Three reactions used in the manufacture of sulphuric acid are shown.
S + O2 -> SO2
2SO2 + O2 -> 2SO3
SO3 + H2O -> H2SO4
Which of these reactions are redox reactions?
I dont understand how it is 1 and 2 only. Where is the reduction and oxidation going on each equation ?
Alright..here's the deal..First and foremost, you have to calculate the oxidation state of the Sulpher species..
1) It goes from 0 in Sulpher (natural element) to +4 i sulpher dioxide...
SO^2...X+(2x-2)=0
and so X=+4.....sulpher is oxidzed
2) It goes from +4 in sulpher dioxide (again) to +6 in sulpher trioxide..
SO^3...X+(-2x3)=0
X=+6..sulpher is oxididzed again
3) No redox reaction
It goes from +6 in sulpher trioxide to +6 in sulphuric acid...
H^2SO^4.....
The H ion is +1..and since there are 2 such ions...the sulphate ion SO^4 must have a charge of -2..
X+(4x-2)=-2
X=+6
Hope that helped..