Need Help With Acetaldehyde-Tollen's Reagent Rxn

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SUMMARY

The reaction between acetaldehyde (CH3CHO) and Tollen's reagent (Ag(NH3)2+) is a redox process resulting in the formation of acetic acid (CH3COOH) and metallic silver (Ag). The oxidation state of silver in the diaminesilver(I) complex is +1, which is reduced to 0 in the final product. The balanced half-reactions are: Reduction: Ag(NH3)2+ + e- → Ag + 2NH3 and Oxidation: CH3CHO + H2O → CH3COOH + 2H+ + 2e-. Combining these yields the complete reaction.

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  • Basic organic chemistry, specifically aldehyde and carboxylic acid reactions
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Homework Statement



CH3CHO (acetaldehyde) + Ag(NH3)2+ (tollen's reagent) -------> ?


Homework Equations




The Attempt at a Solution



I'm guessing that this is a redox reaction. The final product might involve:

CH3COOH + Ag + some other products

I've looked on the internet tirelessly for how to do this and can't find it.
 
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The diaminesilver(I) complex is an oxidizing agent, which is itself reduced to silver metal, which in a clean glass reaction vessel forms a "silver mirror". This feature is used as a test for aldehydes, which are oxidized to carboxylic acids.
Quoted from Wikipedia. I think your answer is correct. Balance it with ammonia/water as necessary.
 
Ok, would anyone know the oxidation state of Ag(NH3)2+ ?

This would really help me do the question.

So far, I have for the final reaction:

2Ag(NH3)2+ + 6H3O+ + C2H4O -----> 2Ag + 4NH3 + 4H2O + CH3CO2-

I'm not sure of the in between two half rxns though.

The reduction half:
(Ag(NH3)2+ + 2H3O+ ---> Ag + 2NH4 + 2H2O + e-) x2

The oxidation half:
C2H4O + 2H3O+ 2e- -----> CH3CO2- + 3H2O

I think I've got the final reaction right but the in between 1/2 reactions wrong.
 
Last edited:
Yep, you know what your own mistake was. You have your mole balance right, but your oxidations and reductions are going backwards.

The (I) in "diaminesilver(I) complex" tells you that the oxidation state of silver is +1. This complex is reduced to silver metal. Since the final product is Ag(s), whose oxidation state is 0, the electrons should be on the reactants side; i.e., the silver must be gaining electrons, and not losing them as you have shown.

Similarly, the acetaldehyde is being oxidized, so it is losing electrons; i.e., the e- should appear on the products side.

So the two half-reactions are
Reduction: [Ag(NH3)2](+) + e(-) -> Ag(s) + 2 NH3
Oxidation: CH3CHO + H2O -> CH3COOH + 2H(+) + 2e(-)

So if you add the two half-reactions and remember to form the ammonium at the end like you did before, you'll get the complete answer.
 

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