Chemistry 3/4 2013 Thread

[clıppy]

Thanks nliu, you're a lifesaver

[barydos]

I just have some electrolysis questions which I haven't quite understood yet:

electrochemical series:

Spoiler

answers:

Spoiler

Thanks in advance!

[psyxwar]

(a) copper (II) bromide

So we can see the standard reduction potential of Cu²⁺ is 0.34V and that of Br₂ is 1.09V. As electrolysis is non spontaneous, 2Br^- is going to be oxidised to Br₂  + 2e^-, and Cu²⁺ is going to be reduced to Cu (giving a negative cell EMF). (water is not a reactant because its E naught (is that how you say it?) is higher than than of Br₂.)

At the anode, Br₂(aq) is being produced (as oxidation occurs here, and this is the product) and at the cathode, Cu(s) is being produced.

It's pretty much the same for all of them. I think with (d), the chloride ion will oxidise preferentially to water (as its a concentrated chloride solution), even though the reaction with water would have a less negative EMF. Might need some clarification with a more experienced user though!

[Alwin]

I just have some electrolysis questions which I haven't quite understood yet:
(Image removed from quote.)
electrochemical series:

Spoiler

answers:

Spoiler

Thanks in advance!

7a. I'll go through the process with you step by step first so hopefully it makes sense, then you can go off and try the rest yourself! If you have any more problems, don't hesitate to post them ^^
Copper(II) Bromide (aq) --> Cu²⁺ and Br^- ions present in the solution. However, as we are dealing with electrolysis we cannot disregard water, H₂O.
So, on the electrochemical series we mark out Cu²⁺, Br^- and H₂O where they react as oxidants or reductants

STEP ONE

Now, we find the strongest oxidant and the strongest reductant as they will react together!

Mark it out on the electrochemical series:

STEP TWO

Now, we can write out the half equations:
Anode (+): 2Br^- (aq) --> Br₂ (aq) + 2e^-
Cathode (-): Cu²⁺ (aq) + 2e^- --> Cu (s)

We now have our products, Br₂ (aq) and Cu (s). You just need to go through the same process over and over ^^

At first you might need to print out some extra electrochemical series(s) to draw all over, but you should get better very quickly if you work at it!
Hope this made sense + good luck!!

EDIT: Always forget to put my pics in spoilers. #myamazingpaintskills LOL

[barydos]

7a. I'll go through the process with you step by step first so hopefully it makes sense, then you can go off and try the rest yourself! If you have any more problems, don't hesitate to post them ^^
Copper(II) Bromide (aq) --> Cu²⁺ and Br^- ions present in the solution. However, as we are dealing with electrolysis we cannot disregard water, H₂O.
So, on the electrochemical series we mark out Cu²⁺, Br^- and H₂O where they react as oxidants or reductants

STEP ONE

Now, we find the strongest oxidant and the strongest reductant as they will react together!

Mark it out on the electrochemical series:

STEP TWO

Now, we can write out the half equations:
Anode (+): 2Br^- (aq) --> Br₂ (aq) + 2e^-
Cathode (-): Cu²⁺ (aq) + 2e^- --> Cu (s)

We now have our products, Br₂ (aq) and Cu (s). You just need to go through the same process over and over ^^

At first you might need to print out some extra electrochemical series(s) to draw all over, but you should get better very quickly if you work at it!
Hope this made sense + good luck!!

EDIT: Always forget to put my pics in spoilers. #myamazingpaintskills LOL

Thank you so much, I always love me some great pictures! haha this helps heaps
Thanks psyxwar as well!

[lzxnl]

(a) copper (II) bromide

So we can see the standard reduction potential of Cu²⁺ is 0.34V and that of Br₂ is 1.09V. As electrolysis is non spontaneous, 2Br^- is going to be oxidised to Br₂  + 2e^-, and Cu²⁺ is going to be reduced to Cu (giving a negative cell EMF). (water is not a reactant because its E naught (is that how you say it?) is higher than than of Br₂.)

At the anode, Br₂(aq) is being produced (as oxidation occurs here, and this is the product) and at the cathode, Cu(s) is being produced.

It's pretty much the same for all of them. I think with (d), the chloride ion will oxidise preferentially to water (as its a concentrated chloride solution), even though the reaction with water would have a less negative EMF. Might need some clarification with a more experienced user though!

From what I see, it's a 1M solution of Cl-. That IS the concentration at which the electrochemical series was created, so chloride would not be oxidised preferentiallly to water.

[psyxwar]

From what I see, it's a 1M solution of Cl-. That IS the concentration at which the electrochemical series was created, so chloride would not be oxidised preferentiallly to water.

Fair enough. I think this stuff about Cl- being oxidised preferentially to water is taught in VCE though, and considering the Nernst equation isn't taught, what's the point? o_o

[lzxnl]

Fair enough. I think this stuff about Cl- being oxidised preferentially to water is taught in VCE though, and considering the Nernst equation isn't taught, what's the point? o_o

This problem arises because VCE chemistry does not address these fine details carefully enough. The point, I think, is to alert students into not following the electrochemical series without thinking.

[BasicAcid]

Just wondering, why is H2O always involved/considered? It only states the solution is present.

[Alwin]

Just wondering, why is H2O always involved/considered? It only states the solution is present.

solutions are aqueous, so water is present as the solvent

If the reaction occurred without water (eg molten state) then we wouldn't consider water.

hope this makes sense

[barydos]

Just wondering, why for electrolysis, the anode is positive and the cathode is negative.
What conditions determine the "charge" of the electrodes?

