Question #ebe51
1 Answer
Explanation:
The thing to keep in mind here is that you're dealing with a redox reaction in which two elements are being oxidized and only one is being reduced.
More specifically, the permanganate anion,
On the other hand, the iron(II) cations,
The unbalanced chemical equation looks like this -- I won't add the states to keep things simple, but keep in mind that the reaction takes place in aqueous solution
#"MnO"_ 4^(-) + overbrace(["Fe"^(2+) + "C"_ 2"O"_ 4^(2-)])^(color(purple)("FeC"_ 2"O"_ 4)) stackrel(color(darkgreen)("H"_ 2"SO"_ 4)color(white)(aaa))(->) "Mn"^(2+) + "Fe"^(3+) + "CO"_ 2#
The reduction half-reaction looks like this
#stackrel(color(blue)(+7))("Mn")"O"_ 4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn")""^(2+)#
The oxidation state of manganese goes from
Since you're in acidic medium, you can balance the oxygen atoms by adding water molecules and the hydrogen atoms by adding protons,
You will have
#8"H"^(+) + stackrel(color(blue)(+7))("Mn")"O"_ 4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn")""^(2+) + 4"H"_2"O"#
The oxidation half-reactions look like this
#stackrel(color(blue)(+2))("Fe")""^(2+) -> stackrel(color(blue)(+3))("Fe")""^(3+) + "e"^(-)#
and
#stackrel(color(blue)(+3))("C")_ 2 "O"_ 4^(2-) -> stackrel(color(blue)(+4))("C")"O"_ 2 + "e"^(-)#
Make sure that the carbon atoms are balanced. If each carbon atom loses one electron, it follows that two carbon atoms will lose a total fo two electrons
#stackrel(color(blue)(+3))("C")_ 2 "O"_ 4^(2-) -> 2stackrel(color(blue)(+4))("C")"O"_ 2 + 2"e"^(-)#
Now, a total of
#"1e"^(-) + 2"e"^(-) = "3e"^(-)#
are being lost in the two oxidation half-reactions and
As you know, in any redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.
To make this happen, multiply the oxidation half-reactions by
#{ (8"H"^(+) + stackrel(color(blue)(+7))("Mn")"O"_ 4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn")""^(2+) + 4"H"_ 2"O" | xx 3), (color(white)(aaaaaaaaaaaaa)stackrel(color(blue)(+2))("Fe")""^(2+) -> stackrel(color(blue)(+3))("Fe")""^(3+) + "e"^(-) color(white)(aaa)| xx 5), (color(white)(aaaaaaaaaaa)stackrel(color(blue)(+3))("C")_ 2 "O"_ 4^(2-) -> 2stackrel(color(blue)(+4))("C")"O"_ 2 + 2"e"^(-) color(white)(a)| xx 5) :}#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(aaaaaaaaaaaaaaaaaaa)#
This is equivalent to
#24"H"^(+) + 3"MnO"_ 4^(-) + color(purple)(5) xx ["Fe"^(2+) + "C"_ 2 "O"_ 4^(2-)] -> 3"Mn"^(2+) + 5"Fe"^(3+) + 10"CO"_ 2 + 12"H"_ 2 "O"#
As you can see, you need
#1 color(red)(cancel(color(black)("mole FeC"_2"O"_4))) * "3 moles MnO"_4^(-)/(color(purple)(5)color(red)(cancel(color(black)("moles FeC"_2"O"_4)))) = color(green)(|bar(ul(color(white)(a/a)color(black)("0.6 moles MnO"_4^(-))color(white)(a/a)|)))#