oxidation reduction question

[ATLATLATL]

guys how do i do this....
wat is the sum of the coefficients in the following redox rxn...

MnO4- + I- + I2 + Mn2+

i suck that these for some reason...poo=(

 

[domonas]

guys how do i do this....
wat is the sum of the coefficients in the following redox rxn...

MnO4- + I- + I2 + Mn2+

i suck that these for some reason...poo=(

I get 5

MnO4(-) + 2I(-) + 8H(+) + 5e(-) --> I2 + Mn(2+) + 4H2O + 2e-

unless you add the coefficients of the H+, H2O, and e-
in which case, the answer is = 23

 

[ATLATLATL]

how did you come up with 8H...i'm just stumped on how to arrive at the answer

 

[Lonely Sol]

guys how do i do this....
wat is the sum of the coefficients in the following redox rxn...

MnO4- + I- + I2 + Mn2+

i suck that these for some reason...poo=(

This is how I would do it

2I- ---->I2 + 2e-
8H+ + 5e- + MnO4- --->Mn2+ + 4H20 (you get 8H+ by finding the balancing the equation)

Now, you need to get equal number of electrons on each side

so,
(2I- ---->I2 + 2e-)*5
(8H+ + 5e- + MnO4- --->Mn2+ + 4H20)*2

Now, we get:

16H+ + 10e- + 2MnO4- + 10I- -----> 5I2 + 10e- + 2Mn2+ + 8H20

electrons cancel out
So you are left with:
16H+ + 2MnO4- + 10I- -----> 5I2 + 2Mn2+ 8H20

So, sum of co-efficent should be : 16+2+10+5+2+8 = 46

*Plz correct me if I am wrong (I am not sure if you are suppose to include H20 and H+ in the sum)

 

[domonas]

This is how I would do it

2I- ---->I2 + 2e-
8H+ + 5e- + MnO4- --->Mn2+ + 4H20 (you get 8H+ by finding the balancing the equation)

Now, you need to get equal number of electrons on each side

so,
(2I- ---->I2 + 2e-)*5
(8H+ + 5e- + MnO4- --->Mn2+ + 4H20)*2

Now, we get:

16H+ + 10e- + 2MnO4- + 10I- -----> 5I2 + 10e- + 2Mn2+ + 4H20

electrons cancel out
So you are left with:
16H+ + 2MnO4- + 10I- -----> 5I2 + 2Mn2+ 4H20

So, sum of co-efficent should be : 16+2+10+5+2+4 = 42

*Plz correct me if I am wrong (I am not sure if you are suppose to include H20 and H+ in the sum)

OMG, i totally forgot to make e- on both sides equal. i do believe the above is correct. Thank you for correcting my error Lonely Sol.

edit: actually, you'd end up with 8 H2O on the right, not 4.