# oxidation reduction question

Discussion in 'DAT Discussions' started by ATLATLATL, May 1, 2007.

1. ### ATLATLATL 2+ Year Member

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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=(

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3. ### domonas Junior Member 5+ Year Member

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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

4. ### ATLATLATL 2+ Year Member

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

5. ### Lonely Sol cowgoesmoo fan! 10+ Year Member

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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)

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