Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Amperage of a Voltaic Cell

  1. Mar 28, 2010 #1
    Hello all,

    I am an AP Chemistry student currently studying electrochemistry. The other day, some friends of mine and I discovered some waste copper oxide laying around the lab. We thought, "Why not reduce all the copper out of this?", and so we set to work.

    We got our experiment authorized by the teacher, and constructed a voltaic cell based on the reduction of copper through the oxidation of zinc. We used nitric acid to make a solution of copper nitrate from the copper oxide, and boiled it down to solid to remove excess HNO3 and dissolved it in distilled water. Zinc nitrate was prepared as the anode solution. A KCl salt bridge was constructed.

    The cell itself works fine, but I have a question about the time this reaction will take to complete. The reaction should complete when all of the copper ions have reduced. There are 2 Faradays of charge transferred per mole, and 96500 Coulombs of charge in a Faraday. Since Amperes measure Coulombs per second, I need to know how many amps the cell draws when short circuiting to calculate how long the reaction should take to reach completion.

    Now, the fundamental question facing me: How do I calculate the amperage a voltaic cell runs at when shorted out? I realize that I could just measure it with my handy-dandy multimeter, but I really would rather learn to calculate it, and thus gain a further understanding of electrochemistry. Any advice is much appreciated.

  2. jcsd
  3. Mar 29, 2010 #2
    Your not going to be able to calculate this so easily. The maximum reaction rate of a global reaction is always limited by the slowest step. In most cases in electrochemistry, the slowest step is ion-migration or mass transport. Do determine your rate of mass transport you would need to take into account not only the properties of your electrolytes but also the geometry of your cell. In other words would basically need to conduct a full CFD simulation of your cell in order to estimate your limiting losses. This is not a easy thing to do. If I were you, I'd stick to the handy-dandy multimeter.
  4. Mar 29, 2010 #3
    Hey, thanks for the reply.

    I thought the rate determining step would relate to the actual oxidation or reduction reactions taking place at the electrodes, so calculation of shorted amperage would at least be feasible. But alas, the mass transfer over the salt bridge is logically much slower; meter it is!

    Just me being insatiably curious, what is a CFD simulation? It sounds like something I should know about if I want a career in chemistry.

    Thanks again,

  5. Mar 29, 2010 #4
    CFD stands for Computational Fluid Dynamics and also can refer to multiphysics problems such that involve chemical reactions, transfer of charge, etc. CFD tends to push the limits of mathematics, something chemists generally despise very much, so you usually won't ever see a chemist doing CFD work.
  6. Mar 30, 2010 #5
    So since the object of this reaction is to plate out copper, increasing the amperage drawn by shorting the cell makes the reaction proceed more quickly. So how can I adjust my salt bridge to accomplish this? It seems that a bridge short in length with a wide diameter would be ideal. Is this correct?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook