[samiam]

Given MnO₄^- + I^- + H20  MnO₂ + I₂ +OH^-

If the coeficent of MnO₄- in balanced equation is 2 what is coeficent of I^-


that I did was break up the equation into two half reaction
Oxidation of I

and Reduction of MnO₄^-
In the reduction equation I put a coefficient of 2 in front of MnO4
than I balanced it as I would in acid(adding oh and h+)
after this i just added equal amount of oh^- to both sides in regards to how many h^+ there was

my  final answer was maleficent for I is 6 and for OH it is 8
did I do this correctly?
thanks!

[Borek]

I guess maleficient means just a a coefficient. 6 doesn't look OK.

[AWK]

Every redox reaction should be checked - if  atoms and charges are balanced.
2MnO₄^- + 6I^- + 4H₂O =  2MnO₂ + 3I₂ + 8OH^-

Count all kind of atoms on both side of equation. If they are equal, and sums of charges on both side are equal, and moreover coefficients have no common divisor then balancing is finished correctly.

[Borek]

I guess maleficient means just a a coefficient. 6 doesn't look OK.

Obviously I was sleepy when I posted. No idea what I meant.