Electrolytic CELL question

[Transformers]

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Hey Guys, in an electrolytic cell with Water...I was wondering what actually happens if you had added two metals, lets say X2+ to the Anode and Y2 to the Cathode....Obviously the External Voltage (which should be greater than the potential between the two combined half reactions) drives the reaction in the reverse direction of the spontaneous reaction....

BUT, when you have WATER in the equation, you have the two half reactions of:

for example, lets say you have HYPOTHETICALLY X2 is at the Cathode and Y2 is at the Anode:

X2+ + 2e- --> X2 E = 5
Y2 + 2e- --> 2Y- E = -10

1.) H20 + 2e --> H2 (g) E = 8
2.) O2 +4e- --> H2O (g) E= -5

So, Im confused....if you have X2+ at the anode and Y2 at the Anode, what do you need to consider when you have these competing reactions 1.) and 2.) due to water???

I mean, since you have an EXTERNAL VOLTAGE...would this drive the the reaction that requires the LEAST chemical potential...so for example, since X2+ had a lower reduction potential than H2O and O2 had a higher potential than Y2...when you combine the two half reactions such that 5 + 5 = 10, thus evolving X2 (reduction) and O2 (oxidation), would these reactions occur in an electrolytic cell? I mean, lets say we assumed that the components of the anode and cathode (X2+ + Y2) proceed spontaneously, then we would produce a potential of (5+10) = 15...and since this is higher than 10, the external voltage battery source would prefer the above reaction right....

Basically, after all this confusing crap...for an electrolytic cell in water, the half reactions that would produce are the ones with the LEAST potential right?...im confused

To make myself clear...why in an NaCl (aq) solution in an electrolytic cell does Chlorine Gas form (oxidation) and Hydrogen Gas (due to reduction) form and why doesnt Sodium metal form INSTEAD of hydrogen gas due to reduction? I get the jist that its harder to reduce sodium than hydrogen, but oxygen is EASIER to oxidize than chlorine and yet chlorine is oxidized...wierd contradiction

Half-reaction E° (V)
Na+ + e− Na(s) −2.71
2 H+ + 2 e− H2(g) ≡ 0
O2(g) + 4 H+ + 4 e− 2 H2O +1.23
Cl2(g) + 2 e− 2 Cl− +1.36

 

[DamienPro]

Hey Guys, in an electrolytic cell with Water...I was wondering what actually happens if you had added two metals, lets say X2+ to the Anode and Y2 to the Cathode....Obviously the External Voltage (which should be greater than the potential between the two combined half reactions) drives the reaction in the reverse direction of the spontaneous reaction....

BUT, when you have WATER in the equation, you have the two half reactions of:

for example, lets say you have HYPOTHETICALLY X2 is at the Cathode and Y2 is at the Anode:

X2+ + 2e- --> X2 E = 5
Y2 + 2e- --> 2Y- E = -10

1.) H20 + 2e --> H2 (g) E = 8
2.) O2 +4e- --> H2O (g) E= -5

So, Im confused....if you have X2+ at the anode and Y2 at the Anode, what do you need to consider when you have these competing reactions 1.) and 2.) due to water???

I mean, since you have an EXTERNAL VOLTAGE...would this drive the the reaction that requires the LEAST chemical potential...so for example, since X2+ had a lower reduction potential than H2O and O2 had a higher potential than Y2...when you combine the two half reactions such that 5 + 5 = 10, thus evolving X2 (reduction) and O2 (oxidation), would these reactions occur in an electrolytic cell? I mean, lets say we assumed that the components of the anode and cathode (X2+ + Y2) proceed spontaneously, then we would produce a potential of (5+10) = 15...and since this is higher than 10, the external voltage battery source would prefer the above reaction right....

