Chemistry Chemistry A Level(Form Six) Chemistry Notes Form Six (A Level)


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Is a branch of science which deals with the study of chemical/physical processes in which electricity is either produced or consumed.

Reduction is the addition of electrons to an atom or ion.
E.g. (i)
( ii)
There is an overall decrease in oxidation state.
Oxidation is the reaction in which electrons are being lost therefore removal of electrons from the atom or ion.
Redox reaction is the reaction in which both oxidation and reduction processes takes place at the same time. In a redox reaction, electrons are transferred between ions or atoms thus electrons are lost and gained in the same reaction.
NOTE: – In a balanced redox reaction, the number of electrons lost should be equal to the number of electrons gained.

From the above reaction (overall reaction)
The substance in a redox reaction which loose electrons is called reducing agent or reductant i.e. Zn the substance in a redox reaction which gains electrons is called oxidizing agent (oxidant) i.e.
NOTE: An oxidizing agent which cause other species to undergo oxidation but itself being reduced.
Not all reactions are redox reaction :-
To identify a redox reaction one should look for the change in oxidation number of an element in the course of the reaction. If there is increase in oxidation number, oxidation has taken place and if there is a decrease in oxidation state, reduction has taken place.
Guidelines or rules for determination of oxidation number of an element in a compound.
1. In a free element, each atom has an oxidation number of zero.
2. For ions consisting of single atom, the oxidation number is equal to the charge of that ion.
3. The oxidation of hydrogen is except in ionic hydrides where the oxidation is . E.g. NaH, KH.
The oxidation state of oxygen is -2 in most compounds except in peroxide where the oxidation state becomes -1 and +2 in oxygen fluoride due to electronegativity of fluorine.
4. The algebraic sum of oxidation number in a neutral compound must be zero and a polyatomic atom must be equal to the ion charge.
E.g.: calculate the oxidation number of the underlined elements in the following.
(i) S
(2×1)+ S + (-2×4) = 0
2 + S – 8 = 0
S = +6
(ii) Cr
2 Cr + (-2 x7) = -2
2 Cr = 12
Cr = +6
(iii) N
N + (-2 x 3) = -1
N – 6 = -1
N = +5

(iv) KMn
1 + Mn + (-2 x4) = 0
1 + Mn – 8 = 0
Mn – 7 = 0
Mn = +7

(v) P
4P= 0
(vi) N
N+ (1x 4) = 0
N = -4

(vii) S
2S + (-2×3) = -2
2S – 6 = -2
2S = 4
S = +2

Disproportion reaction
Is the reaction in which an atom undergoes both oxidation and reduction reaction simultaneously in the same reaction.

Rules for balancing redox reaction
1. By using oxidation number identify which element is oxidized or reduced.

2. Write the half reaction for each process.
3. Balance the atoms that are reduced or oxidized.
4. Balancing number of oxygen atoms
a) In acidic solution
i. Add water molecules to the side that is deficient of oxygen atoms.
ii. Add appropriate number of hydrogen ions to the other side to complete oxygen balance.

b)In basic solution.
i. For every oxygen atom required add OH to the side deficient of oxygen atoms.
ii. Add water molecule to the other side to complete the oxygen balance

5. Balancing the number of hydrogen atom
(a) In acidic solution.
Add hydrogen ions to the side in deficient of hydrogen atoms.
(b) In basic solution.
(i) To every hydrogen atom required add water molecule to the side deficient of hydrogen atom.
(ii) To the other side add hydroxyl ion to complete hydrogen balance.
6. Balance the charge by adding number of electrons to the side which is more positive.
7. Multiply oxidation/reduction by the smallest number to ensure that the numbers of electrons cost/gained are equal.
8. Add the two balanced half reactions by omitting species which appear on both side to obtain the final equation.
9. Check the final result to ensure that species and charge are balanced.
eg. Overall reaction

Example 2
(a) Balance the following redox reaction equation under acidic medium.

