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Thread ID: 111753 2010-08-10 07:13:00 Chemistry Help! xyz823 (13649) PC World Chat
Post ID Timestamp Content User
1126326 2010-08-10 07:13:00 Trying to work out the following question.

Need balanced half equations, fully balanced redox equations and what would be observed.

A freshly prepared solution of sodium hydrogen sulfite is added to a solution on iron (III) chloride.


TIA
xyz823 (13649)
1126327 2010-08-10 07:35:00 A freshly prepared solution of sodium hydrogen sulfite is added to a solution on iron (III) chloride.


That's a question? Looks like statement to me.....
pctek (84)
1126328 2010-08-10 07:37:00 It might help if you put any workings you have so far - otherwise you're likely to get whacked with the DYOFH stick!

:thumbs:
nofam (9009)
1126329 2010-08-10 07:39:00 Fe^3+ + e^- ---> Fe^2+

HSO3^- + H2O ---> SO4^2- + 3H+ + 2e^-

I'm sure you know what to do from here.
Renmoo (66)
1126330 2010-08-10 07:40:00 Final answer that I've got:

2Fe3+ + HSO3- + H2O ----> 2Fe2+ + SO42- + 3H+ + 2e-
Renmoo (66)
1126331 2010-08-10 07:43:00 H2O + HSO3 (-1) -2e = SO4 (-2) + 3H(+1)
Fe (+3) + e = Fe(+2)

H20 + HSO3 + 2 Fe (+3) = SO4 (-2) + 2Fe(+2) + 3H(+1)

Colour change from yellow/brown to pale green

My chemistry is a little rusty though...
user (1404)
1126332 2010-08-10 07:48:00 Cheers guys, just wasn't sure what was actually being reduced and what was being oxidised! Got myself all confused :/ xyz823 (13649)
1126333 2010-08-10 07:49:00 Final answer that I've got:

2Fe3+ + HSO3- + H2O ----> 2Fe2+ + SO42- + 3H+ + 2e-

You don't have any free electrons in a balanced redox equation!
user (1404)
1126334 2010-08-10 07:52:00 It's been 55 years since I did any chemistry, but I'm wondering where did the Sodium go to ? NaHS03 Terry Porritt (14)
1126335 2010-08-10 08:01:00 It's been 55 years since I did any chemistry, but I'm wondering where did the Sodium go to ? NaHS03

Sodium is a spectator ion and takes no part in the reaction. You can add them onto both sides of the equation if you like but it is not necessary in a redox equation. It is the same for the chloride ions.
user (1404)
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