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. 
      
   2. 
      
   3. 
2. The half-reactions are

   

   and

   

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

   

   (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:

   

3. From electrochemical tables, the half reactions are

   

   and

   

   The overall reaction is therefore

   

   with and . From the Nernst equation, we have

   

4. 1. 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:

         

      2. 

      3. 

      4. 

      5. 

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

         

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

         

         or, after simplification,

         

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

      

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