Disproportionation reactions are the reactions in which oxidising and reducing agents are same or the same element from the same compound is getting oxidised as well as reduced. n-factor of a disproportionation reaction can only be calculated using a balanced chemical reaction. We will categorize disproportionation reactions into two types.
(a) Disproportionation reactions in which moles of compound getting oxidised and reduced are same i.e. moles of oxidising agent and reducing agent are same. The n-factor for such compounds is calculated by either the number of mole of electrons lost or gained by
one mole of the compound because in such a case, n-factor of the compound acting as oxidizing agent or as reducing agent would be same.
For example, `2H_2O_2 -> 2H_2O + O_2`
Out of the `2` mole of `H_2O_2` used in reaction, one mole of `H_2O_2` gets oxidised to `O_2` (oxidation state of O changes from `- 1` to `0`) while the other mole of `H_2O_2` gets reduced to `H_2O` (oxidation state of `O` changes from `-1` to `-2`). When `1` mole of `H_2O_2` gets oxidised to `O_2` the half-reaction would be
`O_2^(-2) -> O_2^0 + 2e^(-)`
and when `1` mole of `H_2O_2` gets reduced to `H_2O` , the half-reaction would be `O_2^(2-) + 2e^(-) -> 2O^(-2)`
Thus, it is evident that one mole of `H_2O_2` (which is either getting oxidised or reduced) will lose or gain `2` mole of electrons. Therefore, n-factor of `H_2O_2` as oxidizing as well as reducing agent in this reaction is `2`. Thus,
`underset(n=2)(undersettext(reducing agent)(H_2O_2)) + underset(n=2)(undersettext(oxidising agent)(H_2O_2)) -> underset(n=1)(2H_2O) + underset(n=2)(O_2)`
Or when the reaction is written as
`2H_2O_2 -> 2H_2O + O_2`
where `H_2O_2` ia not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then n-factor calculation can be done in the following manner. Find the number of electrons exchanged (lost or gained) using the balanced equation and divide it by the number of moles of `H_2O_2` involved in the reaction. Thus, the n-factor of `H_2O_2` when the reaction is written without segregating oxidising and reducing agent is `2/2 =1`
`underset(n=1)(2H_2O_2) -> underset(n=1)(2H_2O) + underset(n=2)(O_2)`
(b) Disproportionation reactions in which moles of compound getting oxidised and reduced are not same i.e. moles of oxidising agent and reducing agent are not same.
For example,
`6Br_2 + 12 OH^(-) -> 10Br^(-) + 2BrO_3^(-) + 6H_2O`
In this reaction, the mole of electrons lost by the oxidation of some of the moles of `Br_2` are same as the number of mole of electrons gained by the reduction of rest of the moles of `Br_2`. Of the `6` moles of `Br_2` used, one mole is getting oxidized, losing `10` electrons (as reducing agent) and `5` moles of `Br_2` are getting reduced and accepts `10` moles of electron (as oxidizing agent).
`Br_2 -> 2 Br^(+5) + 10 e^-`
`5Br_2 + 10e^(-) -> 10 Br^-`
Total reaction is : `underset(n=10)(undersettext(reducing agent)(Br_2)) + underset(n=2)(undersettext(oxidising agent)(5Br_2)) -> underset(n=1)(10Br^(-)) + underset(n=5)(2Br^(+5))`
Thus, n-factor of `Br_2` acting as oxidizing agent is `2` and that `Br_2` acting as reducing agent has n-factor `10`.
Or when the reaction is written as
`6Br_2 -> 10Br^( -) + 2Br^(+5)`
where, `Br_2` is not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then for calculating n-factor of compound in such reactions, first find the total number of mole of electrons exchanged (lost or gained) using the balanced equation and divide it with the number of mole of `Br_2` involved in the reaction to get the number of mole of electrons exchanged by one mole of `Br_2`. In the overall reaction, the number of mole of electrons exchanged (lost or gained) is `10` and the moles of `Br_2` used in the reaction are `6`. Thus, each mole of `Br_2` has exchanged `10//6` or `5//3` mole of electrons. Therefore, the n-factor of `Br_2` when the reaction is written without segregating oxidising and reducing agent is `5//3`.
