So I'm a little confused by this question. Oxygen just seems to vanish from the reaction. Does anyone else have this?
What is the sum of the coefficients in the following reaction?

MnO4- + I- -------> I2 + Mn2+

D) 43 (the correct answer)

They're expecting you to know that water (product) and protons (reactants) are in the equations. I- + H+ + MnO4 ------> I2 + Mn2+ + H2O is the real reaction, albeit unbalanced. I don't think the real DAT would ask a question so vague. Poor question imo

So I'm a little confused by this question. Oxygen just seems to vanish from the reaction. Does anyone else have this?
What is the sum of the coefficients in the following reaction?

MnO4- + I- -------> I2 + Mn2+

D) 43 (the correct answer)

This is a redox reaction. So, to find the # of coefficients, you must be able to balance the redox reaction as follows:

1)Write to separate half-reactions:

MnO4- -->Mn2+
I- -->I2

2)Start with balancing the element, which you have the least of (i.e. Mn in case of the first reaction and I in case of the second reaction):

MnO4- -->Mn2+
2I- -->I2

3)Balance Oxygen by adding H2O to the opposite side:

MnO4- -->Mn2+ + 4H2O
2I- -->I2

4)Balance Hydrogen by adding H+ to the opposite side:

MnO4- + 8H+ -->Mn2+ + 4H2O
2I- -->I2

5)Balance charge by adding e- (i.e. The first reaction has 8 + (-1) = +7 charges on the left side, while having 2+ charges on the right side. To balance, add 5e- to the left side, so that 7 + (-5) = +2 = Charge on the right side. For the second reaction, there are 2- charges on the left side and 0 charges on the right side. To balance, add 2e- to the right side):

MnO4- + 8H+ + 5e- -->Mn2+ + 4H2O
2I- -->I2 +2e-

6)Cancel out the electrons by multiplying both reactions with the proper coefficients:

2(MnO4- + 8H+ + 5e- -->Mn2+ + 4H2O)
5(2I- -->I2 +2e-)

We have:

2Mno4- + 16H+ + 10e- -->2Mn2+ + 8H2O
10I- -->5I2 + 10e-

Now you can cancel out the e- from each side.

7)Add each side together to get the final balanced reaction:

2Mno4- + 16H+ + 10I- -->2Mn2+ + 8H2O + 5I2

Now simply add the coefficients:

2 + 16 + 10 + 2 + 8 + 5 = 43

Hope this helps!

They're expecting you to know that water (product) and protons (reactants) are in the equations. I- + H+ + MnO4 ------> I2 + Mn2+ + H2O is the real reaction, albeit unbalanced. I don't think the real DAT would ask a question so vague. Poor question imo

I had a question very similar to this on my DAT. In fact, I recommend you guys to know how to balance redox reactions. There's a good chance you'll see some questions about it!

cool, thanks for the help. I figured it out :)

Lol every book ive ever looked through on redox reactions had this specific example. I think everyone just loves to use it.

Lol

lol

Great write-up nze82.

I see how this goes, but I was wondering how you know when to do a redox balance.

Like this equation you dont have to balance the electrons to get the coefficents.
C4H10 + O2--> CO2 + H20.

I see how this goes, but I was wondering how you know when to do a redox balance.

Like this equation you dont have to balance the electrons to get the coefficents.
C4H10 + O2--> CO2 + H20.

Obviously, this is an oxidation reaction not a redox reaction, which is why you shouldn't balance the same way you balance redox reactions.
One way I use to recognize redox reactions is that most of them involve charged species.
Also, you need to have some general knowledge of various reaction types. In other words, when you see the above reaction you should immediately know it's the oxidation of a carbohydrate.

Thanks I got it! This redox reaction is a real time consumer. =/

wow, nze82 thanks for that little help. It really helped on my DAT this last thursday

wow, nze82 thanks for that little help. It really helped on my DAT this last thursday
You're more than welcome.
Hope it went well!