Some Q's, plz help me :(

[lisafaustina]

1. What happens when a cell is being recharged? like do the electrodes' polarity change.... its not clearly explained in the heinemann book, i dont know what reaction (oxdiation, reduction) occurs at which electrode and such... can somebody clear that up for me plz

2. in electrolysis, if the reaction for the cathode is placed below the reaction for the anode, does a reaction still occur? it doesnt happen in galvanic cells right....

3. in galvanic cells, can water be an oxidant or a reductant? so if you have all these aqueous solutions, do you take water into account when you work out which species is the anode or the cathode?

4. in galvanic cells, if the same half equation is both the oxidation and reduction reaction, does a reaction still occur? eg. Cu(s) + Cu2+(aq) --> Cu(s) + Cu2+(aq) etc

Thank you heaps! :smitten:

[Blakhitman]

1. What happens when a cell is being recharged? like do the electrodes' polarity change.... its not clearly explained in the heinemann book, i dont know what reaction (oxdiation, reduction) occurs at which electrode and such... can somebody clear that up for me plz

No the polarities don't change, however the cathode and anodes exchange positions, so the negative is the cathode and the positive is the anode. And of course, reduction occurs at cathode and oxidation at the anode.

2. in electrolysis, if the reaction for the cathode is placed below the reaction for the anode, does a reaction still occur? it doesnt happen in galvanic cells right....

Yep, in electrolysis it does, however energy is required as the reactions are non-spontaneous.

3. in galvanic cells, can water be an oxidant or a reductant? so if you have all these aqueous solutions, do you take water into account when you work out which species is the anode or the cathode?

Yes, sometimes they may ask why an aqueous solution of a certain metal (e.g Mg) won't deposit on an electrode, and the reason could be (and most likely) that water is reduced preferentially to the metal.
Thank you heaps! :smitten:

Someone correct me if needed.

[Mao]

@Blackhitman, spot on.

4. in galvanic cells, if the same half equation is both the oxidation and reduction reaction, does a reaction still occur? eg. Cu(s) + Cu2+(aq) --> Cu(s) + Cu2+(aq) etc

It is possible. this is the case of electroplating. To show the reaction more clearly, say I'm plating a spoon with copper (because y'know, we don't care about being poisoned or anything), I can write this:

Cu(s,electrode) --> Cu2+(aq) + 2e
Cu2+(aq) + 2e --> Cu(s,spoon)

The overall equation is
Cu(s,electrode) + Cu2+(aq) --> Cu2+(aq) + Cu(s,spoon)
Or, simplified to
Cu(s,electrode) --> Cu(s,spoon)

Note that generally, this is just written as Cu(s) --> Cu(s), however, since the location of these are different, it is still a 'reaction'.

[lisafaustina]

ahh okay. but what about in galvanic cells? it cant happen in galvanic cells right?

[Mao]

ahh okay. but what about in galvanic cells? it cant happen in galvanic cells right?

no, that is electrolysis. An electrochemical reaction (i.e. a galvanic cell reaction) always require a thermodynamically favoured reaction pair (i.e. negative gradient).

[jasoN-]

To add:
EMF = E(oxidant)-E(reductant)
We can see with a galvanic reaction, the EMF value is positive
eg.
Cu^2+ + 2e^- -> Cu  E=0.34
Zn^2+ + 2e^- -> Zn  E=-0.76
EMF = E(oxidant)-E(reductant)
      = 0.34 -(-0.76)
      = +1.10V
The positive value indicates that energy is being released (as electricity)

If we look at an electrolytic reaction (lets say zinc plating copper, so I can use the same example)
Cu^2+ + 2e^- -> Cu  E=0.34
Zn^2+ + 2e^- -> Zn  E=-0.76
EMF = E(oxidant)-E(reductant)
      = -0.76-0.34
      = -1.10V
The negative value indicates that we need to use electrical energy (eg. charger) which has an EMF higher than 1.10V for the reaction to proceed.
This is how forcing electrons gives electrical energy to convert the products back into reactants for recharging.
Anyway the EMF between copper reacting with copper ions is 0, indicating that no energy is released -> does not react

This is just what I think, correct me if I'm wrong.