Start by identifying the oxidation state for each of the element:
#stackrel(color(navy)(bb(+2)))("Fe")stackrel(color(purple)(bb(+6)))("S")stackrel(-2)("O")_4 to stackrel(color(navy)(bb(+3)))("Fe"_2)stackrel(-2)("O")_3 + stackrel(color(purple)(bb(+4)))("S")stackrel(-2)("O")_2+stackrel(color(purple)(+6))("S")stackrel(-2)("O")_3color(white)(-)color(grey)("NOT BALANCED")#
Only three of the four chemicals- #"FeSO"_4#, #"Fe"_2"O"_3#, and #"SO"_2# are directly involved in the redox reaction.
The oxidation state of iron #"Fe"# has increased from
- #color(navy)(+2)# in #stackrel(color(navy)(bb(+2)))(bb("Fe"))stackrel(color(purple)(+6))("S")stackrel(-2)("O")_4# to
- #color(navy)(+3)# in #stackrel(color(navy)(bb(+3)))(bb("Fe")_2)stackrel(-2)("O")_3# by one
and therefore #"Fe"# is oxidized.
The oxidation state of sulfur #"S"# has declined from
- #color(purple)(+6)# in #stackrel(color(navy)(+2))("Fe")stackrel(color(purple)(bb(+6)))(bb("S"))stackrel(-2)("O")_4# to
- #color(purple)(+4)# in #stackrel(color(purple)(bb(+4)))(bb("S"))stackrel(-2)("O")_2# by two
and therefore some of the sulfur atoms have been reduced.
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The total increases in oxidation numbers shall be the same as the sum of decreases in oxidation numbers in a balanced redox reaction.
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The oxidation number increases by #1# for each mole of #"Fe"# atom oxidized and decreases by #2# for each mole of #"S"# oxidized.
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Therefore for each mole of #"S"# reduced, two moles of #"Fe"# atoms shall be oxidized.
Note that sulfur dioxide, #stackrel(color(purple)(bb(+4)))(bb("S")) stackrel(-2)("O")_2#, is the only species containing sulfur atoms of oxidation state #color(purple)(+4)#. Thus all of the reduced sulfur atoms would end up in #"SO"_2#.
The number of moles of sulfur atoms reduced shall therefore equal to the number of #"SO"_2# molecules produced. This number would be slightly smaller than that of #stackrel(color(navy)(+2))("Fe")stackrel(color(purple)(bb(+6)))(bb("S"))stackrel(-2)("O")_4# given that sulfur atoms that were not reduced got eventually into #stackrel(color(purple)(+6))("S")stackrel(-2)("O")_3#.
Add coefficients #2# and #1# in front of #"FeSO"_4# and #"SO"_2#, respectively:
#color(green)(2)color(white)(l)stackrel(color(navy)(bb(+2)))("Fe")stackrel(color(purple)(bb(+6)))("S")stackrel(-2)("O")_4 to stackrel(color(navy)(bb(+3)))("Fe"_2)stackrel(-2)("O")_3 + color(green)(1)color(white)(l)stackrel(color(purple)(bb(+4)))("S")stackrel(-2)("O")_2+stackrel(color(purple)(+6))("S")stackrel(-2)("O")_3color(white)(-)color(grey)("NOT YET BALANCED")#
Deduce coefficients for the rest of the species based on the conservation of iron and sulfur atoms.
#color(darkgreen)(2)color(white)(l)stackrel(color(navy)(bb(+2)))("Fe")stackrel(color(purple)(bb(+6)))("S")stackrel(-2)("O")_4 to color(green)(1)color(white)(l)stackrel(color(navy)(bb(+3)))("Fe"_2)stackrel(-2)("O")_3 + color(darkgreen)(1)color(white)(l)stackrel(color(purple)(bb(+4)))("S")stackrel(-2)("O")_2+color(green)(1)color(white)(l)stackrel(color(purple)(+6))("S")stackrel(-2)("O")_3color(white)(-)color(grey)("BALANCED")#
Take coefficients "#1#" out of the expression:
#2color(white)(l)"FeSO"_4 to "Fe"_2"O"_3 + "SO"_2 + "SO"_3#
