1.

In redox reactions, oxidation and reduction occur simultaneously

2.

Oxidation is defined as the loss of electrons.
ex. Zn ® Zn²⁺ + 2 e^- [oxidation half-equation]
In this reaction, Zn loses 2 electrons to become a Zinc ion, so Zn is oxidised to Zn²⁺ ion.

3.

Reduction is defined as the gain of electrons.
ex. Cu²⁺ + 2 e^- ® Cu [reduction half-equation]
In this reaction, colour ions are reduced to copper atoms after the gain of 2 electrons. Colour change is denoted.

4.

Overall balanced equation for the redox reaction:
Zn (s) + Cu²⁺ (aq) ® Zn²⁺ (aq) + Cu (s)

5.

Oxidising agents (Oxidants) are defined as substances which accept electrons.

Reducing agents (reductants) are defined as substances which donate electrons.

If one substance is oxidised, another must be reduced, hence oxidation and reduction occur simultaneously.
Redox reactions is not only confined to processes like burning, rusting and respiration but also embraces all electron transfer processes.

6.

To determine whether oxidation and reduction has occurred, examine the charges of the species involved in ionic equation.

DECREASE in charge (reduction)
e.g. Cu²⁺ + 2 e^- ® Cu
charge: 2+ 0
INCREASE in charge (oxidation)
e.g. Fe^{2+ ®} Fe³⁺ + e^-
charge: 2+ 3+

However, since no charges exist on atoms in covalent molecules, use of oxidation numbers or oxidation states determines whether a non-ionic substance is oxidised or reduced.

 1.

Reaction of metal with a non-metal
Si + 2 Cl₂ ® SiCl₄

2.

Reaction of metals with water
Ca + 2 H₂O ® Ca(OH)₂

3.

Reaction of metals with acids
Zn + 2 HCl ® ZnCl₂

4.

Reactions at the electrodes during electrolysis
Pb²⁺ + 2 e^- ® ZnCl2 (at the cathode where reduction occurs)
2 Br^- ® Br₂ + 2 e^- (at the anode where oxidation occurs)

1.

The Oxidation Number or Oxidation State of an element in a compound represents the amount of oxidation or reduction necessary to convert on atom of the element from the free (elemental) state (where oxidation state is zero) to that in the compound.

2.

If loss of electrons (oxidation) is required for the change, the oxidation state is POSITIVE.
e.g. Fe ® Fe²⁺ + 2 e^-
Oxidation state: 0 +2
If gain of electrons (reduction) is required for the change, the oxidation state is NEGATIVE.
e.g. S + 2 e^- ® S^2-
Oxidation state: 0 -2

3.

The oxidation number does not represent an actual charge on the atom and should not be confused with ionic charges. It represents the electrical charge the atom would have if the electrons in the compound were assigned according to certain rules.

4.

The convention is that in oxidation states, the sign always PRECEDES the number.
Oxidation state of oxygen in the oxide ion, O^2-, is -2 ; the charge on the oxygen atom the is 2-.
It should now be apparent that oxidation involves an increase in oxidation state while reduction involves a decrease in oxidation state since increasing the oxidation state requires loss of electrons and decreasing the oxidation requires gain of electrons.

1.

The oxidation state of an element in its elemental state is zero.
i.e. O₂, Cl₂, Br₂, P₄, S₈, Ag, Fe, etc

2.

The oxidation state of a monatomic ion is the same as its charge.
e.g. K⁺ is in the +1 oxidation state, Al³⁺ is in the +3 oxidation state and Cl^- is in -1 oxidation state.

3.

Hydrogen is usually +1, except when combined with electropositive metals such as Na, Mg, Ca, where it is in the -1 oxidation state.
Metal hydrides : Na⁺H^-, MgH₂, CaH₂

4.

Oxygen is usually -2, except in peroxides and superoxides, which contain oxygen in a lower oxidation state. Oxygen is positive when bonded to fluorine.

5.

