br2 → 2BrO3¯

Step 2. Balance oxygen by adding water molecules

(6 x H2O because 2BrO3¯ = 6 oxygens ) Br2 + 6H2O → 2BrO3¯

Step 3. Balance hydrogen by adding hydrogen ions

(12 x H+ because 6 x H2O) Br2 + 6H2O → 2BrO3¯ + 12H+

Step 4. Balance the charge by adding electrons

Total charge LHS = Br2 + 6H2O = 0+0 = 0

Total charge RHS = 2BrO3¯+ 12H+ =(-2)+(+12)= +10

10e¯ added to RHS to reduce charge to 0

Br2 + 6H2O → 2BrO3¯ + 12H+ + 10e¯

Ion-electron equation: Br2 + 6H2O → 2BrO3¯ + 12H+ + 10e¯



Example 2. Acidified potassium permanganate (KMnO4) oxidises Fe2+ ions to Fe3+ ions. During

this reaction the MnO4– ions are reduced to Mn2+ ions:

(a) Write an ion-electron equation for the reduction reaction occurring:

Step 1. Balance the main element

(Mn in this case, already balanced) MnO4¯ → Mn2+

Step 2. Balance oxygen by adding water molecules

(4 x H2O because MnO4 = 4 oxygens) MnO4¯ → Mn2+ + 4H2O

Step 3. Balance hydrogen by adding hydrogen ions

(8 x H+ because 4 x H2O) MnO4¯ + 8H+ → Mn2+ + 4H2O

Step 4. Balance the charge by adding electrons

Total charge on LHS = MnO4¯+ 8H+ = (-1)+(+8)= +7

Total charge on RHS = Mn2+ + 4H2O = (+2)+0 = +2

5e¯ added to LHS to reduce charge to 2+

MnO4¯ + 8H+ + 5e¯ → Mn2+ + 4H2O

Ion-electron equation: MnO4¯ + 8H+ + 5e¯ → Mn2+ + 4H2O

Note: The equation above shows that, permanganate, MnO4¯ needs hydrogen ions, H+

in order for this reaction to proceed. The H+ ions are provided by added an acid to

the solution. The term “acidified” is often used when describing oxidising agents eg

acidified dichromate.

(b) Write a balanced redox equation for the overall reaction:

REDUCTION: MnO4¯ + 8H+ + 5e¯ → Mn2+ + 4H2O

OXIDATION: Fe2+ → Fe3+ + e¯

To make a balanced redox equation with no electrons left over, you must multiply to make

the electrons the same on both sides: 5Fe2+ → 5Fe3+ + 5e¯

BALANCED REDOX EQUATION: MnO4¯ + 8H+ + 5Fe2+ → Mn2+ + 4H2O + 5Fe3+



3. Redox Titrations

Example 1. A standard solution of acidified potassium permanganate (oxidising agent) was used to find the concentration of a solution of iron (II) ions (reducing agent). BALANCED REDOX EQUATION: 5Fe2+ + MnO4

- + 8H+ → Mn2+ + 4H2O + 5Fe3+ The end point for this set up is when the solution in the conical flask turns from colourless to purple showing the MnO4

- is in excess. Therefore no indicator is needed for this titration. Titration results:

Titre Volume MnO4

- / cm3

Start volume End volume Titre volume

1 0.0 15.0 15.0

2 15.0 27.1 12.1

3 27.1 39.0 11.9

TRY THIS QUESTION – WORKED SOLUTION IS ON THE NEXT PAGE Question:

Using the information above find the concentration of iron(II) ions in the solution.

