1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Need Help With Acetaldehyde-Tollen's Reagent Rxn

  1. Nov 16, 2007 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant equations

    3. 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.
  2. jcsd
  3. Nov 16, 2007 #2
    Quoted from Wikipedia. I think your answer is correct. Balance it with ammonia/water as necessary.
  4. Nov 22, 2007 #3
    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: Nov 22, 2007
  5. Nov 24, 2007 #4
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook