Equivalent weight of H3PO2 when it disproportionate into PH3 and H3PO3 is :-
Answers
Answered by
81
so frnd ur answer is --
For disproportionation reactions, use a trick to find equivalent mass; E = E1 + E2; where E1 and E2 are equivalent masses of oxidation and reduction half reactions of same element.
Now, find the n-factor(total change in oxidation number per molecule) and then equivalent mass = molar mass/n-factor.
So, when H3PO2 changes into PH3, oxidation state of phosphorous changes from +1 to -3 so as n-factor is 4. Also when H3PO2 changes into H3PO4, oxidation state of phosphorous changes from +1 to +5 so as n-factor is again 4.
Now equivalent mass of H3PO2, E =(M/4) + (M/4) = M/2
hope it helps u!!!☺
For disproportionation reactions, use a trick to find equivalent mass; E = E1 + E2; where E1 and E2 are equivalent masses of oxidation and reduction half reactions of same element.
Now, find the n-factor(total change in oxidation number per molecule) and then equivalent mass = molar mass/n-factor.
So, when H3PO2 changes into PH3, oxidation state of phosphorous changes from +1 to -3 so as n-factor is 4. Also when H3PO2 changes into H3PO4, oxidation state of phosphorous changes from +1 to +5 so as n-factor is again 4.
Now equivalent mass of H3PO2, E =(M/4) + (M/4) = M/2
hope it helps u!!!☺
Answered by
126
Answer:
3M/4
Explanation:
3H3PO2 ------PH3 +2H3PO3
WE have equivalent weight=molecular weight/n factor
for such type of redox reaction the n factor is calculated as
no of electron exchanged/total number of mole used
H3PO2-------PH3
change in oxidation number is from 1 to -3 .Thus 4e- are used
from 2H3PO2 ----2H3PO2
Change in oxidation number is from 1 to +3 therefore electron given is 2e-.for 2 moleculae of H3PO2 thus 4 lectron are used
thus in total 4e- are exchanged
n factor=no of electron/no of mole=4/3
Thus the eq weight=M/(4/3)=3M/4 .
Similar questions