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?
2 Answers
I got:
#2"MnO"_4^(-)(aq) + 6"H"^(+)(aq) + 5"H"_2"O"_2(aq) -> 2"Mn"^(2+)(aq) + 5"O"_2(g) + 8"H"_2"O"(l)#
or
#2"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"(l)#
Oxygen in
DISCLAIMER: LONG ANSWER!
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:
- We'll work backwards to find the half-reactions.
- We'll go back forwards to balance the reaction itself in acidic conditions.
ROUGHLY IDENTIFYING WHICH TWO REACTIONS ARE PRESENT
#"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"(l)#
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
We can simply call that
Thus, we have the following unbalanced ionic equation to decompose:
#color(red)(mathbf("MnO"_4^(-)(aq) + 2"H"^(+)(aq) + "H"_2"O"_2(aq) -> "Mn"^(2+)(aq) + "O"_2(g) + "H"_2"O"(l)))#
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:
#"MnO"_4^(-)(aq) + stackrel("shouldn't be here")overbrace(cancel(2"H"^(+)(aq))) -> "Mn"^(2+)(aq)# (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:
#\mathbf("MnO"_4^(-) -> "Mn"^(2+))#
According to this Pourbaix diagram (remember this?), this can occur, let's say, at
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
That's why "enough"
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:
#"MnO"_4^(-)(aq) + color(red)(8"H"^(+)(aq)) -> "Mn"^(2+)(aq) + 4"H"_2"O"(l)#
Finally, add electrons to balance the charge.
#color(blue)(stackrel(color(red)(+7))("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")(l))#
Indeed, this is a real half-reaction, and it involves
From examining the oxidation states, we see that manganese was reduced as
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:
#\mathbf("H"_2"O"_2(aq) -> "O"_2(g))#
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.
#color(blue)(stackrel(color(red)(+1))("H")_2stackrel(color(red)(-1))("O")_2(aq) -> stackrel(color(red)(0))("O")_2(g) + 2stackrel(color(red)(+1))("H"^(+))(aq) + 2e^(-))#
And from here you can see that oxygen was oxidized, as
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("MnO"_4^(-)(aq) + 8"H"^(+)(aq) + cancel(5e^(-)) -> "Mn"^(2+)(aq) + 4"H"_2"O"(l))#
#5("H"_2"O"_2(aq) -> "O"_2(g) + 2"H"^(+)(aq) + cancel(2e^(-)))#
#"-----------------------------------------------------------"#
#\mathbf(color(blue)(2"MnO"_4^(-)(aq) + 6"H"^(+)(aq) + 5"H"_2"O"_2(aq) -> 2"Mn"^(2+)(aq) + 5"O"_2(g) + 8"H"_2"O"(l)))#
That's good enough.
But... if we add back all the spectator ions and totally soluble ionic compounds, we would get:
#color(blue)(2"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"(l))#
Here are two more answers to your question.