4
$\begingroup$

From what I understand, the standard reduction/oxidation potential table is based on the hydrogen half reaction which makes a reduction and oxidation potential of 0 respectfully.

In a research internship I'm in now, we are looking for more efficient batteries, they are basing potential candidates in respect with magnesium.

My problem is seeing how the 'in respect' part works and how it affects the system. Maybe it's just hard for me to relate to it in my mind so maybe a more common analogy would help?

| improve this question | | | | |
$\endgroup$
  • 2
    $\begingroup$ Just to clarify, are you looking for the equivalent oxidation/reduction potentials with respect to other specific metals instead of hydrogen? $\endgroup$ – user15489 Jun 7 '15 at 18:38
1
$\begingroup$

You cannot define ‘vacuum’ oxidation potentials, or otherwise derive them de novo without having some zero-point. There is, technically, no self-suggesting zero-point around — unlike for example temperature, where $0\,\mathrm{K}$ is equivalent to ‘molecular movement frozen’.

Part of the reason is, that the electron(s) lost or gained during the redox reaction must go or come from somewhere. Therefore, one has chosen the hydrogen cell as a centre of the scale. Why?

  • Hydrogen is the easiest element with the least electrons.

  • It’s simple to build.

  • It actually also comes with a good further distinction: The one between noble and non-noble metals.

You can actually easily recalculate the standard potentials with respect to another half-cell like magnesium: Just add/subtract that cell’s potential with respect to hydrogen to all other potentials. It’s a bit like moving the temperature scale’s zero point from $0\,\mathrm{K}$ to $273\,\mathrm{K} = 0\,\mathrm{°C}$.

| improve this answer | | | | |
$\endgroup$
  • 1
    $\begingroup$ This isn't strictly true. There have been a few definitions of absolute electrode potentials, but they don't really add anything useful outside of specialized applications and are difficult to measure, so they're not generally used. $\endgroup$ – Michael DM Dryden Jun 8 '15 at 4:04

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.