Jump to content
Science Forums

Recommended Posts

Posted

[math]\sqrt[4]{\frac6\pi}[/math]=1.17557500734

 

[math]\sqrt[3]{10}[/math]=2.15443469003

 

[math]\alpha=\left(\frac{e^2}{\hbar c}\right)[/math] fine structure constant

 

[math]re=\left(\frac{\sqrt[4]{\frac6\pi}\times\sqrt[3]{10}^{25}}{\ c^{2}\right)[/math] radius of the electron in cgs units

 

[math]re=\left(\frac{\sqrt[4]{\frac6\pi}\times\sqrt[3]{10}^{19}}{\ c^{2}\right)[/math] radius of the electron in SI units

 

[math]me=\left(\frac{\left(\2\right)^4\times\sqrt[4]{\frac6\pi}\times\sqrt[3]{10}^{-245}\times\left(\pi\right)^2}{\ h^2}\right)[/math] mass of the electron in cgs units

 

[math]me=\left(\frac{\left(\2\right)^4\times\sqrt[4]{\frac6\pi}\times\sqrt[3]{10}^{-296}\times\left(\pi\right)^2}{\ h^2}\right)[/math] mass of the electron in SI units

Posted

Nernst equation? Let me try...

 

[math]{\Delta}G = -W[/math]

 

[math]{\Delta}G = -q{E^{o}}[/math]

 

but since

[math]q = nF[/math]

 

[math]{\Delta}G = -nFE[/math]

 

[math]{\Delta}{G^{o}} = -nF{E^{o}}[/math]

 

For

[math]aA {\,}+{\,} bB {\rightarrow} cC {\,}+{\,} dD[/math]

 

[math]Q = {\frac{{[C]^{c}}{[D]^{d}}}{{[A]^{a}}{^{b}}}}[/math]

 

And because

 

[math]{\Delta}G = {\Delta}{G^{o}} + RT{{Ln}Q}[/math]

 

[math]-nFE = -nF{E^{o}} + RT{{Ln}{\frac{{[C]^{c}}{[D]^{d}}}{{[A]^{a}}{^{b}}}}}[/math]

 

or,

 

[math]E = {E^{o}} - {{\frac{RT}{nF}} {{Ln}{\frac{{[C]^{c}}{[D]^{d}}}{{[A]^{a}}{^{b}}}}}}[/math]

 

Wow! this is some amount of work for something that would take me five seconds on paper!

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
×
×
  • Create New...