MedVision ad

galvanic cells (1 Viewer)

Aerlinn

Member
Joined
Aug 29, 2006
Messages
194
Location
Oz
Gender
Female
HSC
2007
Several pressing questions :)

I was told by one person that it wouldn't matter if the cells aren't under standard conditions (when doing experiment). I've read that the the order in the electrode potentials table can change if you don't! =S Eg. for a particular cell which half cell you connect to the positive/negative terminal might change? (hoping I got that right. Is there a difference between using an anmeter and a voltmeter? We used a voltmeter.) About that question before the last, I'm assuming that you connect half cells to terminals depending on which combination gets a positive voltage, not by examining the electrochemical series (ie. the one lower down's always negative, or something like that), so someone please correct me if i'm wrong

Why should the half cell solutions be comparable in molarity? Eg. All be 1M solutions (I didn't think molarity was measurable...)

If the cell was left to keep going, would the overall reaction continue-- What problems might arise? Some are convinced that you don't need a salt bridge, the reaction runs anyway. But how would you otherwise have charge running through the thing if you don't complete the circuit?

Experiment questions too, yippee :)
Hypothesize why the measured potentials were higher/lower than the calculated values.
It would probably because of it not being under standard conditions? But being more specific... i don't know.
What would a negative reading on the voltmeter indicate? Why?
It's the 'why' part I'm stuck on.

help = :) :wave:
 

xiao1985

Active Member
Joined
Jun 14, 2003
Messages
5,704
Gender
Male
HSC
N/A
It's true that you need the cond to be at standard condition, and at 1M conc for soln.

Take a simple example:

Cu2+ +2e- --> Cu (s)

If you have too much Cu2+, then the reaction is going to favor the product side (LeChatelier's principle), and give you a higher than normal reduction voltage. Conversely, if conc. is too low ...

You will need a salt bridge for the reaction to happen...
Also, if the electrochemical reaction is been left for too long, it will lose voltage overtime, for the similar reason mentioned above (Reactant conc drops. reaction rate drops due to le chatelier)
 

Aerlinn

Member
Joined
Aug 29, 2006
Messages
194
Location
Oz
Gender
Female
HSC
2007
lechatelier's makes sense, but I thought it was something along the lines of the changing order of the reactions listed in the table when you change concentrations around? Dont know EXACTLY how it works, though...

Does loosing voltage mean the reaction has less tendency to occur, or has less energy/ speed, or something else?

Questions i realli need some help with:
1. Hypothesize why the measured potentials were higher/lower than the calculated values.
2. What would a negative reading on the voltmeter indicate? Why?
3. ome questions regarding the electrochemical series:
-When finding species on the series, are there other 'weak' species like NO3-, SO4(2-) which you wouldn't include?
-I've come across several question involving the manganate ion, MnO4-. It's not on the series (that we're given). Or is Mn2+ supposed to be what we're finding?
-I've come across questions involving these half reactions: (especially this one) Mn2+ being converted to MnO4- (and vice versa), and Sn4+ being converted to Sn2+ (and vice versa). How do we predict whereabouts these species are on the series (and relative strengths, etc.), so that we know which species go through half reactions in a cell?
Eg. this one:
Which of the following reactions would be expected to proceed appreciably to the right?
A) 2AG(s) + Cu2+(aq) <---> 2Ag+(aq) + Cu(s)
B) 2H2)(l) + Pb(s) <--> Pb2+ + H2(g) + 2OH-(aq)
C) Sn4+(aq) + Sn(s) <---> 2Sn2+(aq)
D) Fe2+(aq) + Ni(s) <---> Ni2+(aq) + Fe(s)
I answered d, but the answer's c... which doesn't make sense at all. d can happen, c... Fe2+(aq) and Ni(s) are both oxidants...

And this one: A simple galvanic cell based on the reactants MnO4-/H+ and Cu is constructed. In Beaker X is Electrode I in MnO4-/H+ solution, connected to beaker Y by a salt bridge. Beaker Y contins Electrode II in Cu2+ solution. The overall balanced equation representing the reaction occuring in this cell is...
---
Aside from these, what are major errors of experiments involving constructing galvanic cells? I've only listed some.
:)
 

xiao1985

Active Member
Joined
Jun 14, 2003
Messages
5,704
Gender
Male
HSC
N/A
if a reaction has less tendency to happen, of course the E value will decrease. If E value decreases, then the order on standard potential table maybe different or switched.

"does lose ..." correct

"hypothesise..." you tell me. knowing that conc. affact redox like normal reaction would...

"negative reading" that positive terminal and negative terminal is attached wrongly?

" weak species"PO4 3- maybe?

"Mn2+ and Sn" if it's not on the standard potential table, chances are, you won't need to know it (it won't be tested)

ok maybe i was wrong... hmm... it must mean Ni has a lower oxidation potential than Iron (s)... intriguing. I have no answer for you on this one, other than check on standard potential table... If the table doesn't say anything, tell your teacher he's mean.

MnO4/Cu: which one is the oxidant and which one is the reductant?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top