I've got the following question and I'm having doubts with it:

$$\ce{NAD+ + 2e- + H+ -> NADH}$$

Calculate ${E^\circ}'$ for the half cell $\ce{Pt | NADH, NAD+, H+}$ at pH 7, given that $E^\circ = \pu{-0.358 V}$ at $\pu{298 K}$.

It wants the formal potential. So I write the Nernst equation:

$$E = E^\circ - \frac{RT}{nF}\ln\left(\frac{[\ce{NADH}]}{[\ce{H+}][\ce{NAD+}]}\right)$$

so $$E = E^\circ - \frac {RT} {nF} \ln\left(\frac {1} {[\ce{H+}]}\right) - \frac {RT} {nF} \ln\left(\frac{[\ce{NADH}]}{[\ce{NAD+}]}\right)$$
Now if I understand correctly, $$E^\circ - \frac {RT}{nF} \ln\left(\frac {1} {[\ce{H+}]}\right)$$ is the formal potential, I calculated this and got $\pu{-0.56 V}$. But the literature values are different. Can you help?

  • 1
    $\begingroup$ Instead of deleting your old question, you could edit it - the edit will push it into a queue for reopening, and I would probably have reopened it too. Anyway, it's OK. By the way, you can format mathematical and chemical expressions on Chemistry.SE using MathJax; this post contains further details. I've done up a bit and will leave the rest for you. $\endgroup$
    – orthocresol
    Mar 19 '17 at 21:09

With the standard electrode potential assumed to be $\pu{-0.358 V}$, I am getting the same answer as you but I googled the standard electrode potential of the above reaction and I found it to be $\pu{-0.320 V}$. On substituting $\pu{-0.320 V}$ as the standard electrode potential, I got $\pu{-0.529 V}$.


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