Gen Chem question

[ScottW3]

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I've got a couple of last minute chem questions that have to do with oxidation and reduction reactions.

1. If you are asked for the sum of the coefficients of a balanced redox rxn, do you include those from the electrons?
A topscore 3 question asked this and the answer did include them, but I thought in a balanced redox rxn, you cancel the electrons.

2. Kaplan says that "a higher standard reduction potential means a greater tendancy for reduction to occur, while a lower SHE means a greater tendancy for oxidation to occur."
Does this mean higher as in larger number or larger magnitute? I keep getting confused when I try to figure it out. Lets say you have a SHE of -2 for X and .5 for Y. Which gets reduced (X because 2>.5 or Y because +.5>-2) ?

Thanks everyone

 

[jheidenr]

1. pretty sure that the electrons are cancelled in the balanced equation, I don't know whats going on there...

2. For a galvanic or voltaic cell, you would want to add your standard rediction potentional such that you have the greatest positive number, which would indicate a spontaneous reaction (because delta G = -nFE). So if you have two reduction potentials, -2 for X and 0.5 for Y, you would want to flip the -2 around giving you a positive 2 (meaning the X now gets oxidized). By adding 2+.5 you get 2.5 which is a higher value than -2+.5=-1.5. If the addition of the two potentials in not postitve, then the reaction would not be spontaneous, therefore the battery would not work.

So, Y gets reduced and X gets oxidized...I hope that all made sense.

 

[sherry225]

I've got a couple of last minute chem questions that have to do with oxidation and reduction reactions.

1. If you are asked for the sum of the coefficients of a balanced redox rxn, do you include those from the electrons?
A topscore 3 question asked this and the answer did include them, but I thought in a balanced redox rxn, you cancel the electrons.

Thanks everyone

Hum.....can you post the question, I don't have my Topscore with me. I think i did that problem just a couple days ago. Maybe I can help.

 

[sherry225]

If this is the question regarding MnO4- and I2, then the answer was right. And you should cancel out the electrons. Hum...if I remember right, i think you should have 5 e- on the MnO4 equation, and 2e- on the I2 equation. Then balance the e- and cancel them will give you the right answer. However, i could be wrong about which question you are asking...

 

[ScottW3]

Thanks for the help so far. I understand the reduction potential now. As for the TS question, it was question 69 on TS3. I redid it and now I keep getting answer C (35). I don't know how I was getting E before, and I don't know how D (43) is correct.

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

My individual balanced equations are:

10e- + 8H2 + 2MnO4- ---> 2Mn2+ + 8H2O and

10I- ---> 5I2 +10e-

So my full balanced equation is :

10I- +8H2 + 2MnO4- ---> 2Mn2+ + 8H20 + 5 I2

10+8+2+2+8+5=35

I just don't get it. Everything in gchem i get fine except for this and that darn common ion effect.
Any help would be much appreciated

 

[jheidenr]

I think the problem here is that they didn't specify whether to balance the equation in acidic or basic solution. You are doing it in a basic solution which requires adding OH- to balance the oxygens, but if you balance the equation in an acidic solution (ie adding H20 to balance your O's), you get the correct answer:

MnO4- ---> Mn2+
MnO4- ---> Mn2+ + 4H20 (add H20)
8H+ + MnO4- ---> Mn2+ + 4H20 (balance H's)
5e- + 8H+ + MnO4- ---> Mn2+ + 4H20 (balance charge)

I- ---> I2
2I- ---> I2 (balance I's)
2I- ---> I2 + 2e- (balance charge)

2[5e- + 8H+ + MnO4- ---> Mn2+ + 4H20] + 5[2I- ---> I2 + 2e-] =
16H+ + 2MnO4- + 10I ---> 2Mn2+ + 8H2O + 5I2

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