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How to find valency factor for disproportionate reaction

How to find valency factor for disproportionate reaction 

Grade:9

1 Answers

Arun
25750 Points
5 years ago
Consider the reaction:
Hydrogen Peroxide------------> Water + Oxygen(gas)
On balancing it we have
2(Hydrogen Peroxide)----------> 2(Water) + Oxygen(gas)
To see how many electrons were exchanged we break it into a set of ionic equations-one for reduction and one for oxidation (in acidic medium)given below.

Oxidation: Hydrogen peroxide----------> Oxygen(gas) + 2(Hydrogen ion) + 2(electron)

Reduction: 2(Hydrogen ion) + 2(electron) + Hydrogen peroxide--------------> 2(Water)

It is easily seen that 2 electrons were exchanged for two molecules of Hydrogen peroxide. Thus, electrons exchanged per molecule is 2 divided by 2,i.e. 1. Hence,the true valency factor of Hydrogen peroxide in this reaction is 1.

If the breaking up of reaction into ionic reactions is difficult, one can follow an easier way, just calculate the valency factor either for oxidation or for reduction, not both. For example,consider the following reaction;
6(Bromine molecule)(0) + 12(Hydroxide ion)----------> 10(Bromine ion)(-1) + 2(Perbromic ion)(+5) 
(Numbers in brackets are the oxidation states of bromine atom in the molecule preceding them)

To calculate the valency factor of bromine molecule, we see that out of 12 bromine atoms in the bromine molecules, 10 get reduced to -1 oxidation state and 2 get oxidized to +5 oxidation state. So for oxidation, electron exchange=2*5=10. Dividing by total number of bromine molecules, i.e. 6, the Valency Factor of bromine molecule in the above disproportionation reaction is 10/6. If we look only reduction, then total electron exchange= 10*1=10. Dividing by total bromine molecules,i.e. 6, the Valency Factor is again 6. 

Any of the two methods given can be used and they never lead to wrong results, unless the ionic equations or the oxidation states are written wrong.

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