A reaction in which both oxidation and reduction take place simultaneously is called redox (reduction + oxidation) reaction.
In this reaction Ox. No. of hydrogen increases from zero to +1 (oxidation) and that of sulphur decreases from zero to -2 (reduction). In this reaction hydrogen is reducing agent and S is oxidising agent.
In this reaction Ox. No. of Fe increases from +2 to +3 (oxidation) and that of chlorine decreases from zero to (reduction). Hence is reducing agent and is oxidising agent.
In this reaction Ox. No. of hydrogen remains the same, i. e., +1, but the Ox. No. of oxygen in are -1, -2 and 0 respectively. Hence, in this reaction oxygen is undergoing oxidation as well as reduction. Therefore acts both as oxidising and reducing agent. Such a molecule is said to auto-oxidise and the process (reaction) as auto-oxidation. Decomposition of is also an example of auto-oxidation.
In this reaction oxygen is oxidised while nitrogen is reduced. Therefore is acting as both an oxidising and reducing agent.
Such reactions are often called disproportion reactions.
all type of cannizzaro reactions are e. g. , of disproportion.
Intramolecular Redox Reaction: Here in any compound one element is oxidised while the other element is reduced.
Don’t forget these
Loss of electrons = Oxidation
Increase in Ox. No. = Oxidation
Gain of electrons = Reduction
Decrease in Ox. No. = Reduction
Reduction + Oxidation = Redox
Oxidants Oxidize other
Reductants Reduce other
Calculation of Oxidation Number: With the help of the rules mentioned above the oxidation number of any one element in a compound can be calculated. For example;
(i) Oxidation number of chromium in : Let the oxidation number of Cr be x in .Now we know that oxidation numbers of K and O are +1 and -2 respectively, thus
Or 2(+1) + 2 (x) +7 (-2) = 0
Or 2 + 2x -14 = 0
Or x = 6
(ii) Oxidation number of Ta in : Let the oxidation number of Ta be x in . Now we know that oxidation numbers of K and F are +1 and -1 respectively, thus
Or (2)(+1) + x + 7 (-1) = 0
Or 2 + x – 7 = 0
Or x = 5
(iii) Oxidation number of carbon in its various compounds may be determined as follows:
(iv) Oxidation number of nitrogen in its various compounds may be determined as follows:
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