# How are more complex red-ox reactions balanced? The kind of reactions which have 3 or more reactions and multiple products?Also what do we do when more that one species is undergoing oxidation/reduction?

May 18, 2016

I got:

$2 \text{MnO"_4^(-)(aq) + 6"H"^(+)(aq) + 5"H"_2"O"_2(aq) -> 2"Mn"^(2+)(aq) + 5"O"_2(g) + 8"H"_2"O} \left(l\right)$

or

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

Oxygen in ${\text{H"_2"O}}_{2}$ was oxidized, and manganese in ${\text{MnO}}_{4}^{-}$ was reduced.

Normally, we would start with a half-reaction and work towards the final balanced reaction.

I don't know how you got the unbalanced semi-final result, but I guess:

1. We'll work backwards to find the half-reactions.
2. We'll go back forwards to balance the reaction itself in acidic conditions.

ROUGHLY IDENTIFYING WHICH TWO REACTIONS ARE PRESENT

$\text{KMnO"_4(aq) + "H"_2"SO"_4(aq) + "H"_2"O"_2(aq) -> "MnSO"_4(aq) + "K"_2"SO"_4(aq) + "O"_2(g) + "H"_2"O} \left(l\right)$

If you focus on what's similar on either side of the reaction, you should see that sulfate, potassium, and manganese stand out.

color(highlight)(overbrace("KMnO"_4(aq)) + overbrace("H"_2"SO"_4(aq))) + color(red)("H"_2"O"_2(aq)) -> color(highlight)(overbrace("MnSO"_4(aq)) + overbrace("K"_2"SO"_4(aq))) + color(red)("O"_2(g) + "H"_2"O"(l))

This tells me that these compounds have been participating in the same reaction.

REMOVING SPECTATOR IONS

Notice how ${\text{K}}^{+}$ would be a spectator ion, because it is an alkali metal, which, from the solubility rules you should know, renders the resultant ionic compound quite soluble.

${\text{MnSO}}_{4}$ is very soluble in water, so ${\text{SO}}_{4}^{2 -}$ is another spectator ion. Also, ${\text{H"_2"SO}}_{4}$ was there because the solution was under acidic conditions.

We can simply call that $2 {\text{H}}^{+}$, because enough of the sulfuric acid was added to make the conditions reducing enough (we'll look at why soon).

Thus, we have the following unbalanced ionic equation to decompose:

$\textcolor{red}{m a t h b f \left(\text{MnO"_4^(-)(aq) + 2"H"^(+)(aq) + "H"_2"O"_2(aq) -> "Mn"^(2+)(aq) + "O"_2(g) + "H"_2"O} \left(l\right)\right)}$

IDENTIFYING THE HALF-REACTIONS

Note that some of these species should be removed to give the true starting half-reactions, as they were added in the steps required to balance these reactions in the first place.

As we identified earlier, the previously bracketed/unbracketed compounds can be separated into their own separate half reactions:

${\text{MnO"_4^(-)(aq) + stackrel("shouldn't be here")overbrace(cancel(2"H"^(+)(aq))) -> "Mn}}^{2 +} \left(a q\right)$ (1)

"H"_2"O"_2(aq) -> "O"_2(g) + stackrel("shouldn't be here")overbrace(cancel("H"_2"O"(l))) (2)

These are much easier to balance, right? Now, which one is reduction and which one is oxidation? Let's look at how to distinguish them.

BALANCING THE HALF-REACTIONS

(1)

A nice starting half-reaction containing only manganese species derived from (1) is:

$\setminus m a t h b f \left({\text{MnO"_4^(-) -> "Mn}}^{2 +}\right)$

According to this Pourbaix diagram (remember this?), this can occur, let's say, at $\text{pH} = 2$, and involves at least $\text{0.6 V}$.

Recall that on a Pourbaix diagram, going downwards requires adding a negative voltage. Also, to minimize the number of reaction intermediates, we can do this at $\text{pH} = 2$.

