Electrochemistry
6.1 Oxidation and Reduction
Redox reactions
1) Oxidation is:
i. gain of oxygen
ii. loss of hydrogen
iii. loss of electrons
iv. increase in oxidation number
2) Reduction is:
i. gain of hydrogen
ii. loss of oxygen
iii. gain of electrons
iv. decrease in oxidation number
3) Oxidation and reduction always occur simultaneously in a chemical reaction.
Such reaction is called a redox reaction. In a redox reaction, one substance
must be oxidised and another must be reduced.
4) i. An oxidising agent is a substance which oxidises another substance and itself is
reduced.
ii. A reducing agent is a substance which reduces another substance and itself is
oxidised.
5) In short, an oxidising agent undergoes reduction while a reducing agent
undergoes oxidation.
6) Disproportionation is a redox reaction in which both oxidation and
reduction occurs on the same atom. The atom is simultaneously oxidised and
reduced.
Oxidation state (oxidation number) of a substance
1) Oxidation state shows that total number of electrons which have been removed
from or added to an element to get to its present state.
2) i. When electrons have been removed, the oxidation number increases.(positive)
ii. When electrons have been added, the oxidation number decreases.(negative)
3) i. Since removing electrons is an oxidation process, therefore oxidation is the
increase in oxidation number
ii. Since adding electrons is a reduction process, therefore reduction is the
decrease in oxidation number.
3) For example, from V to V²⁺, two electrons have been removed, therefore the
oxidation state of is +2. From V to V³⁺, three electrons have been removed,
therefore the oxidation state is +3. Removing another electron gives:
V³⁺ + H2O → VO²⁺ + 2H⁺ + e⁻
Four electrons have been removed starting from V, therefore the oxidation state
is +4. In all cases, V has been oxidised.
4) Another example, from S to S²⁻, two electrons have been added, therefore the
oxidation state is -2. S is said to have been reduced.
Rules to determine oxidation numbers of a substance
1) All atoms in an atom, molecule or ion can be given an oxidation number.
2) The rules to determine the oxidation number of a substance:
i. All free atoms in elements have an oxidation number of zero.
e.g.
ii. For simple ions, the oxidation number is the same as the charge on the ion.
e.g.
iii. For polyatomic ion, the sum of all the oxidation numbers of the atoms in
the ion is equal to the charge of the ion.
e.g.
iv. For a neutral covalent molecule, the sum of all the oxidation numbers of
the atoms in the molecule is equal to zero.
e.g.
Also, the more electronegative atom is always given a negative oxidation
number while the less electronegative atom is given a positive one.
v. For Group I and Group II elements, their oxidation number are always +1
and +2 respectively. For aluminium, it is always +3.
vi. For hydrogen, its oxidation number is always +1 except in metal hydrides.
For example, NaH, where its oxidation number is -1.
vii. For oxygen, its oxidation number is always -2 except in peroxides and
fluorine compounds. For example, BaO2, where its oxidation number is -1.
viii. For fluorine, its oxidation number is always -1, with no exceptions.
3) A summary:
4) To work out the oxidation number of a particular atom in a molecule/ ion, find
the sum of all the oxidation number of the atoms present and equate it to zero/
charge of the ion. An example:
Balancing redox equations
1) There are two methods to balance complicated redox equations:
i. Using electron half-equations.
ii. Using changes in oxidation number.
2) Using electron half-equations:
i. In this method, the redox equation is divided into two half-equations. One for
oxidation and another for reduction.
ii. Steps(in acidic condition):
- Divide the equation or information given into two half-equations.
- Balance all other elements other than oxygen and hydrogen.
- Balance the oxygen by adding H2O to the appropriate side of the equation.
- Balance the hydrogen by adding H⁺ to the appropriate side of the equation.
- Balance the charge by adding electrons to the appropriate side of the
equation.
