STUFF WE GOTTA KNOW

[StringFever]

Hey u know for redox reactions how for acidic medium you add then , does it work the opposite way for alkaline medium? I've done a few examples adding first and then but is that how you're meant to do it?

Give me an example question and I'll do it for you!

I find the easiest way to do alkaline-based half-equations is to balance it as if it were in an acidic solution (so with H+ ions), and then after that, balance both sides with the same number of OH- ions. In doing so, the H+ and OH- will form water, and all you need to do is clean it up and you have your basic half-equation.

can you give an example please?

OK, let's take Zn to Zn(OH)2.

Step 1: Set up skeleton equation

Zn --> Zn(OH)2

Step 2: Balance like in acidic environment (with H+s and waters)

Zn + 2H2O --> Zn(OH)2

Zn + 2H2O --> Zn(OH)2 + 2H+ + 2e-


Step 3: Balance any H+ with an OH- (ON BOTH SIDES)
Zn + 2H2O + 2OH- --> Zn(OH)2 + 2H+ + 2e- +2OH-

Step 4: H+ and OH- form water
Zn + 2H2O + 2OH- --> Zn(OH)2 + 2H2O + 2e-

Step 5: Clean it up and voila
Zn + 2OH- --> Zn(OH)2 + 2e-

[homghomg1]

Hey u know for redox reactions how for acidic medium you add then , does it work the opposite way for alkaline medium? I've done a few examples adding first and then but is that how you're meant to do it?

half reactions in alkaline conditions are harder to do
i recommend this
http://www.chemguide.co.uk/inorganic/redox/equations2.html
explains how to do it very well, with some good examples
the main point is that sometimes it is a little difficult to know whether to add OH- or H2O first, but if you do it as if it was under acidic conditions and then cancel out the hydrogen ions by adding OH- to each side, it works out
read some of their examples and you'll see what i mean

[/0]

thanks stringfever and homghomg1

[StringFever]

Hey u know for redox reactions how for acidic medium you add then , does it work the opposite way for alkaline medium? I've done a few examples adding first and then but is that how you're meant to do it?

half reactions in alkaline conditions are harder to do
i recommend this
http://www.chemguide.co.uk/inorganic/redox/equations2.html
explains how to do it very well, with some good examples
the main point is that sometimes it is a little difficult to know whether to add OH- or H2O first, but if you do it as if it was under acidic conditions and then cancel out the hydrogen ions by adding OH- to each side, it works out
read some of their examples and you'll see what i mean

Like my example above

[dekoyl]

In an equilibrium reaction: : (the crappy graph below shows the species at equilibrium)

Code: [Select]

A|___
B|___
C|___
 |___:_________
      t
At time t, if we add a little concentration of B, B spikes up but as it establishes equilibrium again, its concentration is higher than before you added it. A gradually goes up and establishes equilibrium. But what happens to C? In a trial exam, it shows that C spikes up as well (but not as much as B) and goes back to being the same concentration again. Is this right?

       

[/0]

That's odd, I would have thought that [C] would decrease since you have a net back reaction o.O

[dekoyl]

Exactly. That's what I thought as well. This was a NEAP trial exam.

LOL learning from trial exams like this will screw me over =\

[kimtywong]

I pretty sure [C] does not spike up, it just decrease right from when more B was added. hmmmm...

[ayu]

wow...thats an excellent way of balancing alkaline solutions. never knew that. thnx .

[dekoyl]

- An equation does not imply that the concentration graphs will start with NO! You must look at the equilibrium constant!

Okay, in relation to this^

If we have , with , how do you know which lines corresponds to which species?

I looked at the concentration of the three to determine it (which was the correct answer) but wasn't sure how to use the equi. constant.

Thanks

Edit: oops stuffed the LaTeX up for the equi constant =\

[/0]

Oops, maybe that was a bit misleading, usually they won't give you enough information to work it out from the equilibrium constant, so you have to try to use this:

"- In , any changes driven by LCP will involve the graphs increasing/decreasing in ratio 1:2:1 for A:B:C
respectively. i.e. the number in front of the chemical will govern how much the graph of that chemical increases/decreases due to LCP."

Which basically means they have to DO something to the equilibrium so you can observe the change.

e.g. If they put in extra reactant A, then graph A will spike up but then graphs A and B will decrease, but B will decrease twice as much as A, so that's how you tell A and B apart. Also, C will increase and that's how you locate C
e.g. If they dilute the solution then all 3 will decrease, but then A and B will increase, with B increasing twice as much as A. C will decrease.

[dekoyl]

Ah! Thank you /0. I understand now

[Over9000]

You know the electrochemical series?
Well if the equations happen in reverse, does that mean the electrovoltage becomes the opposite charge (i.e + to -)

Say for example if
would

[/0]

Hmm I don't think that applies to the electrochemical series, only enthalpies etc.

As long as you have the difference in potentials you can find out how much voltage is produced by a spontaneous reaction or how much voltage is required to drive a non-spontaneous reaction.

[Over9000]

Hmm I don't think that applies to the electrochemical series, only enthalpies etc.

As long as you have the difference in potentials you can find out how much voltage is produced by a spontaneous reaction or how much voltage is required to drive a non-spontaneous reaction.

Thanks, ths electrocehmical stuff is soo hard, I much prefer LPG.