That's why "enough" ${\text{H"_2"SO}}_{4}$ was there: to help reach that acidic $\setminus m a t h b f \left(\text{pH}\right)$ and maximize reaction yield.

Now, balancing this, we "add" water to balance the oxygens:

"MnO"_4^(-)(aq) -> "Mn"^(2+)(aq) + color(red)(4"H"_2"O"(l))

Then we add protons to balance the hydrogens in acidic conditions:

$\text{MnO"_4^(-)(aq) + color(red)(8"H"^(+)(aq)) -> "Mn"^(2+)(aq) + 4"H"_2"O} \left(l\right)$

Finally, add electrons to balance the charge.

$\textcolor{b l u e}{\stackrel{\textcolor{red}{+ 7}}{\text{Mn")stackrel(color(red)(-2))("O"_4^(-))(aq) + 8stackrel(color(red)(+1))("H"^(+))(aq) + 5e^(-) -> stackrel(color(red)(+2))("Mn"^(2+))(aq) + 4stackrel(color(red)(+1))("H")_2stackrel(color(red)(-2))("O}} \left(l\right)}$

Indeed, this is a real half-reaction, and it involves $\pm \text{1.51 V}$.

From examining the oxidation states, we see that manganese was reduced as stackrel(+7)("Mn") -> "Mn"^(2+) (${\text{MnO}}_{4}^{-}$ was the oxidizing agent).

Thus, this is the reduction half-reaction.

(2)

The other one is more tricky, as there are two possibilities. You could go to either water, or oxygen gas. However, since you assume that oxygen gas is produced...

I found a half-reaction here that starts as:

$\setminus m a t h b f \left({\text{H"_2"O"_2(aq) -> "O}}_{2} \left(g\right)\right)$

To balance this, we repeat the process we did before for the reduction half-reaction, except we don't need to "add" water because the oxygens are balanced.

"H"_2"O"_2(aq) -> "O"_2(g) + color(red)(2"H"^(+)(aq))

Since the charge is now unbalanced, let's re-balance it.

$\textcolor{b l u e}{\stackrel{\textcolor{red}{+ 1}}{{\text{H")_2stackrel(color(red)(-1))("O")_2(aq) -> stackrel(color(red)(0))("O")_2(g) + 2stackrel(color(red)(+1))("H}}^{+}} \left(a q\right) + 2 {e}^{-}}$

And from here you can see that oxygen was oxidized, as $\stackrel{- 1}{\text{O") -> stackrel(0)("O}}$ (${\text{H"_2"O}}_{2}$ was the reducing agent).

Thus, this is the oxidation half-reaction.

COMBINING THE HALF-REACTIONS BACK TOGETHER

Finally, we can put these two half-reactions together and get the result by making sure the electrons cancel out (as they were merely an accounting scheme):

$2 \left(\text{MnO"_4^(-)(aq) + 8"H"^(+)(aq) + cancel(5e^(-)) -> "Mn"^(2+)(aq) + 4"H"_2"O} \left(l\right)\right)$
$5 \left({\text{H"_2"O"_2(aq) -> "O"_2(g) + 2"H}}^{+} \left(a q\right) + \cancel{2 {e}^{-}}\right)$
$\text{-----------------------------------------------------------}$

$\setminus m a t h b f \left(\textcolor{b l u e}{2 \text{MnO"_4^(-)(aq) + 6"H"^(+)(aq) + 5"H"_2"O"_2(aq) -> 2"Mn"^(2+)(aq) + 5"O"_2(g) + 8"H"_2"O} \left(l\right)}\right)$

That's good enough.

But... if we add back all the spectator ions and totally soluble ionic compounds, we would get:

$\textcolor{b l u e}{2 \text{KMnO"_4(aq) + 3"H"_2"SO"_4(aq) + 5"H"_2"O"_2(aq) -> 2"MnSO"_4(aq) + "K"_2"SO"_4(aq) + 5"O"_2(g) + 8"H"_2"O} \left(l\right)}$

May 18, 2016