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I am required to write the Nernst equation for the following reaction:

$\ce{O_2 + 4H^+ + 4e^- -> 2H_2O}$

Clearly, here $\ce{O_2}$ is getting reduced from an oxidation number of zero to an oxidation number of $\mathrm{-2}$ in the water molecule.

Usually, the examples of chemical reactions provided in the textbooks(at least in high school textbooks) for writing Nernst equations are of the form: $$\ce{M^{n+}(aq) + ne^- -> M(s)}$$ and the Nernst equation for this electrode reaction is given as: $$E_\ce{(M^{n+}/M)} = E^\circ _\ce{(M^{n+}/M)} - \frac{RT}{nF}\mathrm{ln}\frac{1}{[\ce{M^{n+}}]}$$

But here, in this case, the oxidation state for the $\ce{H^+}$ remains the same on both side of the reaction, rather the oxygen gas gets reduced. And this is what is confusing me.

The solution provided states that the equation should be: $$E = E^\circ+\frac{0.059}{4}\mathrm{log}[\ce{H^+}]^4 \cdot P_\ce{O_2}$$

I specifically didn't understand the reasoning behind writing the part that comes after log.

Please explain.

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    $\begingroup$ You know that Q, the reaction coefficient, equals 1 divided by x, where x equals the partial pressure of oxygen times the 4th power of the hydrogen ion concentration. These are the only two variables. So the second term on the right is approximately -(0.059 mV/4)log(1/x). But -log(1/x) = +log(x). $\endgroup$ – Ed V May 11 at 22:28
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    $\begingroup$ That P you wrote after the log is actually included in the log. It's like the expression of Q but reciprocated and the -ve sign adjusted. $\endgroup$ – Nisarg Bhavsar May 12 at 5:29
  • $\begingroup$ @EdV Thanks, this just made it clear for me. I was confused about whether to take the partial pressure of oxygen into account or not. $\endgroup$ – SmartRadical May 12 at 10:10

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