Author Topic: Half equations (Read 2528 times) Tweet Share

boysenberry · « **on:** December 28, 2009, 07:15:07 am »

As a result of a traffic accident, residents in a Melbourne suburb had to be evacuated when toxic fumes leaked from a container of sodium dithionite $(Na_{2}S_{2}O_{4})$ . The dithionite ion reacts with water according to the equation:

$2S_{2}O_{4}^{2-} (aq) + H_{2}O (l) ---> S_{2}O_{3}^{2-} (aq) + 2HSO_{3}^{-} (aq)$

Write the ionic half equations for the oxidation and reduction reactions that occur when sodium dithionite is mixed with water. I have tried to do this but failed. Would anyone be able to show full compete working out?

vexx · « **Reply #1 on:** December 28, 2009, 01:06:40 pm »

I may as well give this a good as it's practice, but someone correct if i'm wrong ><
$<br />S204^{2-} ---> S203^{2-}$
Add +H20 to the right as there is one less oxygen.
$S204^{2-} ---> S203^{2-} + H20$
And now add +2H to the left
$S204^{2-} + 2H^{+} ---> S203^{2-} + H20$
Charge on the left is +2 + -2 = 0, and charge on the left= -2
So add two electrons to the left side
$S204^{2-} (aq) + 2H^{+} (aq) + 2e^{-} ---> S203^{2-} (aq) + H20 (l)$
I'm quite sure this first half equation is reduction as there is a loss of oxygen.

Okay for the other part, oxidation:

$H20 --> 2HS03^{-}$
Umm, i'm not exactly sure how the S behaves, but i'd assume you can just add S2- to the left side.
$H20 + S^{2-} --> 2HS03^{-}$
Then add 2 waters to the right to balance the two extra oxygens, and then 4 hydrogen ions to the right.
$H20 + S^{2-} --> 2HS03^{-} + 2H20$
$H20 + S^{2-} + 4H^{+} --> 2HS03^{-} + 2H20$
The charge on the left is -2 + 4 = +2
The charge on the right is -2
-2 - +2 = 0, therefore balanced electrons.
$H20 (l) + S^{2-} (aq) + 4H^{+} (aq) --> 2HS03^{-} (aq) + 2H20 (l)$

I'm really unsure now, so can someone please correct me haha?

jimmy999 · « **Reply #2 on:** December 28, 2009, 01:46:07 pm »

The above equation won't work because when you add the equations together, you don't end up with the original equations. Also the oxidation reaction has no electrons in it meaning the equation is not an oxidation equation at all.

First of all, the reduction equation is correct. So we need to fix the oxidation equation. If we look at the first equation, it gives us a hint that there are $2S_2O_4^{2-}$ . This means we could use $S_2O_4^{2-}$ in the reduction and oxidation equations.

Therefore we get

$S_2O_4^{2-} --> HSO_3^{-}$
Balance everything except hydrogen and oxygen
$S_2O_4^{2-} --> 2HSO_3^{-}$
Balance oxygen by adding water
$S_2O_4^{2-} + 2H_2O --> 2HSO_3^{-}$
Balance hydrogen by adding $H^{+}$ and balance the charges
$S_2O_4^{2-} + 2H_2O --> 2HSO_3^{-} + 2H^{+} + 2e^{-}$

Not a particularly easy equation. Had to think for a bit to get it. In fact I've never thought of writing half equations in this way so thanks for this problem

vexx · « **Reply #3 on:** December 28, 2009, 03:24:32 pm »

^ oh thanks for that then.
yeah i wouldn't have thought to use the S2042- twice, so i basically guessed using H20.

Search the forums now!

We have moved!

Author Topic: Half equations (Read 2528 times) Tweet Share

boysenberry

Half equations

vexx

Re: Half equations

jimmy999

Re: Half equations

vexx

Re: Half equations

Recent Posts

User:
Password: