# 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

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1. 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.

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2. Yes ,
3IO3- + 6OH- => 3IO4- + 6e- + 3H2O
2MnO4- + 6e- + 4H2O => 2MnO2 + 8OH-
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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

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3. 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

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4. 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^+.

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5. 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 ?

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6. 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).

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