This question is from the A level chemistry.com site (Redox obvs), which asks you to work out half-equations and then the full equation for the redox reaction. and I can't work out how they've gotten 2e- on the first part of balancing the half-equation for the sulfur compound. I worked out 1.5e-?
Turn the following processes into redox reactions by writing out half-equations and combining them:
S2O32- --> S4O62-, I2 --> 2I-
This is the actual answer...
2S2O32- --> S4O62- + 2e
I2 + 2e --> 2I-
2S2O32- + I2 -->S4O62- + 2I-
BIB: I don't understand how they worked that out?
Thanks in advance!
Simple Redox Question Watch
- Thread Starter
Last edited by Toph; 14-04-2013 at 12:23.
- 14-04-2013 12:22
- 14-04-2013 13:44
I2 + 2e- --> 2I-
2S2O32- --> S4O62- +2e-
If there was not two electrons produced in the second half equation then S4O6 would have a charge of -4.
Because two electrons are needed in the first half equation, and two electrons are produced in the second half equation, they cancel out and you get an overall equation of
I2 + 2S2O32- +2e- --> 2I- +S4062- +2e-
Which is equal too
I2 + 2S2O32- --> 2I- +S4062-