surferlove007

So..I'm taking Chemistry 1411 which as many people know is the intro chemistry class for University.
Well I'm struggling enough as it is...didn't do so hot on the first test. So I have some questions on some Redox Electron equations.
The first question involves balancing Redox equations.
They gave me this:

They told me to enter the coefficients for balancing this sucker.

I realize it involves how many electrons are associated with each combo and such. I just don't understand how to balance such a thing. The gaining electrons, and oxidation states.

Then after that is done...they want me to find the number of Moles of SO_2 that are produced in the formation of one mole of I_2.

n0nsensical

Note other than H2O these are all ionic compounds, K2Cr2O7 and Na2SO3 are really 2K+ + Cr2O7(2-) and 2Na+ + SO3(2-)
Write the half-reactions (looked these up in a table, not sure if you're 'supposed' to do that)
Reduction:
Cr2O7(2-) + 14H+ + 6e- -> 2Cr(3+) + 7H2O
Oxidation:
SO3(2-) + H2O -> SO4(2-) + 2H+ + 2e-

Balance the electrons
Cr2O7(2-) + 14H+ + 6e- -> 2Cr(3+) + 7H2O
3SO3(2-) + 3H2O -> 3SO4(2-) + 6H+ + 6e-

Sum
Cr2O7(2-) + 8H+ + 3SO3(2-) -> 3SO4(2-) + 2Cr(3+) + 4H2O

Get the 8H+ from 8HCl and conveniently there are also 8Cl- in the products. I hope that's everything but I'm not very good at this either. :P

Moles of SO2 in the formation of one mole I2? I don't see any SO2s or I2s. That sounds like a different problem. you need some sort of equation.

surferlove007

It's still really confusing, and of course my friends won't help me unless I pay them for their time, which i know isn't unreasonable.
I put in the answer into the thing, and it still said it was wrong...so blegh I don't know. My teacher's voice was going out when he was teaching this so he was rushing through it with only one example. of course it's due tomorrow at 6:00 p.m.

n0nsensical

I'm getting 1 3 8 2 3 2 4, it says that's wrong or did you put something else? If you have any more questions I can try to help you out

surferlove007

I made a typo, although I wish I had a better understanding of where those numbers came from. The book has a lot of information but it's very confusing.

The next question is "In the process of oxidizing I^- to I_2, \rm {SO_4}^{2-} is reduced to SO_2. How many moles of \rm SO_2 are produced in the formation of one mole of I_2?

n0nsensical

well for every mole of I- oxidized, 1/2 mole of I2 and 1 mole of electrons are produced. This is represented by the half-reaction
2I- -> I2 + 2e-
The reduction of SO4(2-) is represented by
SO4(2-) + 4H+ + 2e- -> SO2 + 2H2O
You get that by balancing SO4(2-) -> SO2 with H+, e-, and H2O so that charge and mass are conserved. Then add the half-reactions to cancel electrons and get an overall balanced reaction. Thats redox in a nutshell, balancing and adding half-reactions.

pig

This thread does my heart good to see. Looks like nonsensical is wearing out the e-chem quite nicely - well done sir, well done.

ghoast: are you comfortable working with balancing equations that don't involve electron transfer? if so, then think about these reactions as being very similar to those reactions, with the main difference being that these have an extra reactant or product: namely electricity. As non has put forth, you've got 'half-reactions' and each one of these half-reactions takes a certain number of moles of electrons in order to go to completion. You can look up the standard form of the half-reactions in Tables from many chemistry books - hopefully yours has one in the Appendix somewhere. One of the reactions will require a certain number of electrons to be pumped into it, and one of them will release a certain number of electrons. These are reductions and oxidations, and they are paired through mass conservation, as well as the conservation of electrical charge.

I don't know if that helps explain where non's numbers are coming from or not, but I'll be happy to fuck up an explanation of it more thoroughly if you like Good luck with the homework.

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surferlove007

I understand how to balance equations that don't involve the electron transfer however the whole when you gain electrons, but you subtract then stuff throws me for a loop. Not to mention how I figure out how many electrons each different element has. I just wish my teacher had explained it better.

His excuse is " Well you guys should remember this from high school." So this should really be a review...not anything new." I think that's a bunch of crap as well as a poor attitude toward teaching this subject.
Chemistry is very, very challenging for me. I work to understand it, but often time it's better to have a one on one session. The homework online is confusing too, it doesn't give the best hints on what to do.

pig

ghoast

ok, take the equations above given by non: these should be available in the appendix in your chem book. i did a quick google, and found some online tutorials, but not the actual electrochem equations. if you have the table, that's a start.

if you take the reduction equation, 2I- -> I2 + 2e-, first notice that the overall charge balances on both sides. I-*2=2negative charges, and the 2e- on the right hand side.

now take the reduction: SO4(2-) + 4H+ + 2e- -> SO2 + 2 H2O ...once again, the charges balance. zero on the left hand side, and overall zero on the right hand side. Now add them:

2I- -> I2 + 2e-
SO4(2-) + 4H+ + 2e- -> SO2 + 2H2O

the 2e- cancel out, leaving
2I- + SO4(2-) + 4H+ -> SO2 + 2H2O + I2

So I'm getting the mole ratio to be (1:1), using the information non gave before. These reaction are typically given as standard reduction reactions. The important thing is that the number of e- have to balance. The half reactions are written such that you will always balance them if you multiply both left and right hand sides by a constant factor. So:

4I- -> 2I2 + 4e-
I- -> (1/2)I2 + e-
etc.

you multiply the equations, essentially finding the lowest common denominator for the number of e-, and then add the half reactions to get the overall balanced reaction.

I don't know if that helps, but I've had some experience with this stuff, and will give what help I can. If you can tell me which part is causing the confusion for you, I might be able to help give you some hints. Free of charge

God, that turned into an awful pun..

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JustJess

Meeeemmm'ries.... all alone in the cheeeemmistry...

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dlish

ahhh ure right JJ.

/threadjack

i loathed chemistry backin high school.. id sit and daydream for the entire hour and then pack up my things and move on when the bell went

the only effort i made was to not come last in the class, which i managed by a paltry 3 marks...

equations dont mean much to me these days..then again they didnt back then! I was more interested in playing rugby or basketball at recess.

sorry..i got carried away..back to normal programming

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