Question #3ef9d

1 Answer
Jul 3, 2017

Answer:

You have got TWO separate oxidation reactions, and I PRESUME that your starting material is #FeS# NOT #FeS_2#.......

Explanation:

#"Ferrous sulfide"# is oxidized to #"ferric oxide"#........

#2FeS(s) +3H_2O rarr Fe_2O_3 +2S^(2-) + 6H^(+) + 2e^(-)# #(i)#

#"Oxygen gas"# is reduced to #"oxide dianion"#

#O_2(g) +4e^(-) rarr 2O^(2-)# #(ii)#

And #"sulfide anion"# is oxidized to #SO_2#.........

#S^(2-) + 2H_2O(l) rarr SO_2(g)+4H^+ +6e^(-)# #(iii)#

And so (purely to eliminate the electrons on each side) let's take........

#(i) +(iii) + 2xx(ii)#.............

#2FeS(s) +H_2O(l)+2O_2(g) rarr Fe_2O_3 +S^(2-) + 2H^(+) +SO_2(g) #

And finally..........

#2FeS(s) +H_2O(l)+2O_2(g) rarr Fe_2O_3(s) +H_2S(g) +SO_2(g)#

Sulfide and #Fe(+II)# are oxidized to give #SO_2# and #Fe(III+)#, oxygen is reduced.............

Do you follow my reasoning........? It's balanced, but whether it corresponds to an actual chemical process is beyond me.......