I have these equations in our book , but I've tried a few times to solve it in half reaction method , but I couldn't yet :
1) KMnO4 + KIO3 => MnO2 + KIO4 ( in a basic solution )
2) MnO4- + Cl- => Mn^+2 + HClO ( in a acidic solution )
I think these are incorrect , but I don't know what to do , because their solutions are wanted from us
thank a lot
Here is a step by step of what I do for half reactions. This is the LONG way of doing it but I use it for two reasons:
1. It requires the student to know oxidations states.
2. The same procedure is used for acid OR basic solution. And I'm going to omit the spectator ions; you can add those.
1. Separate into half equations.
MnO4^- ==> MnO2
IO3^- ==> IO4^-
2. Check to see that you have the same number of atoms of oxidized/reduced materials. Here you have 1 Mn on both sides and 1 I on both sides so things are ok. In the case of something like FeO ==> Fe2O3 you MUST make it read 2FeO => Fe2O3 before proceeding or it will never balance.
3. Using MnO4^- --> MnO2 we assign oxidation states. +7 on left and +4 on right. Add electrons to balance the change in oxdn state.
MnO4^- + 3e ==> MnO2
4. If this is acid solution or basic:
a. if acid solution add H^+ to the appropriate side to balance the charge.
b. if basic solution add OH^- to the appropriate side to balance the charge.
The charge on the left side is 4-, so add 4 OH^- to the right.
MnO4^- + 3e ==> MnO2 + 4OH^-
5. Now add H2O to the appropriate side to balance the H atoms. That will be
MnO4^- + 3e + 2H2O ==> MnO2 + 4OH^-
Now the other one. I won't go through the steps but here is the end result.
IO3^- + 2OH^- ==> IO4^- + 2e + H2O
6. Multiply each half reaction by some number to make the electrons in one half = electrons change in the other half.
7. Add the two half equations and cancel any ions common to each side.
8. Add spectator ions if desired.
You might want to copy this to use it until you have the procedure memorized.
Follow up with any questions here.
Yes ,
3IO3- + 6OH- => 3IO4- + 6e- + 3H2O
2MnO4- + 6e- + 4H2O => 2MnO2 + 8OH-
___________________________________ +
3IO3 + 2MnO4- + H2O => 3IO4- + 2MnO2 + 2OH-
and , the final balanced equation is :
3KIO3 + 2KMnO4 H2O => #KIO4 + 2MnO2 + 2OH-
so , is it possible that we write an ion alongside a complete reaction ?
thank you .. so much
I think the answer is no but you just didn't finish.
You added 5K^+ on the left. 3 of those on the right go with 3KIO4 and the other two go with 2KOH
Actually, there is nothing wrong with having a "mixed" equation in which you have both molecules and ions. However, in your case you added 5K^+ on the left and you added only 3K^+ on the right. So in changing from the ionic to the molecular equation your molecular equation isn't balanced until you add the other two K^+.
Oh ! that was only wrong of typing
So it's not wrong with having a "mixed" equation in which you have both molecules and ions ?
right.
3KIO3 + 2KMnO4 + H2O ==> 3KIO4 + 2K^+ + 2OH^- + 2MnO2
but that is not the usual way of doing it. Usually we have an ionic equation or a molecular equation but a mixed is not wrong (unless your prof says to write a molecular equation in which case you will have points marked off).
To solve these redox equations using the half-reaction method, follow these steps:
1) Identify the oxidation states of each element in the reaction.
2) Divide the overall reaction into two half-reactions: the oxidation half-reaction and the reduction half-reaction.
3) Balance the atoms of each half-reaction, starting with the atoms other than hydrogen and oxygen.
4) Balance the oxygen atoms by adding water molecules (H2O) to the side that lacks oxygen. Then, balance the hydrogen atoms by adding hydrogen ions (H+).
5) Balance the charge in each half-reaction by adding electrons (e-) to the appropriate side.
6) Make sure the number of electrons transferred in each half-reaction is equal and adjust the coefficients to match.
7) Combine the half-reactions, ensuring that the number of electrons is the same in both reactions, to get the overall balanced redox equation.
8) If the reaction takes place in a basic solution, add the same number of hydroxide ions (OH-) to both sides of the equation. Then, combine the resulting hydrogen ions (H+) and hydroxide ions (OH-) to form water (H2O) when possible.
Let's apply these steps to the given equations:
1) KMnO4 + KIO3 -> MnO2 + KIO4 (in basic solution)
First, we identify the oxidation states:
K: +1
Mn: +7 => +4
O: -2
I: +5 => +7
Oxidation half-reaction:
MnO4- -> MnO2
To balance the oxygen, add two water (H2O) molecules to the right-hand side:
MnO4- -> MnO2 + 2H2O
Now, balance the hydrogen by adding four hydrogen ions (4H+):
MnO4- + 8H+ -> MnO2 + 2H2O
Reduction half-reaction:
KIO3 -> KIO4
To balance the iodine atoms, add six electrons (6e-) to the right-hand side:
KIO3 -> KIO4 + 6e-
Since the overall reaction should be balanced, multiply the oxidation half-reaction by six and the reduction half-reaction by one to get an equal number of electrons transferred in both reactions:
6MnO4- + 48H+ -> 6MnO2 + 12H2O
6KIO3 -> 6KIO4 + 36e-
Combine the two half-reactions, canceling out the electrons:
6MnO4- + 48H+ + 6KIO3 -> 6MnO2 + 12H2O + 6KIO4
If this reaction is in a basic solution, add the same number of hydroxide ions (OH-) to both sides:
6MnO4- + 48H+ + 6KIO3 + 6OH- -> 6MnO2 + 12H2O + 6KIO4 + 6OH-
2) MnO4- + Cl- -> Mn^2+ + HClO (in acidic solution)
Here are the oxidation states:
Mn: +7 => +2
O: -2
Cl: -1 => +5
Oxidation half-reaction:
MnO4- -> Mn^2+
To balance the oxygen, add four hydrogen ions (4H+) to the left-hand side:
MnO4- + 4H+ -> Mn^2+
Now, balance the charge in the half-reaction by adding five electrons (5e-) to the left-hand side:
MnO4- + 4H+ + 5e- -> Mn^2+
Reduction half-reaction:
Cl- -> HClO
The chlorine atom is already balanced, so we don't need to add any H2O or H+ ions.
Since the overall reaction should be balanced, multiply the oxidation half-reaction by two and the reduction half-reaction by five to equalize the electrons transferred in both reactions:
2MnO4- + 8H+ + 10e- -> 2Mn^2+
5Cl- -> 5HClO
Combine the two half-reactions, canceling out the electrons:
2MnO4- + 8H+ + 10Cl- -> 2Mn^2+ + 5HClO
This equation is already in acidic solution. If you need to convert it to basic solution, you can follow the additional step mentioned before.
It's crucial to double-check the balanced equations and ensure that the atoms, charges, and electrons are properly balanced throughout the process.