*where*is*the*light*switch -shipwrecks (1 Viewer)

[+:: $i[Q]u3 ::+]

argh.. i thought i understood it.. but i swear, trying to get shipwrecks done okay is n o t h a p p e n i n g...
if anyone has the thickett book (pathways) could u check this up
p419/420
it's got a pic of a beaker with salt water, a zinc anode and iron cathode. pretty familiar eh?
then it says..
anode: zn -> zn2+ & 2e-
cathode: 2h2o & 2d- -> 2oh- + h2(g)
doesn't this have a negative e-value???? >.< argghhh...

 

[mercury]

when Ecell is negative, the reaction is not spontaneous
so u need to apply voltage to it... which is what u do for electrolytical cells.

 

[mercury]

oh i see what u mean

 

[mercury]

ah I see, this is actually two cells connected in series and BOS is blocking me again

edit: may be i shouldn't assume that. in which case, the book may have made a boo boo

 

[+:: $i[Q]u3 ::+]

boo =P

 

[eeyore]

I don't get what you're asking about?

All they're saying is that the zinc will be oxidised preferentially to the iron, hence acting as a sacrificial anode

 

[mercury]

We have

Zn -> Zn 2+ + 2e- E = 0.76
Fe -> Fe 2+ + 2e E = 0.44
H2O + e- -> 1/2 H2 + OH- E = - 0.83

sure, u say that zinc is oxidised preferentially coz then the cell potential would be more positive than oxidising Fe.
Since Zn/H2O Ecell = - 0.07
and Fe/H2O Ecell = - 0.39

but galvanic cells are spontaneous, ie. cell potential must be positive (otherwise u'd need to apply voltage... that'd be electrolytical cell)
In which case, in the presence of Fe, Zn and air + water, the reaction that would dominate would be:

Zn -> Zn2+ + 2e- E = 0.76
1/2O2 + H2O + 2e- -> 2OH- E = 0.40

cell potential = 1.36 > 0 the Fe/O2/H2O one would give Ecell = 0.84

So i don't see why they are chucking in this half equation that yields a non spontaneous reaction as well as ignoring the presence of oxygen. Sure, zinc would be acting as a sacrificial anode... but the cathode reaction is prolly booboo.

 

[+:: $i[Q]u3 ::+]

yep.. its the cathode reaction i'm grumbling about.. and it happens again on the next page =P

 

[eeyore]

That's why you don't bother reading Pathways - it's a bunch a crap

 

[phenol]

The actual reason sique is notice it says salt water solution

the potential of your anode equation is actually dependent on the concentration of amound of zinc 2+ ions in the soln. Since there's like none or near none, the oxidation potential of your anode is actually larger than the standard potential given (which assumes 1M zinc ions). So those two half equations are correct.

If you want to know more about this relationship, it is called the Nernst equation.

Don't think i need to say more for the next set of half equations



you can (really dodgy analogy) think le chatelier, if there's no Zn 2+ ions present, the reaction has larger tendency of going towards the RHS therefore potential is larger than standard.
(this analogy is not flawless, just makes it easier for APhO ppl ok i am being slack, sorry )



P.S. i wanted to go to taiwannnnnn

 

[mercury]

quick reference:
nernst equation: E = E(standard condition) - (RT/nF) lnQ

yep, phenol's right. Q is equilibrium quotient, when Q = K, system at equilibrium, Q > K, equilibrium lies far to the right, Q<K equilibrium lies far to the left. So depending on the salt concentratin of the system, if it gives Q < K, then E will be positive, so reaction would be spontaneous.

I was bugger considering the spontaneity not even considering nerst. in reality both water reactions occur. but the one mentioned in textbook perhaps somehow work a bit better.

 

[+:: $i[Q]u3 ::+]

thanks =) *grumble*

mercury, didn't u hear what phenol said???

Originally posted by phenol
you can (really dodgy analogy) think le chatelier, if there's no Zn 2+ ions present, the reaction has larger tendency of going towards the RHS therefore potential is larger than standard.
(this analogy is not flawless, just makes it easier for APhO ppl ok i am being slack, sorry )

nerst is beyond me. this i understand. LOL...
(it's okay.. i'm used to taking shortcuts.. haha.. we were taught to use *dodgy maths* as well... like dodgy integrating... *shifty*)

 

[mercury]

*puts hand up*

yes yes sique one cannot argue the analogy of an icho medalist.

and after all "this analogy is not flawless, just makes it easier for APhO ppl"

haha, just joking... yeah, phenol's explanation is very good.

 

[hipsta_jess]

yar,
-E= non-spontaneous (ie electrolytic cell)
+E= spontaneous (ie galvanic cell)

 

[Frigid]

given that both cells are exposed to the air why can't the reduction eqn be:

1/2 O2 + H2O + 2e- ---> 2OH- ?

i forget... i remember there's a good reason, but i forget

 

[mercury]

that reaction occurs as well, frigid.
both water reactions occur. in fact, the oxygen/water one theoretically should work better. But in practise the book one works better apparently, can't think of any reasons atm

 

[Frigid]

ohhhhhhhhhhh... worked out why.....

bcoz they're placed in salt water (ie fully submerged)... the diagram is there to show "the equivalent galvanic cell for these two pipes"... it is not the pipes per se, but a graphical equivalent...

so i assume the pipes are fully-submerged and/or underground, where there is a low oxygen concentration... thus the other reduction eqn would dominate. nice.

 

[mercury]

hehe, yeah really depends on oxygen concentration i guess.

 

[inasero]

i think youre both right (yewon and shannon) the system you're talking about is not going to happen unless you impress 0.07V of electricity onto it....so....the cathode reaction should in fact be O2 + 4H+ + 4e- ----->2H2O E0=+1.23V

hehe now u can stop freaking

 

[+:: $i[Q]u3 ::+]

it's the nerst.. the nerst... whatever that was.
read phenol's post. he explained it pretty well. ^^