- Combine two half-equations such that the electrons cancel out each other.
iii. An example:
Q: Balance and complete this equation:
Cr2O7²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺
A: 1) Divide into two equations:
Cr2O7²⁻ → Cr³⁺..........reduction
Fe²⁺ → Fe³⁺................oxidation
2) Balance the atoms:
Cr2O7²⁻ → 2Cr³⁺
Fe²⁺ → Fe³⁺
3) Balance the oxygen:
Cr2O7²⁻ → 2Cr³⁺ + 7H2O
Fe²⁺ → Fe³⁺
4) Balance the hydrogen:
Cr2O7²⁻ +14H⁺ → 2Cr³⁺ + 7H2O
Fe²⁺ → Fe³⁺
5) Balance the charge:
Cr2O7²⁻ +14H⁺ + 6e⁻ → 2Cr³⁺ + 7H2O
Fe²⁺ → Fe³⁺ + e⁻
6) Cancel the electrons and combine the equations:
[ Cr2O7²⁻ +14H⁺ + 6e⁻ → 2Cr³⁺ + 7H2O ] x 1
[ Fe²⁺ → Fe³⁺ + e⁻ ] x 6
the combine....
Cr2O7²⁻ +14H⁺ + 6e⁻ → 2Cr³⁺ + 7H2O
+ 6Fe²⁺ → 6Fe³⁺ + 6e⁻
-----------------------------------------------------------
Cr2O7²⁻ +14H⁺ + 6e⁻ + 6Fe²⁺ → 2Cr³⁺ + 7H2O + 6Fe³⁺ + 6e⁻
same number of electrons on both sides, cancel it....
Cr2O7²⁻ +14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H2O + 6Fe³⁺
is the final answer.
iv. Steps(in alkaline condition):
- Balance the equation as if it happens in an acidic condition first.
- Add OH⁻ to both sides of the equation to react with all the H⁺ to form H2O.
- Cancel the excess H2O on either side of the equation.
Note: If it is obvious enough that OH⁻ must be added in order to
balance the equation, then add OH⁻ instead.
3) Using changes in oxidation number:
i. This method utilises the fact that an increase in certain amount of oxidation
number in a substance must be accompanied by a decrease in same amount
of oxidation number in another substance.
ii. An example:
Q: Balance and complete this equation:
Cr2O7²⁻ + Fe²⁺ → Cr³⁺ + Fe³⁺
A: 1) Calculate the changes in oxidation number:
For Cr: +6 to +3, change is -3
For Fe: +2 to +3, change is +1
2) Balance the changes:
For Cr: -3(1) = -3
For Fe: +1(3) = +3
1Cr2O7²⁻ + 3Fe²⁺ → 1Cr³⁺ + 3Fe³⁺
3) Multiply everything by 2 except Cr2O7²⁻ because there are already two
atoms of chromium inside it.
Cr2O7²⁻ + 6Fe²⁺ → 2Cr³⁺ + 6Fe³⁺
4) Balance the oxygen and hydrogen:
Cr2O7²⁻ + 6Fe²⁺ +14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H2O
Cr2O7²⁻ +14H⁺ + 6Fe²⁺ → 2Cr³⁺ + 7H2O + 6Fe³⁺ is the final answer.
Systematic names of compounds
FeCl2: Iron (II) chloride: ox. no. of 2 Cl atoms is -2 and FeCl2 has overall no charge - ox. no. of Fe is +2
HClO4: Chloric (VII) acid: ox. no. of H is +1, 4 O atoms is -8 and HClO4 has overall no charge so ox. no.
of Cl is +7
NO2: Nitrogen (IV) oxide: ox. no. of 2 O atoms is -4 and NO2 has overall no charge so ox. no. of N is +4
Mg (NO3)2: Magnesium nitrate: this is a salt of the common acid, so they are named without
including the ox. no. of the non-metal
K2SO4: Potassium sulfate: this is a salt of the common acid, so they are named without including the
ox. no. of the non-metal
ClO3-: Chlorate (v) ions: the chlorine atom has a oxidation state +5
Writing ionic equations
