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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: pfnm on October 02, 2009, 04:09:56 AM

Title: Redox
Post by: pfnm on October 02, 2009, 04:09:56 AM
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

Title: Re: Redox
Post by: Borek on October 02, 2009, 05:02:09 AM
While I3- is present in the solution, it is product of another reaction:

I2 + I- <-> I3-

Balance for I2 first.
Title: Re: Redox
Post by: pfnm on October 04, 2009, 07:10:42 AM
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....
Title: Re: Redox
Post by: pfnm on October 04, 2009, 07:15:12 AM
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
Title: Re: Redox
Post by: Borek on October 04, 2009, 07:41:07 AM
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.

Quote
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?
Title: Re: Redox
Post by: pfnm on October 05, 2009, 01:17:38 AM
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
Title: Re: Redox
Post by: AWK on October 05, 2009, 01:49:24 AM
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


Title: Re: Redox
Post by: Borek on October 05, 2009, 03:24:25 AM
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.
Title: Re: Redox
Post by: uofc201 on November 01, 2009, 08:15:55 PM
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



Title: Re: Redox
Post by: Borek on November 02, 2009, 02:46:36 AM
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.