1Hint##n-factor is defined as the total change in oxidation number per mole of the substance
What is the n-factor of H2O2 in the reaction :
2H2O2 → 2H2O + O2
Please explain
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9 Answers
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2H2O2 → 2H2O + O2
change in oxidation number of O frm H2O2 to H2O = 1
change in oxidation number of O frm H2O2 to O2 = 2
SO, N-FACTOR = 1 / ( 1/2 + 1/1 ) = 1 / (3/2) = 2/3
1@aieeee
are u sure ??
i think n factor shld be 2 ( imo ,in such reactions we shld only look for either increase or decrease in ON to calculate the n factor)
1@ yes.... this is a standard method, when ders more thn one varying oxidation states. this is used in finding equivalent mass of such systems.
let M be the molecular weight. of a compound ( here hydrogen peroxide) involved in such a reaction.
we know, equivalent mass = M / n-factor
so, compound hs eq. mass = M/2 and M/1.
so, absolute equivalent mass = M/2 + M/1 = 3M/2 = M / (2/3)→ n-factor.
1357For such reaction
n-factor is calculated as something of like parallel combination in resistors.
i.e. 1/n=1/n1+1/n2
where n1 and n2 are the two different n-factors for reduction and oxidation reaction.
So here n1=2 and n2=2 (H2O2→H2O and H2O2→O2)
1/n=1/2+1/2=1
or n=1
1but sir
while calculating the volume strength of H2O2 the n-factor of H2O2 in reaction
[2H2O2 → 2H2O + O2 ] is used as '2'
1357Also remember n*n-factor=constant
So we have
n*n-factor(H2O2)=n*n-factor(H2O)=n*n-factor(O2)
2*n'=2*1=1*2
or n'=1