Why is the cell potential defined as:

$$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $$

or $$ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} $$

as opposed to $$ E^\circ_{cell} = E^\circ_{anode} - E^\circ_{cathode} $$

or $$ E^\circ_{cell} = E^\circ_{oxidation} - E^\circ_{reduction} $$

A few definitions which relate to the problem:

Cell potential is the measure of the potential difference between two half cells in an electrochemical cell

The potential difference is caused by the ability of electrons to flow from one half cell to the other

The difference between the potential for the reducing agent to become oxidized and the oxidizing agent to become reduced will determine the cell potential. (Chem.Libretexts)

But why is this the case?

Also, why does it then say here that $ E^\circ_{cell} = E^\circ_{reduction} + E^\circ_{oxidation} $ rather than $ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} $


1 Answer 1


It is universally accepted to express potentials as reduction potentials. However, one of the reactions is always going to be oxidation so you need the reverse. Normally cell potential is the SUM of reduction and oxidation potentials but the value you get for the oxidation reaction from tables is actually the reduction potential so you would have to change the sign. So you actually get:

Cell potential = (Reduction potential for the reduction at the cathode) - (Reduction potential for the oxidation at the anode)


Cell potential = (Reduction potential for the reduction at the cathode) + (Oxidation potential for the oxidation at the anode)

because Reduction potential + Oxidation potential = 0 when the same species are involved

  • $\begingroup$ Sorry could you please explain what "Reduction at anode" means? I thought reduction occurs at the cathode and oxidation occurs at the anode? Or are you just saying we treat one of the E0 values as if it were reduction (i.e. flipped) at the anode rather than oxidation? $\endgroup$
    – K-Feldspar
    Commented Oct 9, 2016 at 22:15
  • 2
    $\begingroup$ Yes, by reduction at anode I am referring to the reduction potential for the oxidation reaction which occurs at the anode. For example, if you have the following oxidation half reaction Cu(0) - 2e- -> Cu(2+) you would actually find the value for the reduction Cu(2+) + 2e- -> Cu(0) which can be used in the first equation that I've written. I have edited my answer. $\endgroup$
    – Vlad
    Commented Oct 9, 2016 at 22:23

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