This one doesn't seem to work with Chad's way but... help SDNer
MnO4- + I- + H2O ----> MnO2 + IO3- + OH-
The oxidation reduction run above is to be balanced with lowest whole number of coefficients. What is the coefficient for OH-?
a) 1
b) 2
c) 3
d) 4
e) 5
A question like this probably would not be asked on the DAT bc of its length (i.e. I am assuming you got this from Achiever)
I know Chad suggests a shortcut to doing it; but I generally suggest you just do it the long way so you don't make a mistake and waste more time trying to figure out where the mistake occurred. I.E. do it the way you learned in your general chemistry class. I suggest this because on the DAT the balancing will not be as long or as hard as this one was, or if it is, you would just skip it for the end.
Anyway... I digress.
You start with the half reactions:
1)MnO4- --> MnO2
2)I- --> IO3-
Lets do the first one first (i.e. the one with Manganese)
Step 1, balance the O's with H2O because we can't do anything with the manganese (i.e. they are already balanced)
Mn04- --> MnO2 + 2H2O
Step 2, balance the H's with H+'s
4H+ + MnO4- --> MnO2 + 2H2O
Step 3, balance the e-
3e- + 4H+ + MnO4- --> MnO2 + 2H2O
Done with the first half reaction.
Lets do the second half reaction now.
Step 1, balance the O's wit H2O since the I's are already balanced.
3H2O + I- --> IO3-
Step 2, balance the H's with H+
3H2O + I- --> IO3- + 6H+
Step 3, balance the e-
3H2O + I- --> IO3- + 6H+ + 6e-
Multiply the first half reaction by 2 so the e-'s are equal and then add the half reactions.
8H+ + 2MnO4- + 3H2O + I- --> 2MnO2 + 4H2O + IO3- + 6H+
Cancel the H+ and H2O on both sides to get a simplified equation.
2H+ + 2MnO4- + I- --> 2MnO2 + H2O + IO3-
Now remove the H+ by neutralizing with OH-
Gives your total reaction
H2O + 2MnO4- + I- --> 2MnO2 + IO3- + 2OH-
Hope this helps!
