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Cell EMFs and Electrode Potentials

This section deals with calculating the emf of a chemical cell, based on the the individual half cells' standard redox potentials. This isn't actually very difficult if you follow the rules exactly as given.

Electrode potentials are always written in the same particular conventional form, shown below: -

M+ + e- ® M
x V

It doen't matter whether or not you know that the equation above is written the "wrong" way round to normal, or that it is going to be reversed in the cell, you start from the standard redox potential and apply the rules. Then you always get the right answer. The rules for manipulating these equations, predicting the direction of reaction and the cell emf, or the probablity of the reaction going to completion or being an equilibrium which can be altered by changing the conditions are really very simple indeed.

  1. Always work from the standard redox potentials and the corresponding standard half cell / equation.
  2. For any pair of half cells, the most negative goes in reverse, the most positive goes as written. (Place the species which gets oxidisedon the left hand side, the species which reduced on the right hand side.)
  3. For the correct cell diagram, the most negative half cell is written down first - i.e. on the left, in reverse. The most positive half cell is written down next - i.e. on the right, as written. (Oxidised and reduced as stated above.)
  4. The cell emf is found from the equation E(CELL) = E(RHS) - E(LHS) with the signs as written in the original half cells.
    This is the same as saying (most positive) - (most negative), again with the signs as written in the original half cells. But use the first version of this rule, as it's more reliable in certain types of question, like the first type mentioned
  5. For a cell as given in a question,
    - If E(RHS) - E(LHS) is negative, the cell will not go as written, no reaction will occur.
    - If the emf the emf is positive it will react as given.
    - If the emf is less than about 0.4 V it will usually be an equilibrium, and the positon of equilibrium can be controlled by le Chatelier's Principle.
    - If the emf is less than about 0.1 V, it will probably not react at all.

Here's a few examples to show the rules in action.

Ex 1 - Calculate the cell emf for the following pair of half cells, and draw the resulting cell diagram. Is the resulting reaction likely to occur?

Cu2+ + 2e- ® Cu(s) +0.34 V
Zn2+ + 2e- ® Zn(s) -0.78 V

Rule 1 - We are using the standard half cells

Rule 2 -Reverse the Zn cell, use the Cu cell as written

Zn(s) ® Zn2+ + 2e-
Cu2+ + 2e- ® Cu(s)

Notice that we can clearly see that the Zn will be oxidised, and the Cu2+ reduced if this reaction occurs (which we will discover in a few seconds time.)

Rule 3 - Zn|Zn2+||Cu2+|Cu and that's the cell diagram written out. We miss out the electrons in these diagrams to keep them simple.

Rule 4 - E(CELL) = E(RHS) - E(LHS) with the signs as written in the original half cells

So we get E(CELL) = (0.34) - (-0.78) = 0.34 + 0.78 = 1.12 V.

By putting the numbers in the brackets, we make it easier to work out the signs to use. This is always a good idea when numbers have signs. Don't do bits in your head that might go wrong, write them down and automatically get them right. Some of my friends think I'm rather good at maths, but I don't think I'm anything special, I just don't break the rules. So I don't make mistakes.

Rule 5 - The cell emf is positive and it is greater than 0.4 V - this reaction will go as written in the cell diagram we have drawn.