Example 17: Calculate the electrode potential of given electrode $$\ce{Pt,Cl2 (\pu{1.5 bar}) | 2 Cl- (\pu{0.01 M})}\,; \quad E^\circ_\ce{Cl2/2Cl-} = \pu{1.36 V}$$

Solution: The reaction of electrode is $$\ce{\underset{\pu{1.5 bar}}{Cl2 (g)} + 2 e- -> \underset{\pu{0.01 M}}{2 Cl-}}$$ $$E = E^\circ - \frac{0.0591}{n}\log{\frac{[\ce{Cl-}]^2}{P_\ce{Cl2}}} = 1.36 - \frac{0.0591}{2}\log{\frac{(0.01)^2}{1.5}} = \pu{1.483 V}$$

In the solution, the author has directly used the pressure (which is in bars) in the Nernst equation without first converting it to its SI units. How can this be right?


1 Answer 1


Just as you are supposed to use activity instead of concentration for dissolved components when applying Nernst equation, you are also supposed to use fugacity instead of the partial pressure for the gaseous components. The reason for that will become transparent if you follow the derivation of Nernst equation using chemical potential.

Fugacity is defined with a reference to the standard state with the pressure of $\pu{10^5 Pa}$ or $\pu{1 bar}$. At ordinary pressures fugacity is numerically approximately equal to the pressure expressed in bars or atmospheres ($\pu{1 bar} \approx \pu{0.987 atm}$), so the author of the book is right.

  • $\begingroup$ Oh…I didn't know that. We didn't drive Nernst equation…I'm in class 12, and so I guess that's above my level... Well, thanks man! Also, what's the difference between activity and concentration? $\endgroup$ Commented Jan 3, 2018 at 12:02
  • $\begingroup$ @SenthilArihant No prob; I'd suggest to read en.wikipedia.org/wiki/Thermodynamic_activity $\endgroup$
    – andselisk
    Commented Jan 3, 2018 at 12:07

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