April 28, 2024, 05:33:30 PM
Forum Rules: Read This Before Posting


Topic: Oxidation/eletrons  (Read 3418 times)

0 Members and 1 Guest are viewing this topic.

Offline Duzzy

  • Regular Member
  • ***
  • Posts: 29
  • Mole Snacks: +1/-3
Oxidation/eletrons
« on: May 27, 2012, 05:41:50 PM »
The reaction that occurs is: 6H+ + 5H2C2O4 + 2 MnO4
- ® 10 CO2 + 2Mn2+ + 8H2O.
A number of electrons needed for oxidation of one H2C2O4 molecule is:
A. 1, B. 2, C. 3, D. 5, E. 10

no idea where to start and where to begin. *delete me*

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +469/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: Oxidation/eletrons
« Reply #1 on: May 28, 2012, 03:09:45 AM »
Tart by identifying the oxidation state of Mn on both sides of the equation.
My research: Google Scholar and Researchgate

Offline Duzzy

  • Regular Member
  • ***
  • Posts: 29
  • Mole Snacks: +1/-3
Re: Oxidation/eletrons
« Reply #2 on: May 28, 2012, 03:51:12 AM »
+7

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +469/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: Oxidation/eletrons
« Reply #3 on: May 28, 2012, 04:21:30 AM »
Yes, +7 on the reactants side, now what about the product side? What is the oxidation state of Mn2+?
My research: Google Scholar and Researchgate

Offline Duzzy

  • Regular Member
  • ***
  • Posts: 29
  • Mole Snacks: +1/-3
Re: Oxidation/eletrons
« Reply #4 on: May 28, 2012, 06:45:12 AM »
so +7 on product side and +2 on state side.

Offline Dan

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 4716
  • Mole Snacks: +469/-72
  • Gender: Male
  • Organic Chemist
    • My research
Re: Oxidation/eletrons
« Reply #5 on: May 28, 2012, 09:36:57 AM »
+7 on the reactants side, +2 on products side.

So, how many electrons do you need to go from +7 to +2?
My research: Google Scholar and Researchgate

Offline Duzzy

  • Regular Member
  • ***
  • Posts: 29
  • Mole Snacks: +1/-3
Re: Oxidation/eletrons
« Reply #6 on: May 28, 2012, 09:47:17 AM »
5

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27665
  • Mole Snacks: +1801/-410
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Oxidation/eletrons
« Reply #7 on: May 28, 2012, 10:26:51 AM »
Can you try to think by yourself, or do we have to spoon feed you all the time?

This is only partially correct answer. Yes, 5 electrons per each Mn atom present in the reaction, but in the balanced equation there are two atoms of Mn on each side. This is enough information to calculate numbers of electrons lost by the oxidizer, now just divide it by the number of the oxidized molecules.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Duzzy

  • Regular Member
  • ***
  • Posts: 29
  • Mole Snacks: +1/-3
Re: Oxidation/eletrons
« Reply #8 on: May 28, 2012, 02:21:16 PM »
Ahhhh I get it now. Thanks!

Sponsored Links