help challenging gen chem question

[yahoogoogle]

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2 H2O + 2 e- > H2 + 2 OH- E = -.8V

Na+ + e- > Na E=-2.7V

Cl2 + 2 e- > Cl- E=+1.4V

Based on the reduction potentials given above, which of the follwing would be expected to osccur when electrodes connected to the terminals of a 2V battery are immersed in a solution of NaCl in water?

answer:

E) Chlorine gas will appear at the anode, and hydrogen gas will appear at the cathode.

My reasoning is that to get 2V, we can only use Cl and water reaction where the Cl is a good oxidizing agent(reduction) and should serve at the Cathod end. I just don't see how chlorine gas will show up at the anode end where oxidations normally take place.

 

[Danny289]

2 H2O + 2 e- > H2 + 2 OH- E = -.8V

Na+ + e- > Na E=-2.7V

Cl2 + 2 e- > Cl- E=+1.4V

Based on the reduction potentials given above, which of the follwing would be expected to osccur when electrodes connected to the terminals of a 2V battery are immersed in a solution of NaCl in water?

answer:

E) Chlorine gas will appear at the anode, and hydrogen gas will appear at the cathode.

My reasoning is that to get 2V, we can only use Cl and water reaction where the Cl is a good oxidizing agent(reduction) and should serve at the Cathod end. I just don't see how chlorine gas will show up at the anode end where oxidations normally take place.

Dear yahoo,

first of all, you are absulatly right, and it showes you are following the concepet. and that is good thing. now let me explain this problem to you.
I don't know you have heard this sentence or not; " Chemistry is experimental Science" that means many concepts and explanation comes after result in lab. this case is one of those.
I don't know you have done elctrolyisis in High school or collage or not? in that experiment we add a little( tip of spoon NaCl) to the water, then pass the carrent( electricity) from solution and obtain hydrogyn gas and oxygyn.
the point is if you have just dilute Nacl with very low concentration, you will have H2 and O2 that what you expect. if you have Concentrated Solution ( as above problem ) you are going to have Cl2!
What is explanation?
in low concentarated solution, we know H20 (OH-) will loose electron but in high concentrated soultion anions(cl- and O-2(H20) have competiton for oxidation but Cl- will win because we have more Cl- in solution.
this is what you can see in text books.
but in my opinion(maybe I am wrong) Na+ has very attraction to OH- and dont let H20 (OH-) oxidize.

hopefuly it helped.

 

[yahoogoogle]

Dear yahoo,

first of all, you are absulatly right, and it showes you are following the concepet. and that is good thing. now let me explain this problem to you.
I don't know you have heard this sentence or not; " Chemistry is experimental Science" that means many concepts and explanation comes after result in lab. this case is one of those.
I don't know you have done elctrolyisis in High school or collage or not? in that experiment we add a little( tip of spoon NaCl) to the water, then pass the carrent( electricity) from solution and obtain hydrogyn gas and oxygyn.
the point is if you have just dilute Nacl with very low concentration, you will have H2 and O2 that what you expect. if you have Concentrated Solution ( as above problem ) you are going to have Cl2!
What is explanation?
in low concentarated solution, we know H20 (OH-) will loose electron but in high concentrated soultion anions(cl- and O-2(H20) have competiton for oxidation but Cl- will win because we have more Cl- in solution.
this is what you can see in text books.
but in my opinion(maybe I am wrong) Na+ has very attraction to OH- and dont let H20 (OH-) oxidize.

hopefuly it helped.

thanks for detailed explanantion. the solution says that "a voltage of 2V is not enough to cause the reduction of Na+, so no solid sodium will form. A foltage of 2V will cause the reduction of water, which forms hydrogen gas, and the oxidation of Cl-, which forms cholrine gas.

From AN OX and RED CAT(referring to reduction for cathod and oxidation for anode rule), we know that hydrogen gas is formed at the cathode and that chlorine gas is formed at the anode."

I think that this has little to do with the experiment but simply identification of cathode and anode problem though

 

[Dentalicous1234]

thanks for detailed explanantion. the solution says that "a voltage of 2V is not enough to cause the reduction of Na+, so no solid sodium will form. A foltage of 2V will cause the reduction of water, which forms hydrogen gas, and the oxidation of Cl-, which forms cholrine gas.

From AN OX and RED CAT(referring to reduction for cathod and oxidation for anode rule), we know that hydrogen gas is formed at the cathode and that chlorine gas is formed at the anode."

I think that this has little to do with the experiment but simply identification of cathode and anode problem though

the more positive the reduction potential the more likely the species will be reduced. Cl still has a greater + potential so why isn't it reduced and (using REDCAT) be said to be on the cathode? i understand why Na won't be made but Cl is more positive than H still too.

 

[au5233]

Oh I am so confused!

Ok this is what I think.

Chlorine has the most + E, so we know it will be reduced, thus this will occur at the cathode. But the reduction formula is Cl2 + 2e ---> Cl-, so we are taking chlorine gas AWAY at the cathode (where this reduction process is occuring) to make Cl-. Now, the reverse of this reaction, aka the oxidation, is occuring at the anode. This oxidation process is Cl- ---> Cl2 + 2e. So this is why we are getting chlorine gas at the anode.

Now my confusion is how do we know to use the H20 half reaction and Not the Na half reaction as our second reaction?? Is we look at who is most +E for the reduction, do we look who is least +E for oxidation, which wouldn't that be the Na reaction?

 

[Dentalicous1234]

Oh I am so confused!

Ok this is what I think.

Chlorine has the most + E, so we know it will be reduced, thus this will occur at the cathode. But the reduction formula is Cl2 + 2e ---> Cl-, so we are taking chlorine gas AWAY at the cathode (where this reduction process is occuring) to make Cl-. Now, the reverse of this reaction, aka the oxidation, is occuring at the anode. This oxidation process is Cl- ---> Cl2 + 2e. So this is why we are getting chlorine gas at the anode.

Now my confusion is how do we know to use the H20 half reaction and Not the Na half reaction as our second reaction?? Is we look at who is most +E for the reduction, do we look who is least +E for oxidation, which wouldn't that be the Na reaction?

thanks for the response!
okay the cholrine gas* stuff makes sense. just shows we gotta watch out for the details - I wasn't trying to find the gas i just saw chlorine and was like okay it's being reduced so cathode with chlorine! it's sooo easy to overlook such stuff when your trying to go fast! but the chlorine Gas being at the anode makes sense, anyone else wanna confirm it too?

To answer your question: From the previous posts it sounds like we don't consider Na because we are only supplied with 2V and that's outside the scope of the voltage required for Na so basically we don't have enough voltage to make a reaction for the Na. That's what I'm getting from it but I could be wrong?

anyone wanna confirm this pleaseee? 🙂

 

[au5233]

OK I just spent some more time on electrochemistry and I think it kinda just clicked a lot more in my head now!

I came across a similar problem which used H20, Na and Cl and their explanation was that H20 is easier to reduce tham Na (b/c more positive E) so that is the reason why this is our second oxidat/reduc reaction with chlorine.

Good question, I'm glad I came across your posts because I think I understand this whole section a lot better now. Best of luck studying!