September 29, 2024, 02:22:13 AM
Forum Rules: Read This Before Posting


Topic: reduction potential and pH questions  (Read 2190 times)

0 Members and 1 Guest are viewing this topic.

Offline alanjz

  • Regular Member
  • ***
  • Posts: 24
  • Mole Snacks: +1/-1
reduction potential and pH questions
« on: March 29, 2013, 10:10:09 PM »
Could someone help me understand where these answers came from?

Co3++e- --> Co2+ EO = 1.82 V
Co2++2e- --> Co EO = -0.28 V
Given these standard reduction potentials, what is the standard reduction potential for Co3++3e- --> Co?
I just did 1.82 + -0.28=1.54 but the answer is 0.42 V.

A solution of 0.10 M NaZ has a pH=8.90. What is the Ka of HZ?
The answer was 1.6x10-5
I really wasn't sure where to start so if an explanation is too much to ask for, at least a hint would be nice.

Thanks.

Offline Corribus

  • Chemist
  • Sr. Member
  • *
  • Posts: 3530
  • Mole Snacks: +541/-23
  • Gender: Male
  • A lover of spectroscopy and chocolate.
Re: reduction potential and pH questions
« Reply #1 on: March 30, 2013, 12:09:09 AM »
Question 1: Electric standard potentials are not additive.  The reason is because adding one electron and then adding two more in separate steps has a different energetic cost than adding all three at the same time.  You need to convert the electric potentials into quantities that ARE additive.  Can you think of what that might be?

Question 2: You should start by writing out some balanced equations or reactions.  For instance - what is Ka for a weak acid?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline alanjz

  • Regular Member
  • ***
  • Posts: 24
  • Mole Snacks: +1/-1
Re: reduction potential and pH questions
« Reply #2 on: March 30, 2013, 09:48:01 AM »
Thanks a lot! I think I figured them out.
For 1, I converted them to ΔG, added them, then converted back to standard reduction potentials so (1)(96500)(1.82)+(2)(96500)(-0.28)=121590 then 121590/(3*96500)=0.42
For 2, I just set up the equation for Kb:
[OH-][HZ]/[Z-] Solved for Kb then put that under 10-14 to get Ka

Sponsored Links