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Solutions to the Practice Problem on Electrochemistry

  1. (The steps are numbered sequentially so the numbers don't correspond directly to those from the procedure taught in class.)
    1. tex2html_wrap_inline249
      tex2html_wrap_inline251
    2. tex2html_wrap_inline253
      tex2html_wrap_inline255
    3. tex2html_wrap_inline257
  2. The half-reactions are

    displaymath259

    and

    displaymath261

    The spontaneous overall reaction is the one with a positive value of tex2html_wrap_inline263 . Since the tex2html_wrap_inline265 term of the Nernst equation is generally a small correction, we guess

    displaymath267

    (If we have guessed wrong, we will get a negative voltage and we will know that the spontaneous reaction in fact runs in the opposite direction.) We compute the electrochemical potential from the Nernst equation:

    eqnarray54

  3. From electrochemical tables, the half reactions are

    displaymath269

    and

    displaymath271

    The overall reaction is therefore

    displaymath273

    with tex2html_wrap_inline275 and tex2html_wrap_inline277 . From the Nernst equation, we have

    eqnarray88

      1. One of the half-reactions is the standard hydrogen electrode. Thus we only need to balance the other half-reaction to start with. The other half-reaction turns iodate ions into iodine:

        displaymath279

      2. displaymath281

      3. displaymath283

      4. displaymath285

      5. displaymath287

        Now that we know that iodate is reduced, the other half-reaction must be the oxidation of hydrogen:

        displaymath289

      6. If we multiply the second half-reaction by five and add this to the first half-reaction, we get

        displaymath291

        or, after simplification,

        displaymath293

    1. We need the activities of all the reactants and products. The activity of iodine is tex2html_wrap_inline295 . We started out with 0.04mol/L of iodate ions but each iodine formed requires the reaction of two iodate ions so tex2html_wrap_inline297 . The initial pH is 4 so we started out with tex2html_wrap_inline299 . Again, we need two hydrogen ions per iodine so at the time of measurement, tex2html_wrap_inline301 . The activity of hydrogen gas is 1. Therefore, by rearrangement of the Nernst equation,

      displaymath303

      Referring back to the balancing process, we see that tex2html_wrap_inline305 . Therefore

      eqnarray160



Marc Roussel
Wed Dec 11 12:34:02 MST 1996