[teletubbies_95]

Omg haven't been on this thread for a long time!
@Anonymiza:
So essentially , the anode is positive because the it is giving up electrons to the positive cathode of the galvanic cell . The way how I remember these concepts is that "+" always attracts with "+". So the positive terminal of the galvanic cell is attracted to the anode , thus the anode has to be positive.
On the other hand, as the cathode is inputting electrons , thus becoming negatively charged. Thus, the anode of the galvanic cell is negative as well. Remember "-" and "-" attract.

I hope that helps !

[lzxnl]

Just wondering, why for electrolysis, the anode is positive and the cathode is negative.
What conditions determine the "charge" of the electrodes?

Omg haven't been on this thread for a long time!
@Anonymiza:
So essentially , the anode is positive because the it is giving up electrons to the positive cathode of the galvanic cell . The way how I remember these concepts is that "+" always attracts with "+". So the positive terminal of the galvanic cell is attracted to the anode , thus the anode has to be positive.
On the other hand, as the cathode is inputting electrons , thus becoming negatively charged. Thus, the anode of the galvanic cell is negative as well. Remember "-" and "-" attract.

I hope that helps !

Erm...last time I checked...positive charges attract negative charges. I don't see how your logic works there. For a galvanic cell, the positive electrode is the site of reduction, i.e. where the electrons are headed.

Don't think of them as "positive" and "negative" electrodes. Think of them as electrodes with higher and lower electric potential, so the positive electrode has higher electric potential. Generally, positive charges move from higher to lower electric potential spontaneously; negatives do the opposite. Makes sense; electrons leave the negative terminal of the battery to go to the positive. So, negative electrode = electrode at lower potential. ETC.

The way I think about it is like this. To create an electrolytic cell, you need to meet two requirements.
1. Voltage of the power source is higher than the voltage of the galvanic cell that would otherwise be created.
2. Positive terminal of battery (assuming we have a battery) is connected to the positive electrode, negative terminal connected to negative electrode.

So what happens? Let's assume that initially the positive electrode of the battery and the positive terminal were connected first. Therefore they are at the same potential. Then, we connect the negative terminal to the negative electrode. As the potential drop across the battery is larger than the potential difference from the positive electrode to the negative by definition, the negative terminal of the battery is actually at a lower potential than the negative electrode. Therefore, electrons flow from the negative terminal to the negative electrode, i.e. the negative electrode is the cathode.
Now, as the voltage across the battery is constant, electrons are drawn from the positive electrode to the positive terminal of the battery to replace the electrons lost at the negative terminal. This way, a current flows where the electrons come from the positive electrode and go to the negative.

Alternatively, if the negative terminal is connected first to the negative electrode, when the positives are connected, the positive terminal of the battery is at higher potential than the positive electrode as the battery's voltage is bigger than the cell's voltage. This means the electrons will flow from the positive electrode to the positive battery terminal. Et cetera.

This is what the books really mean by "external power sources drive a non-spontaneous reaction". You get a larger power source to override the potential difference.

[teletubbies_95]

Haha! Sorry for my wording! There was no logic involved in that ! It's just my way of remembering it!
Sorry if I confused anyone!
and btw thanks for the amazing explanation once again!

[zvezda]

Erm...last time I checked...positive charges attract negative charges. I don't see how your logic works there. For a galvanic cell, the positive electrode is the site of reduction, i.e. where the electrons are headed.

Don't think of them as "positive" and "negative" electrodes. Think of them as electrodes with higher and lower electric potential, so the positive electrode has higher electric potential. Generally, positive charges move from higher to lower electric potential spontaneously; negatives do the opposite. Makes sense; electrons leave the negative terminal of the battery to go to the positive. So, negative electrode = electrode at lower potential. ETC.

The way I think about it is like this. To create an electrolytic cell, you need to meet two requirements.
1. Voltage of the power source is higher than the voltage of the galvanic cell that would otherwise be created.
2. Positive terminal of battery (assuming we have a battery) is connected to the positive electrode, negative terminal connected to negative electrode.

So what happens? Let's assume that initially the positive electrode of the battery and the positive terminal were connected first. Therefore they are at the same potential. Then, we connect the negative terminal to the negative electrode. As the potential drop across the battery is larger than the potential difference from the positive electrode to the negative by definition, the negative terminal of the battery is actually at a lower potential than the negative electrode. Therefore, electrons flow from the negative terminal to the negative electrode, i.e. the negative electrode is the cathode.
Now, as the voltage across the battery is constant, electrons are drawn from the positive electrode to the positive terminal of the battery to replace the electrons lost at the negative terminal. This way, a current flows where the electrons come from the positive electrode and go to the negative.

Alternatively, if the negative terminal is connected first to the negative electrode, when the positives are connected, the positive terminal of the battery is at higher potential than the positive electrode as the battery's voltage is bigger than the cell's voltage. This means the electrons will flow from the positive electrode to the positive battery terminal. Et cetera.

This is what the books really mean by "external power sources drive a non-spontaneous reaction". You get a larger power source to override the potential difference.

When you say that after connecting the positive terminal to the positive electrode, the two are therefore at the same potential, what do you mean? Why therefore? Would you be able to explain further?