Chapter 20: Tricks and Tips

Chapter 20: Tricks, Tips and Common Mistakes

1. Trying to balance redox reactions by inspection
   • Yes, sometimes you can do this. But it generally doesn't work, especially in acidic or basic solution where you'll have to add water or protons to get the reaction to balance. Use the half-cell method outlined in section 20.1 of Chang.

2. If you add OH^- to balance reactions in basic solution you must add OH^- to BOTH sides of the reaction!
   • We need to do this, otherwise we would unbalance the O's and H's that we so carefully balanced in the previous step. This is the single most common mistake that people make in balancing redox reactions.

     Worst of all, if you forget to do this, there is almost no way you can recover from your mistake. If you check your reaction at the end and it does not balance, do not start changing stoichiometric coefficients willy-nilly. That won't save you. Instead, go back through each step of the process and make sure you balanced each kind of atom (and then charges) in turn.

     Another common error is adding up the charges incorrectly. Remember to multiply your charges by any stoichiometric coefficients. For example, 2 Cr³⁺ has six charges because there are two 3+ ions!

3. Don't multiply E⁰ values when you balance a redox reaction.
   • We have these two reactions:
     • A⁵⁺ + 5 e^- A(s) E⁰ = -2.00 V

       B(s) B⁺ + e^- E⁰ = +0.50 V

     The net reaction is obtained by multiplying the bottom half-cell reaction by five to give

     A⁵⁺ + 5 B(s) A(s) + 5 B⁺

     The E⁰_cell for this reaction is

     E⁰_cell = E⁰_ox + E⁰_red

     E⁰_cell = +0.50 + (-2.00) = -1.50 V

     We do NOT multiply the E⁰ value of the second half-cell reaction by five even though we multiplied the reaction coefficients! Remember: E⁰ is an intensive property and does not depend on the amount of stuff we have present!

4. Mixing up the anode and cathode

   Oxidation Is Loss of electrons

   Reduction Is Gain of electrons

   "OIL RIG"

   Oxidation occurs at the anode, which we always write on the left of a cell diagram.

   Reduction occurs at the cathode.

   Oxidation and reduction are in alphabetical order, so are anode and cathode...

5. What do I do with the Pt in this cell diagram...most cell diagrams have only two things on either side, and this one has THREE!?"
   • The standard hydrogen electrode (SHE) consists of a piece of Pt metal where the following reaction takes place:

     2 H⁺(aq) + 2 e^- H₂(g)

     The platinum does not react in this reaction...its only role is to act as a place for the reaction to occur. In other words some electrodes are "passive" in that the electrode material itself is not reduced or oxidized. Thus, if this were hooked up to a Cu²⁺/Cu half-cell the cell diagram would be

     Pt(s) | H₂ | H⁺ | KCl(sat'd) | Cu²⁺ | Cu(s)

     Another example where we see this is in the electrochemical cell for the production of Na(l) and Cl₂(g) from molten NaCl. The graphite and iron electrodes are not consumed...they merely exist to serve as an electrical connection between the circuit and the NaCl melt. Graphite and iron are chosen because they do not undergo chemical reaction with any of the other cell components or products.

6. Which of these 2 species is easier to reduce (or oxidize)?
   • We can always reason this out if we remember that G⁰ = -nFE⁰.

     Let's ask which is easier to oxidize, Ca(s) or Zn(s). We can look up the reduction potentials and we find:

     • Ca²⁺ + 2 e^- Ca(s) E⁰ = -2.87 V

       Zn²⁺ + 2 e^- Zn(s) E⁰ = -0.76 V

     We're interested in the oxidation, so the oxidation potential of Ca(s) is +2.87 V and Zn(s) is +0.76 V.

     G⁰ = -nFE⁰, so the one with the largest E⁰ will be the most spontaneous reaction. Therefore, Ca(s) is easier to oxidize because there is a greater thermodynamic driving force for the reaction.

     The easier it is to oxidize a species, the better reducing agent it is (more negative E⁰_red)

     If we had asked which is easier to reduce, Zn²⁺ or Ca²⁺, we again apply G⁰ = -nFE⁰. Both reactions are non-spontaneous and we'd have to apply a voltage to get the reaction to go in the desired direction. We'd have to supply less voltage to make Zn(s) than we would for Ca(s), so Zn²⁺ is more easily reduced.

     The easier it is to reduce a species, the better oxidizing agent it is (more positive E⁰_red)

7. Don't memorize Table 20.2
   • Learn how to use your equations. If you know these simple equations:

     G⁰ = -nFE⁰

     G⁰ = -RTlnK

     You can derive ALL of Table 20.2. For example, if E is negative, that makes G negative, which means the reaction is spontaneous. If K is >1, then lnK is >1 and the reaction is spontaneous. If G is postive, then E must be negative. Try it and see.

8. Don't confuse log and ln.
   • These do not give the same answer. It is true that lnx = 2.303logx, however. But be careful not to confuse the ln with log, particularly when using the Nernst equation etc. It's an easy mistake to make in a rush on an exam.....

9. Watch those units in your electrolysis calculations!
   • Don't be sloppy about units, particularly in electrolysis calculations. Always write them out and make sure they cancel properly in order to to get the right answer.

     Remember that you need to convert time to seconds in these calculations because current is expressed in amperes and 1 ampere = 1 C/s (coulomb per second).

     Also remember to watch the number of electrons transferred in the reduction process. While calculating the charge required to reduce something is not too difficult, it is easy to forget to account for multi-electron reductions. For example, it takes three times as many electrons to reduce Al³⁺ to Al(s) as it does to reduce Na⁺ to Na(s). Said another way, it takes three times as long to reduce a mole of Al³⁺ than a mole of Na⁺at the same current.