Redox Equations: How to Determine Right Side of Equation

[campa]

Hi,

I did a question once and in that Mno4- -> Mno2 but in another question it was Mno4- -> Mn2+
How can these equations defer and is there a way to know what the right side of the half equation is going to be when the left is given?

 

[gravenewworld]

With some redox reactions there will be no way in hell you could eyeball a reaction and figure out what the correct equation is with everything blanaced properly. Do these steps for redox reactions:

1) Split the reaction into half reactions
2) Balance non O and non H's using coefficients
3) Balance the O's with H20
4) Balance the H's with H+
5) Balance the charges with electrons.
6) Balance the number of electrons in each half reaction by multiplying the half reactions by least common multiple of the number of electrons in each half reaction.
7) Combine half reactions.
8) Cancel out electrons and like terms if any from both sides (or combine like items with each other).

In the end you eq. should be balanced.



For MnO4- ----->Mn2+ you basically have a half reaction, you don't need to find half reactions for this.So follow the steps. The # of Mn's on both sides are balanced. Next step. Balance the O's with H20. Thus you need 1 H20 on the right

MnO4- ------> Mn2+ + H20


Balance H's with H+. You have 2 H's on the right and 0 on the left, so add 2H+'s to the left

MnO4- +2H+ ------> Mn2+ + H20

Balance Charges. You got off lucky here, no need to go further. you have -4+2 on the left and +2 on the right, so you have +2 charge on both sides. You're done.

 

[campa]

Yes thanks but my question is that if the question asks us to determine what Mno4- ->
becomes in this case what should you choose Mn2+ or Mno4-

 

[gravenewworld]

let me correct myself first, i read MnO4- as MnO 4- instead of MnO4 1-. You would still just follow the same steps though.

If you are given just MnO4- ----> ? You can not predict what the product will be unless you are told what conditions you are working with. if you look at the half rxn eq. you have

MnO4- ==> MnO2

MnO4-(aq) + 8H+(aq) + 5e- ==> Mn2+(aq) + 4H2O(l)

The 2nd obviously involves some sort of acid. While the first can probably be obtained by doing it in non basic or acidic conditions, i.e. in just water since

H2O + 2MnO4- + (?)e -> 2MnO2 + (?)OH- not balanced

The product you get will all depend on the conditions you work with.

 

[campa]

thanx that's what I was talking about so is there a website where I can get a list of these equations ex- half equations in acidic conditions and half equations for basic conditions