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. $\endgroup$ Mar 19, 2017 at 21:09

1 Answer 1


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 $\pu{-0.529 V}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.