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### Topic: Redox  (Read 33232 times)

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#### pfnm

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##### Redox
« 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

#### Borek

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##### Re: Redox
« Reply #1 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.
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#### pfnm

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##### Re: Redox
« Reply #2 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-

First:

IO3- > I3-

(adding waters, H+, OH, electrons, cancelling all except electrons )

Reduction:

9H2O + 3IO3- + 18e- > I3- +18OH-

Oxidation:

27I- > 9I3- + 18e-

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....

#### pfnm

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##### Re: Redox
« Reply #3 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

#### Borek

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##### Re: Redox
« Reply #4 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?
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#### pfnm

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##### Re: Redox
« Reply #5 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

#### AWK

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##### Re: Redox
« Reply #6 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

« Last Edit: October 05, 2009, 02:33:52 AM by AWK »
AWK

#### Borek

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##### Re: Redox
« Reply #7 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.
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#### uofc201

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##### Re: Redox
« Reply #8 on: November 01, 2009, 08:15:55 PM »
I am trying to balance the following reaction:

IO3- (aq) + I-  I2 (aq)

I am having no luck though. According to the answer key it should be:

IO3- + 5I- + 6H+  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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