Basically, after all this confusing crap...for an electrolytic cell in water, the half reactions that would produce are the ones with the LEAST potential right?...im confused

To make myself clear...why in an NaCl (aq) solution in an electrolytic cell does Chlorine Gas form (oxidation) and Hydrogen Gas (due to reduction) form and why doesnt Sodium metal form INSTEAD of hydrogen gas due to reduction? I get the jist that its harder to reduce sodium than hydrogen, but oxygen is EASIER to oxidize than chlorine and yet chlorine is oxidized...wierd contradiction

Half-reaction E° (V)
Na+ + e− Na(s) −2.71
2 H+ + 2 e− H2(g) ≡ 0
O2(g) + 4 H+ + 4 e− 2 H2O +1.23
Cl2(g) + 2 e− 2 Cl− +1.36

You're right with your supposition here, where it would seem easier to oxidize water than chloride ions. However, the standard state rule isn't the only thing governing at the anode.

First off, the standard electrode potentials for the half reactions of water and chloride ions are relatively close to each other, where a mixture of oxygen and chlorine gases could be liberated at the anode.

Secondly, this is an aqueous solution, say a 15% NaCl (aq) where only 15% of the solution is NaCl by mass. This would significantly decrease the potential needed to oxidize chloride ions.

Thirdly, as the solution turns more and more alkaline due to hydroxide ions produced at the cathode when liberating the hydrogen gases, it becomes a lot harder for water (an extremely weak electrolyte) to get oxidized. (Note: O2(g) + 4&#8202;H+ + 4&#8202;e <&#8722;-> 2&#8202;H2O +1.23 is not as simple as it appears here. In order to liberate O2, H2O needs to get ioinized to H+ and OH- where OH- is further oxidized to O2.)

With these factors involved, getting water oxidized to oxygen is no longer the ideal 1.23 V. Depending on the experimental conditions chosen, the voltage needed can be relatively higher than 1.36 V, making the chloride anion (instead of water) discharge at the anode.

 

[Transformers]

Then, USUALLY...would my SUPPOSITION apply to the fact that at the CATHODE...hydrogen gas will evolve pretty much all the time (compared to any cation we add say Na+ earlier, but now lets say a compound such as MgCl2 or KCl or BaF) while at the ANODE...since O2 is REALLY hard to oxidize and is not reflective in its electrochemical value, we would prefer the other anion in solution, that being chloride or fluoride ion for example which would instead evolve as F2 and Cl2 as opposed to O2..?

I know its weird beacuse we just compare numbers, but I am going to ALWAYS choose that Hydrogen Gas will evolve from the cathode (getting reduced because its EASY as hell to reduce with a E = 0.00 V) while at the Anode, the anion of the salt we Add (MX), I will always say that X- will get oxidized...

 

[DamienPro]

Then, USUALLY...would my SUPPOSITION apply to the fact that at the CATHODE...hydrogen gas will evolve pretty much all the time (compared to any cation we add say Na+ earlier, but now lets say a compound such as MgCl2 or KCl or BaF) while at the ANODE...since O2 is REALLY hard to oxidize and is not reflective in its electrochemical value, we would prefer the other anion in solution, that being chloride or fluoride ion for example which would instead evolve as F2 and Cl2 as opposed to O2..?

I know its weird beacuse we just compare numbers, but I am going to ALWAYS choose that Hydrogen Gas will evolve from the cathode (getting reduced because its EASY as hell to reduce with a E = 0.00 V) while at the Anode, the anion of the salt we Add (MX), I will always say that X- will get oxidized...

You're pretty much right with the assumption for gas evolving from the cathode. However, I highly doubt fluorine gases are going to be liberated there at the anode because the standard electrode potentials for the half reactions of water and fluoride ions are relatively far apart. Your textbook should have an example quoting aqueous sodium sulfate. If sulfate anion is having a hard time getting oxidized, it will be even harder for fluoride ions to do so.


But, I believe questions of this sort would be cut and dry if they ever appeared in the MCAT.

 

[Vihsadas]

Please remember to post content specific questions in the Study Question Q&A forum. I'll move this thread there for you now. 🙂 Good luck on test day.