3
$\begingroup$

Is it possible for $n$ in $\Delta G=-nFE$ to ever be negative? Trying to figure out where I went wrong on some thing and want to rule that out.

$\endgroup$
8
$\begingroup$

No. This result occurs from the Nernst equation.

The "n" is the number of electrons transferred. If no electrochemical reaction occurred, then n = 0.

Otherwise "n" is positive.

Having a negative number of electrons transferred would be impossible.

$\endgroup$
  • 4
    $\begingroup$ And indeed, there is a direct relationship between Nernst and $\Delta G = -n F E$ as indicated in the link in my answer (i.e., here) or in many other sources. $\endgroup$ – Geoff Hutchison Apr 20 '15 at 12:10
  • 10
    $\begingroup$ @ADG I think you're being incredibly rude. I teach physical chemistry. I don't need to look this up on Google. $\endgroup$ – Geoff Hutchison Apr 20 '15 at 12:10
  • 1
    $\begingroup$ Could it be that both these standard equations can be derived from each other? Backtracing the chemwiki derivation? $\endgroup$ – ChemExchange Apr 20 '15 at 17:00
  • 1
    $\begingroup$ Indeed, derivations can go both directions. $\endgroup$ – Geoff Hutchison Apr 20 '15 at 18:09
  • 5
    $\begingroup$ @ADG I've been responding to flags for 5 minutes and I've run into a multitude of rude, obnoxious, and otherwise non-constructive comments that you've left. It's going to be a matter of shape up or ship out here. Some radical changes need to be made immediately. Every user here deserves your utmost respect and courtesy and no one needs your grandstanding. $\endgroup$ – jonsca Apr 20 '15 at 23:20
1
$\begingroup$

Is it possible for $n$ to ever be negative? Trying to figure out where I went wrong on some thing and want to rule that out.

$\Delta G$ is the maximum non expansion work and hence should be equal to external force under consideration times the distance for a reaction to be at equilibrium, hence we need to find that external force.

I know that from electrostatics that for a charge in a uniform electric field the force is $\bar F=q\bar E$. for $q=-1.6\times10^{-19}\ \mathrm C$. When we take our system to be $1\ \mathrm{mol}$ of electrons, we have a total of $-1.6\times10^{-19}\ \mathrm C\times6.02\times10^{23}\approx -96500\ \mathrm C$ and we have that $1\ \mathrm F=96500\ \mathrm C$. So total force is: $$\Delta G=W_\text{max,add}=\bar F_\text{ext}\cdot \bar d$$ $$\Delta G=\bar F_\text{ext}\cdot \bar d=q_\text{total}\cdot E\cdot 1\ \mathrm m=(-n) _\mathrm F\cdot E _\mathrm V=-nFE_\mathrm J$$

Now think, can the number of moles of electrons transferred be negative?

$\endgroup$
-1
$\begingroup$

$n$ cannot be negative.

$$\Delta G = -nFE$$

Transfer of $\ce{e-}$ cannot be negative.

$\endgroup$
  • $\begingroup$ This answer has been flagged as low quality, probably because (1) you didn't make the effort to write full sentences, including punctuation or (2) you did not provide a clear answer. Please make changes or the answer risks being deleted. $\endgroup$ – Buck Thorn Mar 11 at 3:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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