The halogens have a -1 oxidation state in binary halides whether ionic or covalen, when bonded to a less electronegative element. Fluorine is always -1, but the other halogens may exhibit positive oxidation states.

6.

The oxidation states of all the elements in a neutral compound add to zero. In the case of polyatomic ions (e.g. SO₄^2-, C₂O₄^2-), the sum of the oxidation states of elements equals the charge on the ion.
O.N. of S = +6
SO₄^2- : O.N. = (-2)4 + 6 = -2

7.

In compounds of two different atoms, negative oxidation states are assigned to the more electronegative elements.
e.g. In HCl, oxidation state of H is +1 since Cl is more electronegative than H. In F2O, the oxidation state of F is -1 and that of oxygen is +2 since fluorine is more electronegative of the two.

1.

A redox reaction involves two half-equations; one representing oxidation, the other reduction.

2.

The two half-equations are added to give the overall redox reaction, in such a way that the number of electrons lost by the reducing agent (in the oxidation half-equation) equals the number of electrons gained by the oxidising agent (in the reduction half-equation).

3.

If an element undergoes an increase in oxidation state then it is a REDUCING AGENT (the substance oxidised), conversely if the element undergoes a decrease in oxidation state, then it is the OXIDISING AGENT (the substance reduced).

1.

Write the net ionic equation, without trying to balance it.

e.g. the rxn between iron(II) sulphate and potassium manganate (VII) in acid solution
Fe²⁺ + MnO₄^- ® Fe³⁺ + Mn²⁺

2.

Write the two partial equations : an oxidation half-equation and a reduction half-equation.

[O] : Fe²⁺ ® Fe³⁺
(+2) (+3)
[R] : MnO₄^- ® Mn²⁺
(+7) (+2)

3.

Balance the ATOMS on each side of the partial equation.
In acid solution, H⁺ ions and H₂O molecules may be added
In alkaline solution, OH- ions and H₂O molecules may be added.
MnO₄^- + 8 H⁺ ® Mn²⁺ + 4 H₂O

4.

Balance the two half-equations electrically by adding ELECTRONS to the appropriate side of the equation so that the charges on both sides of the half-equations are equal.
[O] involves loss of electrons, so the electrons are added to the product side.
[R] involves gain of electrons, so the electrons are added to the reactant side.

[O] : Fe²⁺ ® Fe³⁺ + e^-
[R] : MnO₄^- + 8 H⁺ + 5 e^- ® Mn²⁺ + 4 H₂O

5.

Multiply each half-equation by an appropriate number so that the electrons lost in the oxidation equal the electrons gained in the reduction.

[O] : 5 Fe²⁺ ® 5 Fe³⁺ + 5 e^-
[R] : MnO₄^- + 8 H⁺ + 5 e^- ® Mn²⁺ + 4 H₂O

6.

Add the two half-equations and eliminate electrons, ions or water molecules that appear on both sides of the equation:

MnO₄^- + 5 Fe²⁺ + 8 H⁺ ® 5 Fe³⁺ + Mn²⁺ + 4 H₂O ^-

7.

Check that the equation is balanced on both sides for each element and that the net charges are equal on both sides.

The final equation is:
MnO₄^- + 5 Fe²⁺ + 8 H⁺ ® 5 Fe³⁺ + Mn²⁺ + 4 H₂O ^-

Try this #1: C₂O₄^2- + MnO₄^- ® CO₂ + Mn²⁺ + H₂O in acid medium
Try this #2: Cl₂ ® ClO^- + Cl^- + H₂O in alkaline medium

1.

Main types of redox titrations are based on the use of the following reagents:
(i) potassium permanganate
(ii) potassium dichromate
(iii) sodium thiosulphate

2.

Potassium permanganate functions as an oxidising agent in acidic conditions.
MnO₄^- + 8 H⁺ + 5 e^- ® Mn²⁺ + 4 H₂O
One mole of MnO₄^- ions can accept 5 moles of electrons.

