CHEMISTRY COMMUNITY

Posted: **Fri Feb 05, 2016 8:54 pm**

So from the course reader, $E=E^\circ -\frac{RT}{nF}ln\frac{[P]}{[R]}$ and at 25 degrees Celsius, $E_{cell}=E^\circ -\frac{0.0592}{n}LOG_{10}Q$

but in the practice midterms, they use this formula $E=E^\circ -\frac{0.0257}{n}ln\frac{[P]}{[R]}$ . can someone explain where 0.0257 comes from?

Posted: **Fri Feb 05, 2016 9:15 pm**

The 0.0257 comes from plugging in and solving for (R*T)/F at 25 degrees Celsius. 0.0257 is used when using ln as opposed to log10. If you want to convert to log, you multiply 0.0257 and ln(10). The product of these two is 0.0592 and can be found in the variation of the Nernst equation that uses the log function.

Posted: **Sat Feb 06, 2016 2:27 pm**

Oh, ok. That makes much more sense now. Thank you!

CHEMISTRY COMMUNITY

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