May 11, 2024, 09:26:17 PM
Forum Rules: Read This Before Posting


« previous next »
Pages: [1]   Go Down

Topic: Standard Reduction Potential Question  (Read 3107 times)

0 Members and 1 Guest are viewing this topic.

Offline Steven90

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Standard Reduction Potential Question
« on: May 16, 2009, 04:06:29 AM »
The standard reduction potential for the half reaction below at pH 6?

O2+2H2O+4e --> 4OH- E=+0.40V

My Attempt:
pH=6 pOH=8
[OH]=10^-8
E=E^nod + 0.059/4 lg 1/(10^-8)^4
=0.4+0.48
=0.88

But the actual answer is 0.64 which is (0.4+0.24), can anyone points out the mistake for me?
Logged

Offline leve

  • Regular Member
  • ***
  • Posts: 20
  • Mole Snacks: +1/-0
Re: Standard Reduction Potential Question
« Reply #1 on: May 16, 2009, 10:07:57 AM »
The equation is

E = E0 - 0.0592/n logQ

What is Q? Do you have enough information for Q?
« Last Edit: May 16, 2009, 10:22:29 AM by leve »
Logged

Offline Steven90

  • Very New Member
  • *
  • Posts: 2
  • Mole Snacks: +0/-0
Re: Standard Reduction Potential Question
« Reply #2 on: May 17, 2009, 03:35:16 AM »
The only non-standard condition here is the pH so I just calculated the Q by using "pH+pOH=14" and I get the [OH-] as 10^-8
Logged

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27676
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Standard Reduction Potential Question
« Reply #3 on: May 17, 2009, 05:32:17 AM »
pO2?

Could be it has to be assumed 1, but it is not 100% clear to me.
Logged
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info
Pages: [1]   Go Up
« previous next »
Sponsored Links