[confusedcollegestudent]

When balancing redox equations, I'm confused about how to figure out how many electrons to add.
In a problem like MnO₄^-+I^- MnO₂+I₂, I know it would split into
I I₂ and MnO₄ MnO₂
And after balancing the equation, you would get:
2I^- I₂+2e^-
MnO₄^-+4H⁺+3e^- MnO₂+2H₂O
Except in Br₂ BrO₃^-+Br^- (in basic solution)
I don't understand how to solve this type of problem and why it is solved the way it is.
What I had done is the following:
Br₂ 2Br^-
Br₂+6H₂O 2BrO₃^-+12H⁺
At the end, I would convert the H⁺ and OH^-, which I will add, into H₂O and cancel.
Except I was apparently wrong with what I had done and should have gotten an answer of:
Br₂ 2BrO₃^-+10e^-
Why are we adding 10 electrons to the right side instead of the left?
I thought that adding electrons was supposed to correct the charge imbalance? So wouldn't 2 electrons be added to Br₂?
I don't understand why in the second case, the added number of electrons is based off of the oxidation number instead of the charge?
Can someone explain the difference to me because I'm so confused.

[Nobby]

You have to check the oxidation numbers of each product and educt.

do you know how to calculate it for BrO₃^-?

[confusedcollegestudent]

You have to check the oxidation numbers of each product and educt.

do you know how to calculate it for BrO₃^-?

Yeah, O₃ produces a -6, and the Br would have a +5 because the overall charge is -1.
But I don't understand why we're doing that when in the problem with the MnO₄^-, the 4H⁺ and the 3e^- would negate the charge imbalance.
What is the difference between the two equations and the process of solving them?
Like.. why do we look at the oxidation number for BrO₃^-, but we don't for MnO₄^-?

[Nobby]

We do also for MnO₄^-

You have to get it for MnO₂ and MnO₄^-

So what are the oxidation numbers for each and the difference between?

[confusedcollegestudent]

We do also for MnO₄^-

You have to get it for MnO₂ and MnO₄^-

So what are the oxidation numbers for each and the difference between?

MnO₂ is -4+4 = 0
And MnO₄^- is -8+7 = -1
But wait, okay, so I tried to go through it again, and the answer appears to be right?
I got 3Br₂+6OH^- BrO₃^-+3H₂O+5Br^-
Does it matter if we balance the atoms first and then add electrons, or if we add electrons and then balance the atoms?

[Borek]

Sounds to me like you are mixing two approaches. You EITHER add electrons to balance change in the oxidation number OR you balance atoms first and then add electrons to balance the charge.

Br₂ 2BrO₃^-+10e^-

This is not balanced. Try to balance oxygen/water/H⁺ and see what you get then.

[Nobby]

We do also for MnO₄^-

You have to get it for MnO₂ and MnO₄^-

So what are the oxidation numbers for each and the difference between?

MnO₂ is -4+4 = 0
And MnO₄^- is -8+7 = -1
But wait, okay, so I tried to go through it again, and the answer appears to be right?
I got 3Br₂+6OH^- BrO₃^-+3H₂O+5Br^-
Does it matter if we balance the atoms first and then add electrons, or if we add electrons and then balance the atoms?

No you compare MnO₂ and MnO₄^-

You alredy calculated +4 for Mn in MnO₂ and +7 for Mn in Permangante

The difference is 3 electrons

Next step is to balance the Oxygen on both sides. You use water for adding and H⁺ for removing

Try this now