TPR electrochemistry question

[rcmp1234]

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Two inert electrodes connected to a power source are placed into an aqueous solution of sodium chloride. In the resulting electrolytic reaction, which of the following species forms at the anode?
a) Cl2
b) O2
c) H2
d) Na

Na+ +e = Na(s) E=-2.71V
Cl2 + 2e = 2Cl- E=+1.42V
2H2O + 2e = H2 +2OH E=-0.83V
O2 + 4H3O+ + 4e = 6H2O E=+1;23V

I'm having trouble understanding why the answer is b) and not a). I vaguely recall from high school that it has something to do with competition between the two reactions, but can someone please explain step by step why it is O2 ?

thanks alot

 

[Jengreef]

never mind

 

[dsoz]

Two inert electrodes connected to a power source are placed into an aqueous solution of sodium chloride. In the resulting electrolytic reaction, which of the following species forms at the anode?
a) Cl2
b) O2
c) H2
d) Na

Na+ +e = Na(s) E=-2.71V
Cl2 + 2e = 2Cl- E=+1.42V
2H2O + 2e = H2 +2OH E=-0.83V
O2 + 4H3O+ + 4e = 6H2O E=+1;23V

I'm having trouble understanding why the answer is b) and not a). I vaguely recall from high school that it has something to do with competition between the two reactions, but can someone please explain step by step why it is O2 ?

thanks alot

Reverse the two reactions and you get

2Cl- --> Cl2 + 2e- E=-1.42
6H2O --> O2 +4H3O+ + 4e- E=-1.23

Between the -1.23 and -1.42, which is more likely to happen. If you had to pay $1.42 or $1.23 for exactly the same item, which would you choose? Probably the cheaper of the two. In the reaction, the least negative is the cheaper one.

HTH

dsoz

 

[Jengreef]

Ok, I felt like I was saying something wrong in my first explanation, so I edited it out. I think I have it now, though (lol).

Oxidation occurs at the anode for negative species to lose electrons and become neutral.

Reduction occurs at the cathode for positive species to gain electrons and likewise become neutral.

Since we're interested in the anode, we eliminate choice C and D which are cations.

Between Cl2 and O2, like dsoz pointed out, you want the lesser negative value, but you have to first change the reduction potentials given for Cl2 and O2 into oxidation potentials. Remember, we're not reducing Cl2 or O2, but oxidizing them (assume the default potential given is always reduction). You convert reduction to oxidation potential by changing the sign of the reduction potential.

After this we can see that it costs less energy to oxidize O2 than Cl2, so your answer is b.

As a side note, you'll notice that oxidizing either of these is unfavorable due to the negative oxidation potentials. This is fine because that's the whole point of the electrolytic cell is to drive redox reactions in unfavorable directions in order to separate elements in a compound into their various components.

Hopefully I didn't make any incorrect statements here this time. 😛

 

[rcmp1234]

that makes so more sense to me now, thanks!!

 

[sciencebooks]

Ok, I felt like I was saying something wrong in my first explanation, so I edited it out. I think I have it now, though (lol).

Oxidation occurs at the anode for negative species to lose electrons and become neutral.

Reduction occurs at the cathode for positive species to gain electrons and likewise become neutral.

Since we're interested in the anode, we eliminate choice C and D which are cations.

Between Cl2 and O2, like dsoz pointed out, you want the lesser negative value, but you have to first change the reduction potentials given for Cl2 and O2 into oxidation potentials. Remember, we're not reducing Cl2 or O2, but oxidizing them (assume the default potential given is always reduction). You convert reduction to oxidation potential by changing the sign of the reduction potential.

After this we can see that it costs less energy to oxidize O2 than Cl2, so your answer is b.

As a side note, you'll notice that oxidizing either of these is unfavorable due to the negative oxidation potentials. This is fine because that's the whole point of the electrolytic cell is to drive redox reactions in unfavorable directions in order to separate elements in a compound into their various components.

Hopefully I didn't make any incorrect statements here this time. 😛

Everything seems correct with a quick read through, 🙂.