September 28, 2024, 01:30:18 PM
Forum Rules: Read This Before Posting


Topic: Oxidation #s  (Read 3519 times)

0 Members and 1 Guest are viewing this topic.

qfruit

  • Guest
Oxidation #s
« on: February 28, 2006, 12:58:23 AM »
I'm looking at an exam question and it says that

2SO42-  --> S2O82-

is an oxidation half reaction.


When I work out the charges, they come out equal on both sides. I know that the gain of oxygens may also be a clue, but my question is...do you include the balancing 2 in the first sulphate when you figure out oxidation numbers? I did when I tried to figure out this question, but in the answer explanation, they did not.

Thanks.

Offline mike

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 1245
  • Mole Snacks: +121/-35
  • Gender: Male
Re:Oxidation #s
« Reply #1 on: February 28, 2006, 01:03:53 AM »
I think maybe you have miscalculated, try again.
There is no science without fancy, and no art without facts.

qfruit

  • Guest
Re:Oxidation #s
« Reply #2 on: February 28, 2006, 01:25:03 AM »
I'm sure. If I include the 2 that balances the sulphate, both sulphurs come out as 7+. Without the 2, the sulphate comes out as 6+, like it says in the answers, so then it would be oxidizing...but I thought you do include the balancing 2 when you do these questions.

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7978
  • Mole Snacks: +555/-93
  • Gender: Male
Re:Oxidation #s
« Reply #3 on: February 28, 2006, 01:28:26 AM »
2SO42-  --> S2O82- + 2e-
Persulfate is obtained by electrochemical oxidation
AWK

Sponsored Links