1. Half reaction equation

2. Balance under acidic medium

3. Balanced charged.

4. Overall reaction equation

(b) Balance the following redox reaction equation under basic medium

Therefore overall reaction :-
eg 2.

1.Half reaction
2.Half reaction under basic medium
3.Half reaction under basic medium
4. Balanced of charged
Overall reaction equation
Example 3
Balance the following redox reactions according to the media given.

3. Solution

SnO2 is an oxidant
CrO is an reductant

Hydrogen peroxide (H2O2 ) has both oxidizing and reducing power. However its reaction depends on the state of the second reagent whether it has to be reduced or oxidized,
(i) H2O2 as an oxidizing agent
Like any oxidizing agent H2O2 is capable of accepting electrons when treated with any electron donating species and itself being reduced to water.

(ii)H202 as reducing agent
H2O2 is capable of supplying electrons when treated with electron accepting species and itself being oxidized to oxygen gas.


It is applied on permanganometry titration. These are titration in which potassium permanganate is the tit rants and indicators are used. This is because permanganate itself acts as an indicator. It changes its colour due to the formation of manganese (Mn 2+) from manganese (Mn 7+) [in acidic media]

Oxidizing power of permanganate depends on the media used.
a) In neutral media or weakly alkali medium.
b) In acidic medium

Dilute. Sulphuric acid (H2SO4)is only used HCl cannot be used because will oxidize it to chlorine gas thus is interfering the power of permanganate

NOTE :– Concentration. and are also not used because they are also oxidizing agents.
c) In strong alkali medium.
n = number of electrons transferred per mole.
Basicity is the number of H+ per 1 molecule of acid when dissolved in H2O.

1.A standardization of in acidic solution gave the following data:
0.16g of potassium salt i.e. needed solution. What is the molarity of the solution

(i) To find molarity of K2 C2 04 .2H20
Concentration = 0.032gL-1
202g = 1 mol
0.16g = x
X = 7.92 × 10-4 moles
Concentration = 0.032 g dm-3
Molarity= 2.038 x 10-4

(ii) To find molarity of KMno 4 from reaction equation :-

= 7.92 x 10-4 moles
From the reaction equation :-

2 moles of KMnO4 = 5 moles of
2. Calculate the percentage of iron from the following data in a sample of iron wire:
1.4g of the wire was dissolved in excess dilute sulphuric acid and the solution was made up to 250cm3. 25cm3 of this solution required for oxidation. [If it was concentration. , it could have oxidized Fe to Fe3+ since it is a STRONG oxidizing agent]
……………………………………. (1)
Reaction of and
1 mole of Fe →1 mole of FeSO4
? x mole →0.0249 moles
X = 0.0249 moles of Fe
n = 0.025 moles
3. 100cm3 of H2O2 solution was diluted to 1 dm3 of solution. 25cm3 of this solution, when acidified with dilute H2SO4 reacted with 47.8cm3 of solution. Calculate the concentration of the original H2O2 solution.

Ratio is 5: 2
= 0.0956M
0.0956 moles → 1000cm3
X → 100cm3
X = 9.5 x 10-3 moles
MD = 0.0956
VD = 1000cm3
Mc =?
VC = 100
4. 25cm3 of sodium oxalate (Na2C2O4) solution acidified with dilute sulphuric acid and heated to 80°c is titrated with standard KMnO4 solution of concentration 3gl-1. 26.4cm3 of KMnO4 solution was required for complete reaction.
a.Calculate the concentration of oxalate ions in Na2C2O4 in gl-1
b.Hot acidified Na2C2O4 reacts with MnO2 according to the equation.
What volume of sodium oxalate (Na2C2O4) will be required to react with 0.1g of MnO2.
Conc. = x M.M

x = 1.149 x 10-3 moles of C2O4
(ii) Iodometry titration
These are titrations in which iodine is produced by using of starch as an indicator. All iodometry
titrations use KI as source of Iodine.

a)If the oxidant is acidified KMnO4
……………… (1)

