Consider the following set of half-reactions
|Fe2+ + 2e– Fe
||E°red = –0.44 V
|Zn Zn2+ + 2e–
|E°ox = +0.76 V
|Zn + Fe2+ Zn2+ + Fe
||E°cell = 0.32 V
This is valid only if all species have concentrations of 1 M.
What if not all the species are at Standard Conditions?
Le Châtelier's principle gives us some ideas. Look at the overall reaction. Consider the reaction to be at equilibrium (Ecell = 0 V). If we add more Fe2+ to the reaction mixture, The equilibrium will shift to the right to use up the excess Fe2+. This will, of course cause a positive cell potential to be measurable.
Conversely, if we removed Fe2+ from the reaction mixture, the equilibrium would shift to the left to try to replace some of the missing Fe2+. This would result in a measurable negative potential (reaction goes left).
If the reaction were not at equilibrium then these preceding two changes would me additive to the measured voltage (say standard voltage). So, if we add Fe2+ to a mixture at Standard Conditions, then the voltage would be more positive and if we removed some Fe2+ from a mixture at Standard Conditions the measured voltage would be lower (less positive).
The reverse results would be found for addition or removal of Zn2+ (a product). More Zn2+ would give a lower positive potential and vice versa.
Now we try to quantify the previous hand-waving discussion.
Recall the relation: DG = DG° + RT ln Q (for non-standard conditions)
Substitute the new definition of DG and we get
–nFE = –nFE° + RT ln Q
(This is called the Nernst Equation
Thus, we can calculate the cell potential of any cell, given only the initial conditions (to calculate Q) and the reduction potentials (to calculate E°).
Now, looking back at the reaction cell discussed at the beginning of this section we can calculate the voltage for the situation where the concentrations are not standard. For example, consider [Fe2+] = 0.1 M and [Zn2+] = 1.9 M.
For this reaction, we also know that there are two electrons transferred for each equation (n = 2).
Although the reaction is 90% complete (from Standard Conditions initial) the cell potential has only dropped by a small amount (0.04 V). This is normal. For example, the batteries in your flashlight will put out almost full voltage until the last traces of chemical are almost used up at which point the voltage will drop off rather sharply. This is an especially useful characteristic of cells powering electronic equipment, (like a calculator or CD player, etc.) which require a certain minimum (and dependable) voltage to operate successfully.
We can also use the Nernst equation to calculate such things as equilibrium constants.
At equilibrium, Q = K and Ecell = 0 so the Nernst equation becomes
What is the equilibrium constant for the reaction of copper metal with bromine to form copper(II) ions and bromide ions in aqueous solution at 25°C?
|Br2 + 2e– 2 Br–
||E°red = 1.09 V
|Cu Cu2+ + 2e–
|E°ox = –E°red = –0.34 V
|Cu + Br2 Cu2+ + 2 Br–
||E°cell = 0.75