`underset(n=5//3)(6Br_2) -> underset(n=1)(10Br^(-)) + underset(n=5)(2Br^(+5))`
Disproportionation reactions are the reactions in which oxidising and reducing agents are same or the same element from the same compound is getting oxidised as well as reduced. n-factor of a disproportionation reaction can only be calculated using a balanced chemical reaction. We will categorize disproportionation reactions into two types.
(a) Disproportionation reactions in which moles of compound getting oxidised and reduced are same i.e. moles of oxidising agent and reducing agent are same. The n-factor for such compounds is calculated by either the number of mole of electrons lost or gained by
one mole of the compound because in such a case, n-factor of the compound acting as oxidizing agent or as reducing agent would be same.
For example, `2H_2O_2 -> 2H_2O + O_2`
Out of the `2` mole of `H_2O_2` used in reaction, one mole of `H_2O_2` gets oxidised to `O_2` (oxidation state of O changes from `- 1` to `0`) while the other mole of `H_2O_2` gets reduced to `H_2O` (oxidation state of `O` changes from `-1` to `-2`). When `1` mole of `H_2O_2` gets oxidised to `O_2` the half-reaction would be
`O_2^(-2) -> O_2^0 + 2e^(-)`
and when `1` mole of `H_2O_2` gets reduced to `H_2O` , the half-reaction would be `O_2^(2-) + 2e^(-) -> 2O^(-2)`
Thus, it is evident that one mole of `H_2O_2` (which is either getting oxidised or reduced) will lose or gain `2` mole of electrons. Therefore, n-factor of `H_2O_2` as oxidizing as well as reducing agent in this reaction is `2`. Thus,
`underset(n=2)(undersettext(reducing agent)(H_2O_2)) + underset(n=2)(undersettext(oxidising agent)(H_2O_2)) -> underset(n=1)(2H_2O) + underset(n=2)(O_2)`
Or when the reaction is written as
`2H_2O_2 -> 2H_2O + O_2`
where `H_2O_2` ia not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then n-factor calculation can be done in the following manner. Find the number of electrons exchanged (lost or gained) using the balanced equation and divide it by the number of moles of `H_2O_2` involved in the reaction. Thus, the n-factor of `H_2O_2` when the reaction is written without segregating oxidising and reducing agent is `2/2 =1`
`underset(n=1)(2H_2O_2) -> underset(n=1)(2H_2O) + underset(n=2)(O_2)`
(b) Disproportionation reactions in which moles of compound getting oxidised and reduced are not same i.e. moles of oxidising agent and reducing agent are not same.
For example,
`6Br_2 + 12 OH^(-) -> 10Br^(-) + 2BrO_3^(-) + 6H_2O`
In this reaction, the mole of electrons lost by the oxidation of some of the moles of `Br_2` are same as the number of mole of electrons gained by the reduction of rest of the moles of `Br_2`. Of the `6` moles of `Br_2` used, one mole is getting oxidized, losing `10` electrons (as reducing agent) and `5` moles of `Br_2` are getting reduced and accepts `10` moles of electron (as oxidizing agent).
`Br_2 -> 2 Br^(+5) + 10 e^-`
`5Br_2 + 10e^(-) -> 10 Br^-`
Total reaction is : `underset(n=10)(undersettext(reducing agent)(Br_2)) + underset(n=2)(undersettext(oxidising agent)(5Br_2)) -> underset(n=1)(10Br^(-)) + underset(n=5)(2Br^(+5))`
Thus, n-factor of `Br_2` acting as oxidizing agent is `2` and that `Br_2` acting as reducing agent has n-factor `10`.
Or when the reaction is written as
`6Br_2 -> 10Br^( -) + 2Br^(+5)`
where, `Br_2` is not distinguished as how much of it functions as oxidizing agent and how much as reducing agent, then for calculating n-factor of compound in such reactions, first find the total number of mole of electrons exchanged (lost or gained) using the balanced equation and divide it with the number of mole of `Br_2` involved in the reaction to get the number of mole of electrons exchanged by one mole of `Br_2`. In the overall reaction, the number of mole of electrons exchanged (lost or gained) is `10` and the moles of `Br_2` used in the reaction are `6`. Thus, each mole of `Br_2` has exchanged `10//6` or `5//3` mole of electrons. Therefore, the n-factor of `Br_2` when the reaction is written without segregating oxidising and reducing agent is `5//3`.
`underset(n=5//3)(6Br_2) -> underset(n=1)(10Br^(-)) + underset(n=5)(2Br^(+5))`