The iodine produced is titrated with to pale yellow, then add starch and continue titrating until the blue-black colour is discharged.
Starch is not added at the beginning because the concentration of iodine is large. Iodine react with starch to form the blue black complex hence more volume of is required to discharge the blue black complex.
Reaction with Na2S2O3
When equation (1) and (2) are combined
1: 5
1: 5
b) If the oxidant is acidified K2Cr2O7

Mole ratio will be
1: 6

c) If the oxidant is acidified KIO3
The mole ratio is
1: 6

Metals have a small tendency to dissolve in solution of their ions producing cations leaving their valency electrons on the metal rod. The metal acquires a negative potential which prevents further release of cations and equilibrium is established
ne number of electron (s)
As a result the region of solution very close to the rod suffers an increase in charge while the rod carries a layer of negative charge (electrons). Then an electric double layer is set up and this layer is known as “Helmholtz double layer”
Whenever there is a separation of negative and positive charges we should be able to measure the voltage i.e. voltage between the electrode and surrounding solution and this is called Electrode potential.

Electrode potential – is the potential difference formed between an electrode and its hydrated ions.
Magnitude of electrode potential
This depends on the position of equilibrium of reversible reaction. The further to the right the greater is the electron density on the surface of metal and larger is the Potential Difference (P.D) between metal and solution. The opposite is true.
For a given metal the position of equilibrium forward or backward depends on the concentration of solution into which the electrode is dipped

If the concentration of the solution is high the equilibrium will lie towards the left i.e. tendency of zinc rod to dissolve decrease and vice versa.
For different metals placed in solution contain same concentration of their ions at the same temperature i.e. the positions of equilibrium is governed by the overall energy change forming hydrate ions from the metal.
Electrode potential involves the following stages
(i)Atomization of electrode
(ii) Ionization of gaseous atoms
(iii)hydration of ions
Atomization ionization hydration
Zn(s) → Zn (g) → Zn2+ Zn (aq) 2+
Energy Energy Energy
If the total energy is low, electrode dissolve more easily and its equilibrium will move forward hence large potential (electrode potential) value
Metals with large electrode potential release electrons easily and are good reducing agents

I. Metal – metal ion electrodes
This consists of metal dipped into its soluble salts.
II.Gaseous electrode
This consists of platinum to which a gas at 1 atm, and 25°c is bubbled and dipped in ions of gas at a given concentration.
III.Metal – insoluble salt electrode
This consists of metal dipped into its insoluble salts at
IV. Redox electrodes
It consists of platinum dipped in cations having different oxidation states at a given concentration at 25oc.

Measurements of electrode potential
We can always measure the electrode potential for an electrode in combination with standard electrode which is standard hydrogen electrode (SHE)
It was conventionally agreed that electrode potential for hydrogen is zero.
When electrode potential of any element is measured against hydrogen electrode at 1 atm and 1M concentration is called standard electrode potential.
Salt Bridge
Is an inverted U-tube that contains on electrolyte (e.g. ) which connects the solution of two half cells to balance the charge and to complete the circuit.

The standard redox potential is actually reduction potential. All elements below hydrogen have negative reduction potential and positive oxidation potential hence strong reducing agents.
All elements above hydrogen have positive reduction potential and negative oxidation potential hence strong oxidizing agents.

When writing the cell description the hydrogen electrode is placed on the left conventional to determine the polarity of the right hand electrode.

Functions of salt bridge
I.To complete the circuit
II.To balance the charge

Consider the following arrangements for determine electrode potential fo
I. =

The zinc electrode is negative.
Cell reaction:

The copper electrode is positive. (It can easily be reduced)
Cell reaction:

Application of standard electrode potential
I. Construction of electrochemical cells.
II. Prediction of occurrence of chemical reaction
III. Determination of of a solution without meter
IV. Replacement of elements in the electrochemical series.
Electrolytic cell = produce electrical power through chemical reaction in electrochemical cells.

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