# Question 66324

Jul 6, 2017

Potassium permanganate is the oxidizing agent and oxalic acid is the reducing agent.

#### Explanation:

Start by assigning oxidation numbers to all the atoms that take part in the reaction.

$\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")_ 2 stackrel(color(blue)(+3))("C")_ 2 stackrel(color(blue)(-2))("O")_ 4 + stackrel(color(blue)(+1))("K") stackrel(color(blue)(+7))("Mn") stackrel(color(blue)(-2))("O")_ 4 + stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_ 4 -> stackrel(color(blue)(+4))("C") stackrel(color(blue)(-2))("O")_ 2 + stackrel(color(blue)(+2))("Mn") stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_ 4 + stackrel(color(blue)(+1))("K")_ 2 stackrel(color(blue)(+6))("S") stackrel(color(blue)(-2))("O")_ 4 + stackrel(color(blue)(+1))("H")_ 2stackrel(color(blue)(-2))("O}}$

Now, notice that the oxidation number of carbon goes from $\textcolor{b l u e}{+ 3}$ on the reactants' side to $\textcolor{b l u e}{+ 4}$ on the products' side. Since the oxidation number of carbon is increasing, you can say that carbon is being oxidized.

On the other hand, the oxidation number 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. Since the oxidation number of manganese is decreasing, you can say that manganese is being reduced.

Now, the compound that contains the element that is being oxidized acts as a reducing agent because it reduces the compound that is being reduced.

Similarly, the compound that is being reduced acts as an oxidizing agent because it oxidizes the compound that is being oxidized.

In your case, oxalic acid, ${\text{H"_2"C"_2"O}}_{4}$, will act as a reducing agent because it reduces potassium permanganate, ${\text{KMnO}}_{4}$, to manganese(II) sulfate, ${\text{MnSO}}_{4}$.

Similarly, potassium permanganate will act as an oxidizing agent because it oxidizes the oxalic acid to carbon dioxide, ${\text{CO}}_{2}$.

The oxidation half-reaction looks like this

stackrel(color(blue)(+3))("C") _ 2"O"_ 4^(2-) -> 2stackrel(color(blue)(+4))("C") "O"_ 2 + 2"e"^(-)

The reduction half-reaction loos like this--keep in mind that the reaction takes place in acidic medium, so use water molecules and protons, ${\text{H}}^{+}$, to balance the atoms of oxygen and of hydrogen, respectively.

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

To balance this redox reaction, multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$, this will give you equal numbers of electrons lost in the reduction half-reaction and gained in the oxidation half-reaction.

{(color(white)(aaaaaaaaaaa)stackrel(color(blue)(+3))("C") _ 2"O"_ 4^(2-) -> 2stackrel(color(blue)(+4))("C") "O"_ 2 + 2"e"^(-)" " xx 5), (8"H"^(+) + stackrel(color(blue)(+7))("Mn") "O"_ 4^(-) + 5"e"^(-) -> stackrel(color(blue)(+2))("Mn")""^(2+) + 4"H"_ 2"O"" " xx 2) :}

You will end up with

{(color(white)(aaaaaaaaaaaaa)5stackrel(color(blue)(+3))("C") _ 2"O"_ 4^(2-) -> 10stackrel(color(blue)(+4))("C") "O"_ 2 + 10"e"^(-)), (16"H"^(+) + 2stackrel(color(blue)(+7))("Mn") "O"_ 4^(-) + 10"e"^(-) -> 2stackrel(color(blue)(+2))("Mn")""^(2+) + 8"H"_ 2"O") :}

Add the two half-reactions to get

$16 \text{H"^(+) + 5"C"_ 2"O"_ 4^(2-) + 2"MnO"_4^(-) + color(red)(cancel(color(black)(10"e"^(-)))) -> 10"CO"_2 + color(red)(cancel(color(black)(10"e"^(-)))) + 2"Mn"^(2+) + 8"H"_2"O}$

You will end up with

$5 \text{H"_ 2"C"_ 2"O"_ 4 + 2"KMnO"_ 4 + 3"H"_ 2"SO"_ 4 -> 10"CO"_ 2 + 2"MnSO"_ 4 + "K"_ 2"SO"_ 4 + 8"H"_ 2"O}$

Jul 6, 2017

The oxalate ion was the reducing agent, and the permanganate ion was the oxidizing agent.

