# Question ebe51

Jul 8, 2016

${\text{0.6 moles MnO}}_{4}^{-}$

#### 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, ${\text{MnO}}_{4}^{-}$, will be reduced to manganese(II) cations, ${\text{Mn}}^{2 +}$.

On the other hand, the iron(II) cations, ${\text{Fe}}^{2 +}$, will be oxidized to iron(III) cations, ${\text{Fe}}^{3 +}$, and the oxalate anion, ${\text{C"_2"O}}_{4}^{2 -}$, will be oxidized to carbon dioxide, ${\text{CO}}_{2}$.

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

${\text{MnO"_ 4^(-) + overbrace(["Fe"^(2+) + "C"_ 2"O"_ 4^(2-)])^(color(purple)("FeC"_ 2"O"_ 4)) stackrel(color(darkgreen)("H"_ 2"SO"_ 4)color(white)(aaa))(->) "Mn"^(2+) + "Fe"^(3+) + "CO}}_{2}$

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 $\textcolor{b l u e}{+ 7}$ on the reactants' side to $\textcolor{b l u e}{+ 2}$ on the products' side, which is why we say that manganese is being reduced.

Since you're in acidic medium, you can balance the oxygen atoms by adding water molecules and the hydrogen atoms by adding protons, ${\text{H}}^{+}$.

You will have

$8 \text{H"^(+) + stackrel(color(blue)(+7))("Mn")"O"_ 4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn")""^(2+) + 4"H"_2"O}$

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

${\text{1e"^(-) + 2"e"^(-) = "3e}}^{-}$

are being lost in the two oxidation half-reactions and ${\text{5e}}^{-}$ are being gained in the reduction half-reaction.

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 $5$ and the reduction half-reaction by $3$. Add the three half-reactions to get

$\left\{\begin{matrix}8 {\text{H"^(+) + stackrel(color(blue)(+7))("Mn")"O"_ 4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn")""^(2+) + 4"H"_ 2"O" | xx 3 \\ color(white)(aaaaaaaaaaaaa)stackrel(color(blue)(+2))("Fe")""^(2+) -> stackrel(color(blue)(+3))("Fe")""^(3+) + "e"^(-) color(white)(aaa)| xx 5 \\ color(white)(aaaaaaaaaaa)stackrel(color(blue)(+3))("C")_ 2 "O"_ 4^(2-) -> 2stackrel(color(blue)(+4))("C")"O"_ 2 + 2"e}}^{-} \textcolor{w h i t e}{a} | \times 5\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a}}$

$24 \text{H"^(+) + 3"MnO"_ 4^(-) + color(red)(cancel(color(black)("15e"^(-)))) + color(purple)(5) xx ["Fe"^(2+) + "C"_ 2"O"_ 4^(2-)] -> 3"Mn"^(2+) + 5"Fe"^(3+) + color(red)(cancel(color(black)("15e"^(-)))) + 10"CO"_ 2 + 12"H"_ 2"O}$

This is equivalent to

$24 \text{H"^(+) + 3"MnO"_ 4^(-) + color(purple)(5) xx ["Fe"^(2+) + "C"_ 2 "O"_ 4^(2-)] -> 3"Mn"^(2+) + 5"Fe"^(3+) + 10"CO"_ 2 + 12"H"_ 2 "O}$

As you can see, you need $3$ moles of permanangate anions to oxidize $\textcolor{p u r p \le}{5}$ moles of ferrous oxalate, which means that $1$ mole of ferrous oxalate would require

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)|)))#