someone correct my work please? fun, fun chemistry

posted by Rienna

Okay. I'm very, very confused.

The initial question is: predict the products of the electrolysis of a 1 mol/L solution sodium chloride.

This is my work:

The four different possible equations are:
2H2o + 2e- -------> H2 + 2OH- (V= -0.83)
O2 + 4H+ + 4e- -------> 2H2O (V= 1.23)
Na+ + e- ---------> Na (V= -2.71)
Cl2 + 2e- -----> 2Cl- (V= 1.36)

So the reaction of O2 and the reaction of Cl2 are the reduction reactions, and the reaction of 2H20 and Na are the oxidation reactions.

[ could someone tell me if I'm using the right 2 equations for water? ]

Then I combine the pairs of half reactions:

Cl2 reaction with the 2H2o reaction: (2.19)
Cl2 reaction with the Na reaction: (4.07)
O2 reaction with the 2H2o reaction: (2.06)
O2 reaction with the Na reaction: (3.94)

[i used the equation E cell= Ecathode-Eanode. However, I'm confused because its an electrolysis reaction, and I'm supposed to get negative numbers.]

And then I know that the reaction of 02 with 2H2O requires the least amount of energy, so...

O2 + 4H+ + 4e- ---------> 2H2O (v=1.23)
H2 + 2OH- -----> 2H2O + 2e- (flipped) (V= +0.83)

2H2 + o2 + 4OH- ----------> 6H2O (+0.40)

ANNNDDD that what I did. And I don't get the right answer, as my products are supposed to be hydrogen and oxygen. Could someone PLEASE correct my work?

Water is reduced in preference to Na^+ at the cathode. Water is also prreferentially oxidized relative to the chloride at the anode.
2H2O + 2e ==> H2 + 2OH^-
2H2O ==> O2 + 4H^+ + 4e

Multiply equation 1 by 2 and add.
4H2O + 4e ==> 2H2 + 4 OH^- (cathode)
2H2O ==> O2 + 4H^+ + 4e (anode)
6H2O ==>2H2 + 4 OH^- + 4H^+ + O2
combine ions
6H2O ==> 2H2 + 4H2O + O2
net equation: 2H2O ==> 2H2 + O2

I just reread your answer and although you did a lot of unnecessary work I think your error was that you didn't multiply the equations involving water (the last set you wrote) to make the electons equal. Therefore, the final equation wasn't balanced and that's why the H^+ and OH^- didn't combine and add out to give 2H2O on the left.

Respond to this Question

First Name

Your Answer

Similar Questions

  1. English

    1. This book is really fun. 2. This book is real fun. 3. This book is great/good fun. (I have a question. You said #1 and #2 are right, but that #2 is not right. What about 'good' fun?
  2. Chemistry

    I have a question about buffers. Part A So it starts with 20ml 0.1 sodium acetate and 25ml 0.1 acetic acid. Calculate ph of buffer is 4.74 because the acid and conjugate base have the same molarity correct?
  3. English

    1. I liked the work very much because I could help orphans who needed love. 2. I liked the work very much because I could help children who had no father and mother. 3. I liked the work very much because I could help orphans who needed …
  4. English

    1. I saw something very funny. 2. I saw something very fun. (Are both expressions correct?
  5. chemistry

    sodium is manufactured commercially by electrolysis of molten sodium chloride. explain why it is not possible to use sodium chloride solution for this.
  6. Chemistry

    Your Open Question Show me another » Chemistry help needed... Confused?
  7. Chemistry

    Please help! I have spent hours trying to work these three problems. How many grams of solid barium sulfate form when 27.1 mL of 0.160 M barium chloride reacts with 53.4 mL of 0.065 M sodium sulfate?
  8. Chemistry

    Predict the products of electrolysis in an aqueous solution of FeCl2. Please help. I don't understand how the answer is OH-, O2, H+, H2.
  9. Mental Health

    If the correlation between one’s interest in statistics and being a “fun date” was –0.70, it would mean that: a. The higher someone’s interest was in statistics, the more likely it would be that he or she is a fun date b. …
  10. Chemistry

    Hello this is my first experiment I've ever done before and it was fun but I gotta figure out what NaCl produced and what it yielded in percentage. Can someone double check my work. Na2CO3+2HCl=2NaCl+H2O+CO2 Mass of empty crucible …

More Similar Questions