1didnt read ur one bro just saw the post where u mentioned that u thought it was 4...anywaz the main funda is it gave me an opportunity to recall the concept....
What is the n-factor of Fe2(SO4)3 in the following reaction:
Fe2(SO4)3 --> FeSO4 + SO2 ?
-
UP 0 DOWN 0 1 26
26 Answers
1
1@ Ishan ..you are making a mistake here
In 4e + 4H+ + Fe2(SO4)3 --> 2FeSO4 + SO2 + 2H2O
2 S atom in Reactant will not change their state only one S atom will change .
So n factor will be = 2*1 +2*1
=4
1It would be good if someone stated the exact rules for determining the n-factor.....Why is this confusion arising?
1wel the correct answer is 2.
(Fe+3)2 + 2e → 2(Fe+2)
electrons taken by one mole of Fe2(SO4)3 is 2.
24ans is indeed 4...
for n factor we need to balance only the oxidised and reduced elements..which has been given by manish sir [1]
1GUYZZZ ... Manish Sir....where am i wrong ?????
S atoms are getting converted to both states in FeSO4 and SO2. so the net change in oxidation state per S atom is 2-4 = -2
So for 3 S atoms .. it's 3 x (-2) = -6
106no aveek and ishan u are wrong.
See the method I had given was mine (for verification only)
The answer is indeed 4
In Fe2(SO4)3 the OS (oxdn state) of Fe is +3 which reduces to +2 and in SO42-, S has +6 . Out of 3-Sulphurs only one of them gets reduced to SO2. So corresponding ON change = 6-4 = 2
So n-factor = 2*change in Fe(as there are 2moles per molecule) + 2*change in S (which are not reduced) + change in S(which is reduced) = 2*1 + 2*0 + 2 = 4
1357Fe2(SO4)3→2FeSO4+SO2
Only one of the 3 S in the reactants is getting reduced.
This may help you solving it
1Balancing will need more time and simply waste the importance of oxidation state which requires less time and is much easier
1for 1 mole change is of 4 so for 2 moles (after balncg) change is of 8 ....thus n=8
49Fe2(SO4)3, S=+6, Fe=+3, Fe2=+6
SO2, S=+4
FeSO4, Fe=+2, 2Fe=+4
n factor=6-4=2
1@Ishan.....u r right dude.
And there's no difference between your method and mine.
1n factor = 8
see both Fe and S are getting reduced hence they both will get added as ashish said
In Fe2(SO4)3 the OS of Fe is +3 and in FeSO4 the OS of Fe is +2
hence n=1*2 (as the change is considered per mole of compound and it contains 2 Fe atoms)
i.e n=2
similarly in the reactant OS of S is +8 and it gets reduced to +6
hence n=2*3=6
as both the n are for reduction hence nnet=6+2=8
1I eagerly wait for anyone to prove me wrong......NISHANT/Rajiv SIR DER ????
1U have to consider SO42- ion here as a whole that is doing the work.
Let x be ox. state of S.
Then x - 8 = -6
x= +6
Now you have Fe2(SO4)3 . So there are 3 S atoms. So...ox state of each S atom is +6/3 = +2.
1Where you guys went wrong was that you had not considered S being oxidised . And Asish i don't know why but the method Akash has given (as you have posted) is a pure crap. If you HAVE to balance then what's the use of N-factor ?????
1ANSWER IS 8 AS IS GIVEN IN THE PROBLEM DUDE.
Go on with a simpler logic :
Fe2(SO4)3 : the oxidation state of Fe = +3
FeSO4 : ox. state of Fe = +2
ox state of S in Fe2(SO4)3 = +2
ox. state of S in SO2 = +4
Hence n-factor = | 2(2-3) | + | 3(2-4) |
= 2+6
= 8
(Ans. )
106yeah.. i know
but this was a question from aakash aits 2011 amd they had given answer as 8 (so doubted)
simply balance the equation:
4e + 4H+ + Fe2(SO4)3 --> 2FeSO4 + SO2 + 2H2O
so n-factor = 4
additional questions:
in above:
find n-factor of FeSO4 and SO2 ans: 2 and 4 respectively
FeS2 --> Fe3+ + SO2 n factor of FeS2
11See
Fe goes from +3 to +2 therefore change is +1*2 since 2 moles
Now S goes from + 6 to +6 for 2 mole
" " +6 to +4 for 1 mole therefore change is 2
Now there has to be a O2 at the product for balancing
Thereofore O goes from -2 to -2 for 10 moles
Therefore O goes from -2 to 0 for 2 moles n=4
Therefore LCM(4,4)=4
106ans given (8) while i think (4)
@subho: both Fe and S are getting reduced. So shouldnt the changes of both be added?