July 02, 2022, 01:14:42 PM
Forum Rules: Read This Before Posting


Topic: A problem of electrochemistry.  (Read 5493 times)

0 Members and 1 Guest are viewing this topic.

Offline Telesforo

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
A problem of electrochemistry.
« on: April 03, 2010, 11:43:33 AM »
An exercise I have tried to solve is this:

"Show electrode semiractions that take place during the electrolysis of a solution 1,0 M of KOH."

My proceeding to solve this problem:
At the cathode, the negative pole, occurs the reduction.
Species wich can undergo reduction are K+ and H2O.
It is necessary however to see what semielement algebraically has the HIGHEST potential of reduction.
Possible semireactions:

1) K+ + e-  -->  K
2) 2 H2O + 2 e-  -->  2 OH- + H2 .

Since [K+] e [OH-] are 1 M, respective potentials are:

1) E(K+/K) = E°(K+/K) = -2,93

2) E(H2O/H2) = E°(H2O/H2) = -0,83 .

Therefore at the cathode the reduction of water to hydrogen takes place.
This is right.  8)
We now consider the anode, the positive pole, which the oxidation takes place at.  
Here the oxidizable species are H2O and OH-.
It is necessary to see however what semielement algebraically has the LOWEST potential of reduction.
Possible semireactions:

3) 2 H2O --> 4 H+ + O2 + 4 e-

4) 4 OH- --> O2 + 2 H2O + 4 e- .

Semielement potentials in this case:

3) E(O2/H2O) = E°(O2/H2O) + 0,0591/n log [H+]^4 =
= 1,23 + 0,0591/4 * (-56) = 0,40 .

(note that [H+]=10^-14 M, since [OH-] is 1 M; therefore [H+]^4 is 10^-56)

4) E(H2O/OH-) = E°(H2O/OH-) = 0,40
(since  [OH-] is 1 M).

My question:
Why my book is sure that the only possible reaction at the anode is the 4), if he could according to the calculations happen in the same way also the 3) (water oxidation)?  ???
The oxygen overvoltage does not enter there because oxygen forms in both cases.

Where is the explanation?

Thank you for your answer.
Be clear and precise please!
« Last Edit: April 03, 2010, 12:08:49 PM by Telesforo »

Offline Grundalizer

  • Full Member
  • ****
  • Posts: 257
  • Mole Snacks: +19/-31
Re: A problem of electrochemistry.
« Reply #1 on: April 03, 2010, 04:09:26 PM »
I'm not following on
Quote
since [OH-] is 1 M; therefore [H+]^4 is 10^-56

Offline Telesforo

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Re: A problem of electrochemistry.
« Reply #2 on: April 03, 2010, 04:39:57 PM »
At standard ambient temperature and pressure, the product of [H3O+] and [OH-] is a constant named Kw = [H3O+][OH−] = 10^-14 M^2.
Therefore, if [OH-]=1 M, [H3O+] is 10^-14 M, and [H3O+]^4=(10^-14)^4=10^(-14*4)=10^-56 M^4.

Offline FreeTheBee

  • Regular Member
  • ***
  • Posts: 87
  • Mole Snacks: +12/-1
Re: A problem of electrochemistry.
« Reply #3 on: April 05, 2010, 10:00:59 AM »
The two reactions are basically the same thing, since [OH-] and [H+] are linked by Kw. For example if you add 4 protons on either side of reaction 4 you see that it becomes the same as 3.

4OH- + 4H+ -> O2 + 4e + 2H2O + 4H+
4H2O -> O2 + 4e + 2H2O + 4H+
2H2O -> O2 + 4e + 4H+

Next, fill in the Nernst equation for both reactions. Since oxygen comes in the same way in both reactions and we are only comparing the two, we can set the concentration to some random value. Let's choose [O2]=1M. This a ridiculous number but it makes life easy when writing the Nernst equation.

for half reaction 3
E = E0 + 2.3RT/nF log(Ox/Red)
E = 1.23 + 2.3RT/4F log([H+]^4)
E = 1.23 + 2.3RT/F log([H+])
E = 1.23 + 0.059 log([H+])
E = 1.23 - 0.059 pH

for half reaction 4
E = 0.4 + 2.3RT/4F log(1/[OH-]^4)
E = 0.4 + 2.3RT/F log(1/[OH-])
E = 0.4 - 2.3RT/F log([OH-])
E = 0.4 - 0.059 log([OH-])
E = 0.4 + 0.059 pOH
note pOH + pH = 14 at room temperature
E = 0.4 + 0.059(14-pH)
E = 0.4 + 0.83 - 0.059 pH
E = 1.23 - 0.059pH

So you can see it is the same. If you look at the reactions and consider the pH of your solution, reaction 4 would describe the process better I'd say. Since most protons produced will form water when they meet one of the many OH-.

Offline Grundalizer

  • Full Member
  • ****
  • Posts: 257
  • Mole Snacks: +19/-31
Re: A problem of electrochemistry.
« Reply #4 on: April 05, 2010, 03:23:35 PM »
Then I'd guess that since they are the same thing, two neutral water molecules meeting at the anode are the likely case for reduction, since trying to force negatively charged hydroxide ions close enough to form O2 and electrons would be much harder.

Offline FreeTheBee

  • Regular Member
  • ***
  • Posts: 87
  • Mole Snacks: +12/-1
Re: A problem of electrochemistry.
« Reply #5 on: April 05, 2010, 04:57:14 PM »
Your reasoning is correct but it would be the opposite in this case. The anode is the electrode at which oxidation occurs, in this case oxygen evolution. In electrolysis it is also the positive electrode, hence it would attract hydroxyl ions.

Offline Telesforo

  • New Member
  • **
  • Posts: 3
  • Mole Snacks: +0/-0
Re: A problem of electrochemistry.
« Reply #6 on: April 07, 2010, 03:49:32 PM »
Compliments! A thousand thanks!
You are a clear and professional mind!

Sponsored Links