How Do You Balance the Redox Reaction Involving C2H5OH and I3-?

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SUMMARY

The discussion focuses on balancing the redox reaction involving ethanol (C2H5OH) and triiodide (I3-), which produces iodide (I-), carbon dioxide (CO2), formate (CHO2-), and iodoform (CHI3). Participants emphasize the necessity of understanding oxidation states and the rules for determining them to approach the problem effectively. The consensus is that the reaction can be split into two separate half-reactions, but it presents complexities that lead to multiple valid balancing equations, indicating that there is no unique solution. Ultimately, the reaction can be balanced in a general sense, but it does not yield a definitive stoichiometric equation.

PREREQUISITES
  • Understanding of oxidation states and their determination
  • Familiarity with redox reactions and half-reaction method
  • Knowledge of acidic solution behavior in redox chemistry
  • Basic concepts of stoichiometry in chemical reactions
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  • Study the rules for determining oxidation numbers of elements
  • Learn the half-reaction method for balancing redox reactions
  • Explore the concept of multiple equilibria in chemical reactions
  • Research the properties and reactions of iodoform (CHI3) and formate ion (CHO2-)
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Chemistry students, educators, and anyone involved in redox reaction analysis and balancing, particularly in organic chemistry contexts.

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Homework Statement



C2H5OH(aq) + I3-(aq) → I-(aq) + CO2(g) + CHO2-(aq) + CHI3(aq)

Homework Equations





The Attempt at a Solution



I am stuck on how to get C to combine with I in CHI3. Thanks.
 
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You mean in terms of oxidation numbers? Which is more electronegative - carbon, or iodine?
 
I am trying to balance it in acidic solution. Sorry I forgot to add that. I do not know which is more electronegative. We haven't learned that.
 
Please elaborate then on what your problem is - I don't understand what you mean by "how to get C to combine with I". Show - step by step - how you approach the problem.
 
To be able to solve the problems of redox balancing, you need to know at least two things:

1.) Some prefixed rules in determining the oxidation number of some select few elements.

2.) Oxidation states and most stable oxidation states of various elements.

If you know these two things, balancing is an easy affair.
 
Most redox problems are easy... But this one is different. I solved every other one with no problem but I'm not sure how to do the half reactions for this one.
 
Can you see the structure of CHI3 - Iodoform. You can then apply the rules you may have learned to determine the oxidation number on C.
 
How should I split the half reactions?
 
HINT: Both half reactions uses Ethanol at reactant side.

HINT: Can you break this question into two different redox reactions? I am pretty sure there are multiple redox reactions going on. (Why?)
 
Last edited:
  • #10
Thank you. I'll try that.
 
  • #11
AGNuke is right - this reaction can't be balanced in an unambiguous way.
 
  • #12
I tried splitting this into:
C2H5OH + I3- > CO2 +I-
C2H5OH + I3- > CHO2- + CHI3

That didn't work, so I tried:

C2H5OH + I3- > CO2 + CHI3
C2H5OH + I3- > CHO2- + I-

I also couldn't solve it like this. Could someone show me what my half-reactions should be and I am sure I can take it from there. Thanks
 
  • #13
CHO2- = Formate ion?

From what I tried, this equation can't be solved. I tried very hard but I can't solve it.
 
  • #14
So have I. I also believe it can't be solved but it's in my textbook. I don't know what else to do.
 
  • #15
Is this is written CHO2- or CHO2-?
 
  • #16
It can't be balanced because it is in fact sum of two reactions, each can be balanced separately:

2C2H5OH + 9I3- -> 18I- + CO2 + 3CHI3 + 9H+

2C2H5OH + 8I3- -> 15I- + HCOO- + 3CHI3 + 8H+

Now, you can add them side by side:

4C2H5OH + 17I3- -> 33I- + CO2 + HCOO- + 6CHI3 + 17H+

Technically this is a balanced reaction, but this approach doesn't yield unique equation, as any linear combination of these reactions will still look balanced, for example:

136C2H5OH + 595I3- -> 1173I- + 51CO2 + 17HCOO- + 204CHI3 + 595H+

but no real mixture will follow this stoichiometry.
 
  • #17
Ah. Infinite solutions. I made a silly mistake in determining the equation no. 1. XD
 
  • #18
I will post the equation in a different format to avoid confusion:


C2H5OH(aq) + I3-(aq) → I-(aq) + CO2(g) + CHO2-(aq) + CHI3(aq)
 
  • #19
cleared Borek! Couldn't see this up
 

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