The Nernst equation tells us that a half-cell potential will change by 59 millivolts per 10-fold change in the concentration of a substance involved in a one-electron oxidation or reduction; for two-electron processes, the variation will be 28 millivolts per decade concentration change. Thus for the dissolution of metallic copper
Cu(s) → Cu2+ + 2e–
the potential
E = (– 0.337) – .0295 log [Cu2+]
becomes more positive (the reaction has a greater tendency to take place) as the cupric ion concentration decreases. This, of course, is exactly what the
Le Châtelier Principle predicts; the more dilute the product, the greater the extent of the reaction.
Electrodes with poise
The equation just above for the Cu/Cu
2+ half-cell raises an interesting question: suppose you immerse a piece of copper in a solution of pure water. With
Q = [Cu
2+] = 0, the potential difference between the electrode and the solution should be infinite! Are you in danger of being electrocuted? You need not worry; without any electron transfer, there is no charge to zap you with. Of course it won't be very long before some Cu
2+ ions appear in the solution, and if there are only a few such ions per liter, the potential reduces to only about 20 volts. More to the point, however, the system is so far from equilibrium (for example, there are not enough ions to populate the electric double layer) that the Nernst equation doesn't really give meaningful results. Such an electrode is said to be un
poised. What ionic concentration is needed to poise an electrode? I don't really know, but I would be suspicious of anything much below 10
–6 M.
The Nernst equation works only in dilute ionic solutions
Ions of opposite charge tend to associate into loosely-bound ion pairs in more concentrated solutions, thus reducing the number of ions that are free to donate or accept electrons at an electrode. For this reason, the Nernst equation cannot accurately predict half-cell potentials for solutions in which the total ionic concentration exceeds about 10
–3 M.
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