Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: pfnm on October 02, 2009, 04:09:56 AM
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Reaction of iodide ions: with iodate ions in basic aqueous solution to form triiodide ions. Extract the appropriate half equations and balance it in basic conditions.
I- (aq) + IO3- > I3-
Here's how I've been working it out:
Extracting reduction half equation:
6H+ + IO3- + 6e- > I3- + 3H2O
Oxidation:
3I- > I3- + 2e- (multiply by two to get equal stochiometry)
6I- > 2I3- + 6e-
But at the stage where I am going to add oxidation and reduction I am stuck because both oxidation, and reduction half reactions contain I3- as a product. Can I3- be produced by both reduction of IO3 and oxidation of I- at the same time? How does one split this into half reactions?
Thank you
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While I3- is present in the solution, it is product of another reaction:
I2 + I- <-> I3-
Balance for I2 first.
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But if I extract half reaction
I2 > I3-
It appears a reduction half reaction just as is
IO3- + I- > I3-
Another step I've tried based on your advice:
First:
IO3- > I3-
(adding waters, H+, OH, electrons, cancelling all except electrons )
Reduction:
9H2O + 3IO3- + 18e- > I3- +18OH-
Oxidation:
27I- > 9I3- + 18e-
Added:
9H2O + 3IO3- + 27I- > I3- + 9I3- + 18OH-
Final result appears to balance but am I on the wrong track, I've seen different results? I don't want to divide because of the lone I3- in the reactants....
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Appears final result is:
3H2O + IO3- + 8I- = 3I3- + 6OH-
I still can't see where I've gone wrong, any help appreciated, maybe its a matter of redoing this equation a few times
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First of all - it goes in low pH, so not OH- produced, but H+ consumed.
Honestly, I have problems following what you wrote. I don't think you have followed my advice - as you were balancing all the time using I3-. Note, that you final reaction (while incorrect in terms of H+/OH-) can be rewritten as
3H2O + IO3- + 5I- -> 3I2 + 6OH-
Difference - three I- on both sides - can be treated as a spectator, as it doesn't take part in the redox reaction itself, iodides react with the produced iodine AFTER redox took place.
What I have proposed was not meant to be the only correct approach. Reaction can be balanced using I3-, it is just easier to ignore them doing redox.
But if I extract half reaction
I2 > I3-
Extract from what? Iodine was not betwen reactants.
It appears a reduction half reaction just as is
IO3- + I- > I3-
Have you tried to assign oxidation numbers to see if you have reduction or oxidation?
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What I mean is,
I have only seen redox equations with two products, and two reactants, and from those I use oxidation numbers to 'extract' into one reduction half reaction, and one oxidation half reaction.
But with
I- + IO3- > I3-
There are two reactants, and only one product.
But yes I can see now that the I- is a spectator ion.
And also yep I did not know what I could ignore the I3 ion.
As for the OH-: according to this problem, we must balance the half reactions in basic conditions (so adding OH- to each side of each half reaction), I hope this is clear, thanks for the help
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Almost all is OK but correct your errors. Then just add both reactions. The final reaction is a sum of two reactions
6H+ + IO3- +5I- > 3I2 + 3H2O
and
3I- + 3I2 = 3I3-
Reaction of iodide ions: with iodate ions in basic aqueous solution to form triiodide ions. Extract the appropriate half equations and balance it in basic conditions.
I- (aq) + IO3- > I3-
Here's how I've been working it out:
Extracting reduction half equation:
6H+ + IO3- + 6e- > 1/3I3- + 3H2O (correct mass balance)
Oxidation:
3I- > I3- + 2e- (multiply by two to get equal stochiometry)? 2 x 2 = 4 not 6
6I- > 2I3- + 6e-
But at the stage where I am going to add oxidation and reduction I am stuck because both oxidation, and reduction half reactions contain I3- as a product. Can I3- be produced by both reduction of IO3 and oxidation of I- at the same time? How does one split this into half reactions?
Thank you
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As for the OH-: according to this problem, we must balance the half reactions in basic conditions (so adding OH- to each side of each half reaction)
Kinda stupid, reaction
IO3- + 5I- + 6H+ -> 3I2 + 3H2O
requires presence of H+ and is even used for volumetric determination of small amounts of strong acids.
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I am trying to balance the following reaction:
IO3- (aq) + I- :rarrow: I2 (aq)
I am having no luck though. According to the answer key it should be:
IO3- + 5I- + 6H+ :rarrow: 3I2 + 3H2O
I seem to be doing it wrong. It's either because it's late and I am not thinking straight or there is something flawed with my method.
Any help would be grand.
Carl
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http://www.chembuddy.com/?left=balancing-stoichiometry&right=balancing-redox
http://www.chembuddy.com/?left=balancing-stoichiometry&right=half-reactions-method
http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method
And show us what you did, we will be able to point out where you went wrong.