Here's another way to do it. What I would do (if I were to be given this unbalanced equation) is to identify the phases and "trim off the fat".

The goal here is to identify the main participants without having to rederive the reaction.

IDENTIFYING THE MAIN REACTION PARTICIPANTS

(Here, I will provide the balanced reaction anyway.)

$\textcolor{red}{5} \text{H"_2"C"_2"O"_4(aq)+color(red)(2)"KMnO"_4(aq)+color(red)(3)"H"_2"SO"_4(aq) -> color(red)(10)"CO"_2(g)+color(red)(2)"MnSO"_4(aq)+"K"_2"SO"_4(aq)+color(red)(8)"H"_2"O} \left(l\right)$

Each of the aqueous compounds here are easily soluble in water because they are potassium-based (alkali-metal-based), except for manganese(II) sulfate, which if you look it up, has a high solubility of over $\text{52 g/100 mL water}$ at ${25}^{\circ} \text{C}$.

So, really, the main participants in this reaction are ${\text{H"_2"C"_2"O}}_{4}$, ${\text{KMnO}}_{4}$, ${\text{CO}}_{2}$, and ${\text{MnSO}}_{4}$, while $\text{H"_2"O}$ is the solvent, and the ${\text{H"_2"SO}}_{4}$ is what sets up the acidic conditions.

WHAT WOULD THE BALANCED NET IONIC REACTION BE?

Now, just to check, if we separate the compounds into ions, and cancel out spectator ions, we get:

$10 \text{H"^(+)(aq) + 5"C"_2"O"_4^(2-)(aq) + cancel(2"K"^(+)(aq)) + 2"MnO"_4^(-)(aq)+6"H"^(+)(aq) + cancel(3"SO"_4^(2-)(aq)) -> 10"CO"_2(g)+2"Mn"^(2+)(aq) + cancel(2"SO"_4^(2-)) + cancel(2"K"^(+)(aq)) + cancel("SO"_4^(2-)(aq)) + 8"H"_2"O} \left(l\right)$

This gives the net ionic equation that would be obtained from balancing the reaction from scratch using half-reactions (as Stefan has done):

$\textcolor{g r e e n}{16 \text{H"^(+)(aq)) + color(purple)(5"C"_2"O"_4^(2-)(aq)) + color(green)(2"MnO"_4^(-)(aq)) -> color(purple)(10"CO"_2(g))+color(green)(2"Mn"^(2+)(aq) + 8"H"_2"O} \left(l\right)}$

There, now that we can see which elements get oxidized or reduced more easily.

FINDING THE TWO OXIDIZED/REDUCED SPECIES

One trick to identify oxidation is to try looking for what gains oxygen atoms, and vice versa for reduction.

Filter out the following two half-reactions (currently with unbalanced charge!), noting that the oxalate to carbon dioxide half-reaction is easier to spot:

$\textcolor{g r e e n}{\overbrace{16 {\text{H"^(+)(aq) + 2"MnO"_4^(-)(aq))^("charge = 14"^(+)) -> overbrace(2"Mn"^(2+)(aq) + 8"H"_2"O"(l))^("charge = 4}}^{+}}}$

$\textcolor{p u r p \le}{\overbrace{5 \text{C"_2"O"_4^(2-)(aq))^("charge = 10"^(-)) -> overbrace(10"CO"_2(g))^("charge = 0}}}$

This is what we were looking to spot from the very beginning, and with practice, you can see this without going through the above.

Now, we can hopefully spot that:

• the manganese atom was reduced (by oxalate), from within stackrel(color(blue)(+7))("Mn")"O"_4^(-)# to being $\stackrel{\textcolor{b l u e}{+ 2}}{{\text{Mn}}^{2 +}}$.

This reduction can easily be seen by noticing that the number of oxygens decreased.

• the carbon atom was oxidized (by permanganate), from within ${\stackrel{\textcolor{b l u e}{+ 3}}{\text{C"_2"O}}}_{4}^{2 -}$ to within ${\stackrel{\textcolor{b l u e}{+ 4}}{\text{C""O}}}_{2}$.

This oxidation is less obvious, since the number of oxygens per carbon atom has not changed, but by knowing what was reduced, the OTHER thing must be oxidized!

This indicates that the oxalate ion was the reducing agent, and the permanganate ion was the oxidizing agent.