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Redox
The term REDOX stands for RED UCTION- OXID ATION. Oxidation can be defined as gain of oxygen or loss of hydrogen. Reduction can be defined as loss of oxygen or gain of hydrogen.
The most important definition is given in terms of electrons. OXIDATION is LOSS of ELECTRONS REDUCTION is GAIN of ELECTRONS
One way of accounting for electrons is to use OXIDATION NUMBERS.
e.g. Fe2+^ needs to gain two electrons for it to become neutral iron atom therefore its oxidation number is +2.
Using oxidation numbers it is possible to decide whether redox has occurred. Increase in oxidation number is oxidation. Decrease in oxidation number is reduction.
We can apply a series of rules to assign an oxidation state to each atom in a substance.
Oxidation number
The oxidation number of an atom shows the number of electrons which it has lost or gained as a result of forming a compound
Oxidation Number Rules
- The oxidation number of an uncombined element is 0.
- Certain elements have fixed oxidation numbers. All group 1 elements are +1. All group 2 elements are +2. Hydrogen is always +1 except in hydrides. Fluorine is always –1. Oxygen is always –2 except in peroxides, superoxides and when combined with fluorine. Chlorine is always –1 except when combined with fluorine and oxygen.
- The sum of oxidation numbers in a compound is always 0.
- The sum of oxidation numbers in an ion always adds up to the charge on the ion.
Examples
- The oxidation number of S in H 2 SO
- H 2 S O
- 2 x +1? 4 x -2 =
- +2? -8 =
- +2 +6 -8 =
- S 2 O 2. The oxidation number of S in S 2 O 8 2- -? 8 x -2 = - -? -16 = -
- +14 -16 = -
- S = +
- The oxidation number of Cl in NaClO
- Na Cl O
- +1? 3 x -2 =
- +1? -6 =
- +1 +5 -6 =
- Mn O 4. The oxidation number of Mn in MnO 4 - -? 4 x -2 = - -? -8 = -
- +7 -8 = -
- Mn = +
Oxidation number and redox reactions
When a redox reaction occurs an electron transfer takes place and so the oxidation numbers of the substances involved changes.
Consider the following reaction: 2HOBr + 2H+^ + 2I-^ Br 2 + I 2 + 2H 2 O
Reactants Products Species Oxid’n No Species Oxid’n No H in HOBr +1 Br in Br 2 0 O in HOBr -2 I in I 2 0 Br in HOBr +1 H in H 2 O + H+^ +1 O in H 2 O - I-^ -
The table shows us that the oxidation number of Br goes from +1 to 0, so it is reduced. The iodine goes from -1 to 0, so this is oxidised.
Another example 3NaOCl 2NaCl + NaClO 3
Reactants Products Species Oxid’n No Species Oxid’n No Na in NaOCl +1 Na in NaCl + O in NaOCl -2 Na in NaClO 3 + Cl in NaOCl +1 Cl in NaCl - Cl in NaClO 3 + O in NaClO 3 -
In this reaction the Cl in NaOCl is oxidised in one reaction to +5 and in another reaction is reduced to -1. Such an occurrence is called disproportionation.
Disproportionation takes place a particular species undergoes simultaneous oxidation and reduction.
Half Equations
When a redox reaction occurs, one substance gains electrons and one substance losed electrons. These two processes can be considered separately.
Using the example of magnesium and copper sulphate:
Electron gain Cu2+(aq) + 2e-^ Cu(s)
Electron loss Mg(s) Mg2+(aq) + 2e-
These are called half equations.
Constructing Half Equations
Half equations can be constructed as follows: a) Add H 2 O molecules to balance any oxygen atoms b) Add H+^ ions to balance any hydrogen atoms c) Add electrons to balance any charge in the equation.
NB – To write a balanced half equation you may only add; H 2 O molecules H+^ ions OH-^ ions (not usually done) Electrons
e.g. Construct a half equation for: NO 3 -^ NH 4 +
a) balance oxygen atoms with water N O 3 -^ NH 4 +^ + 3 H 2 O
b) balance hydrogen atoms with hydrogen ions NO 3 -^ + 10H+ N H 4 +^ + 3H 2 O
c) balance the charges using electrons 8e-^ + NO 3 -^ + 10 H + NH 4 +^ + 3H 2 O
Further example.
Construct a half equation for: Cr 2 O 7 2-^ 2Cr 3+
a) balance oxygen atoms with water Cr 2 O 7 2-^ 2Cr 3+^ + 7H 2 O
b) balance hydrogen atoms with hydrogen ions 14H+^ + Cr 2 O 7 2- 2Cr3+^ + 7H 2 O
c) balance the charges using electrons 6e-^ + 14 H +^ + Cr 2 O 7 2-^ 2 Cr 3+^ + 7H 2 O
