Dec 14, 2009
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Q1. CN- + OH- ---> CNO- + H2O + e-

Q2. MnO4- + I- + H2O -----> MnO2 + IO3- + OH-

Could you balance the above reaction?
 

sfoksn

10+ Year Member
Sep 4, 2006
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For the second one, I think it's just a redox rxn.

so you got MnO4- -> MnO2, so if you balance that half rxn, it would be

3e- + 4H+ + MnO4- -> MnO2 + 2H2O

you do same for the I- -> IO3-,

3H2O + I- -> IO3- + 6H+ + 6e-.

Then you multiple the MnO4 half rxn coefficients by 2, to match up the 6 e's from the other half rxn.

Then add them up, to get

2H+ + I- + 2MnO4- -> IO3- + 2MnO2 + H2O.

May have done something wrong as I was doing it on the computer w/o writing them out, but the idea should be good :)
 

sfoksn

10+ Year Member
Sep 4, 2006
761
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Oh, and you do the same for the first equation.

CN- -> CNO-, and OH- -> H2O.

If you balance them, you shoudl get

2H+ + 2OH- + CN- -> H2O + CNO-

Please let me know if this is the right way to approach the problem.

Thanks for posting good problems, kpark