0.1 mol l-1

acidified

permanganate

solution (purple)

25 cm3 Fe2+

solution

(colourless)



WORKED SOLUTION: Question: Using the information above find the concentration of iron(II) ions in the solution. (a) Find the average titre value

Average titre value = 12.1+11.9

2 = 12.0 cm3

Rough titre of 15.0 cm3 is ignored because it is not concordant, ie not within 0.2 cm3

(b) Find the unknown concentration

Step 1 – find the number of moles of the 0.1 moll-1 permanganate solution: n = c x v

= 0.1 x 0.012

= 0.0012 moles

Step 2 – find the number of moles of the iron(II) solution Mole ratio from balanced redox equation MnO4

- : 5Fe2+ 0.0012 moles MnO4

- : 0.0012 x 5 Fe2+ = 0.006 moles Fe2+ ions Step 3 – find the concentration of the 25 cm3 iron(II) solution c = n/v

= 0.006 / 0.025

= 0.24 moll-1

EXTENSION QUESTION – TRY THIS QUESTION – ANSWER AT THE END OF THE LESSON: The iron(II) solution was made by dissolving five iron tablets in water. The solutions and rinsings were added to a 250 cm3 standard flask and made up to the graduation mark with distilled water. Find the mass of iron(II) in one iron tablet. WATCH: Ms Adams Chemistry – Redox (18 mins) https://youtu.be/vxEavA8nPNY

https://youtu.be/vxEavA8nPNY



You should have now revised:



3. Redox Titrations

To learn more about proteins, try the following online resources:

BBC Bitesize: https://www.bbc.co.uk/bitesize/guides/z2r44wx/revision/1

Read revision pages (optional) and TRY THE END TOPIC TEST

https://www.bbc.co.uk/bitesize/guides/ztkdd2p/test

TRY THE END TOPIC TEST (also includes chromatography)

Scholar: Log in through GLOW

Higher Chemistry Chemical Changes and Structure Topic 6: Oxidising and

Reduction Agents

Read through the exercises and TRY THE END TOPIC TEST

ALSO Higher Chemistry Chemistry in Society Topic 8: Volumetric Analysis

Read through the exercises and TRY THE END TOPIC TEST

Evans2 chem web: https://www.evans2chemweb.co.uk/login/index.php#

Username: snhs password: giffnock

Select any teacher revision material CfE Higher Unit 1: Chemical

Changes and Structure Oxidising and Reducing Agents

ALSO

Unit 3: Chemistry in Society Chemical Analysis

Learning Outcomes

Further Reading

https://www.bbc.co.uk/bitesize/guides/z2r44wx/revision/1

https://www.bbc.co.uk/bitesize/guides/ztkdd2p/test

https://www.evans2chemweb.co.uk/login/index.php



ANSWER TO EXTENSION QUESTION: The iron(II) solution was made by dissolving 5 iron tablets in water. The solutions and rinsings were added to a 250 cm3 standard flask and made up to the graduation mark with distilled water. Find the mass of iron(II) in 1 iron tablet. Step 1 – scale up the number of moles in 25 cm3 to total number of moles in 250 cm3 iron(II) solution number of moles of iron(II) in 25 cm3 = 0.006 (from question above)

number of moles of iron(II) in 250 cm3 = 0.006 x 10 = 0.06 moles Fe2+ ions

Step 2 – find the mass of iron in 5 tablets mass = n x gfm =

= 0.06 x 55.8 = 3.348 g in 5 tablets

Step 3 – find the mass of iron in 1 tablet 1.348 g in 5 tablets, therefore 1 tablet = 3.348 / 5 = 0.6696g or 670mg



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Check your understanding – Answers the questions below in you class jotter



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Use the online learning link above if you would like to extend your knowledge on this topic.

For more practise questions, use your Revision Questions for Higher Chemistry “Blue book”:

Redox reactions page 105 Q1-5

Writing ion-electron equations page 107 Q1-2

Oxidising and reducing agents page 108 Q1-3

Redox titrations page 111 Q1-9

Answer to Redox Revision Questions

1. A

2. D

3. A

4. C

5. D

6. D

7. C

8. B

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EXTENSION WORK

ANSWERS CAN BE FOUND ON THE NEXT PAGES FOR YOU TO CHECK YOUR

WORK AS YOU GO ALONG.



12. (a)

(b)

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higher chemistry: unit 2 chemical changes and structure

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