Acidic KMnO₄ solution is used mainly in redox titrations with iron (II) salts, ethanedioates (oxalates) and hydrogen peroxide solution.

Ionic half-equation for iron(II) ions donating electrons is: Fe²⁺ ® Fe³⁺ + e^-
One mole of MnO₄^- ions will oxidise 5 moles of Fe²⁺ ions

Ionic half-equation for oxidation of ethanedioate (oxalate) ions is: C₂O₄^2- ® 2 CO₂ + 2e^-
2 moles of MnO₄^- ions will oxidise accept 5 moles of C₂O₄^2- ions

Ionic half-equation for hydrogen peroxide acting as a reducing agent is:
H₂O₂ ® 2 H⁺ + O₂ + 2 e^-
2 moles of MnO₄^- ions will oxidise accept 5 moles of H₂O₂ .

3.

Potassium Dichromate acts as an oxidising agent in acidic conditions.
Cr₂O₇^2- + 14 H⁺ + 6 e^- ® 2 Cr³⁺ + 7 H₂O
One mole of K₂Cr₂O₇ will oxidise 6 moles of Fe²⁺ ions.

K₂Cr₂O₇ has one particular advantage over KMnO₄ in that it can be used to estimate iron(II) ions in the presence of chloride ions (without oxidising the chloride ions to chlorine).
KMnO₄ will oxidise both the iron(II) ions and the chloride ions: 2 Cl^- ® Cl₂ + 2 e^-

4.

Na₂S₂O₃ reacts as a reducing agent towards iodine.
2 S₂O₃^2- ® S₄O₆^2- + 2 e^-
Simultaneously, the iodine (potassium iodide solution) is reduced by accepting the electrons: I₂ + 2 e^- ® 2 I^-
Þ I₂ + 2 S₂O₃^2- ® 2 I^- + S₄O₆^2-
Þ One mole of S₂O₃^2- ions will reduce 1/2 moles of I₂.

Standard sodium thiosulphate solution is used to estimate iodine or, more frequently, any oxidising agent with liberates iodine quantitatively from KI solution.

Some common oxidising agents which would liberate iodine from iodide ions include dichromate (VI), iodate (V), bromate (V), manganate (VII), chromate (VI), chlorate (I), chlorine and bromine.
Br₂ (aq) + 2 I^- (aq) ® 2 Br^- (aq) + 1 I₂ (aq)
i.e. 1 mole Br₂ º 1 mol I₂

The quantity of iodine liberated by the oxidising agent is determined by titratiions with standard Na₂S₂O₃ solution using starch as indicator.

Thus the stoichiometric ratio between the oxidising agent Br₂ and Na₂S₂O₃ may be derived: 1 mole Br₂ º 1 mol I₂ º 2 mol S₂O₃^2-

5.

Try this #3: A standard solution is made by dissolving 1.02 g of K₂Cr₂O₇ and making up to 250 cm³. A 25.0 cm³ portion is added to an excess of KI and dilute sulphuric acid. The iodine liberated is titrated with sodium thiosulphate solution. 18.50 cm³ of thiosulphate solution are needed. Calculate the concentration of the thiosulphate solution in mol dm^-3. [0.113]

Try this #4: A sample of 4.25 g of copper (II) sulphate-5-water is dissolved and made up to 250 cm³. A 25.0 cm³ portion is added to an excess of KI. The iodine liberated required 18.00 cm³ of a 0.0950 mol dm^-3 solution of sodium thiosulphate for reduction. Calculate the percentage by mass of copper in the crystals. [25.6%]

Try this #5: 12.0g of iron(II) ammonium sulphate crystals were made up to 250 cm³ of acidified aqueous solution. 25.0 cm³ of this solution required 25.50 cm³ of 0.0200 mol dm^-3 K₂Cr₂O₇ for oxidation. Calculate x in the formula FeSO₄.(NH₄)₂SO₄.xH₂O. [6]

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