[petstar21]

MnO4- + Br- ---> MnO2 + BrO3-

I set MnO4- ---> MnO2
and Br- ---> BrO3-
and solved that way but i didnt get the right answer...


H20 + H3PO3 + I(iodine)2----> H3PO2 + IO3- +H+
How would you do it since there are 3 things?


Cu + HNO3 ---> Cu(NO3)2 + NO + H2O
I dont know what to set --> to what bc there are 3??

-- I dont know how to do it when theres three reactants/ products and i dont completely understand how to do them anyway, please *delete me*
also,

Fe + O2 + H2O---> Fe(OH)3

--is this one even possible?? its supposed to be the rusting of iron or something! ive worked so hard!

[cliverlong]

MnO4- + Br- ---> MnO2 + BrO3-

 << snip rest >>

Your struggle starts here because you don't have a meaningful equation.

Where do you get your equation from? are you trying to make your own examples?

I would suggest thinking along these lines

MnO₄^-  --> Mn²⁺

2Br^- --> Br₂

Then add some H⁺ to the left hand side so it "soaks up" any liberated O into water on right hand side.

Try to balance those and see what happens - there's more to do after that point but that will come later.

[Borek]

Your struggle starts here because you don't have a meaningful equation.

Where do you get your equation from? are you trying to make your own examples?

I would suggest thinking along these lines

MnO₄^-  --> Mn²⁺

MnO₄^- -> MnO₂ is a valid half reaction that can be used to balance the equation.

[petstar21]

I got the answer to the first eqaution....

but is the answer to the second one
5H3PO3 + 2I2 +H2O ---> 5H3PO3 +2IO-3 +2H+

I worked hard but i dont have the answer to check it lol!

[Borek]

5H3PO3 + 2I2 +H2O ---> 5H3PO3 +2IO-3 +2H+

Iodine is not balanced. Could be a typo, just like H3PO3 on the right side.

[petstar21]

uch I have no idea how to do it

and the one below it with copper IM COMPLETELY LOST ON.
I have no idea how to split it up at all. PLEASE HELp

[Borek]

You were pretty close.

If you have no idea what is being oxidized and what is being reduced always start with sytematic assignement of oxidation numbers to all elemenets in all reagents. This is less work than you may think - most abundant elements usually don't change their oxidation states.

[cliverlong]

Your struggle starts here because you don't have a meaningful equation.

Where do you get your equation from? are you trying to make your own examples?

I would suggest thinking along these lines

MnO₄^-  --> Mn²⁺

MnO₄^- -> MnO₂ is a valid half reaction that can be used to balance the equation.

Aha !

I wasn't sure if it was - which is why I suggested the change to Mn²⁺

From my quick working MnO₂ should lead to easier numbers.

Clive

[cliverlong]

I got the answer to the first equation....

<< snip >>

Can you post what you think is the correct answer and explain the equation?

That will be the key to all such problems.

Clive

[The_Student]

MnO4- + Br- ---> MnO2 + BrO3-

I set MnO4- ---> MnO2
and Br- ---> BrO3-
and solved that way but i didnt get the right answer...


H20 + H3PO3 + I(iodine)2----> H3PO2 + IO3- +H+
How would you do it since there are 3 things?


Cu + HNO3 ---> Cu(NO3)2 + NO + H2O
I dont know what to set --> to what bc there are 3??

-- I dont know how to do it when theres three reactants/ products and i dont completely understand how to do them anyway, please *delete me*
also,

Fe + O2 + H2O---> Fe(OH)3

--is this one even possible?? its supposed to be the rusting of iron or something! ive worked so hard!

Does your teacher give you steps on how to do these equations? If not then i can give you the steps to solve REDOX reactions.