Could someone, please, explain to me how the standard Hydrogen Electrode works and how it is used to measure electrode potentials? Also, why does it have a potential value of zero?

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    $\begingroup$ Zero is by definition. $\endgroup$ – Mithoron Apr 11 '15 at 11:23
  • $\begingroup$ @Hexacoordinate-C Why did you approve the edit when you then manually revert it back to the original state? $\endgroup$ – Martin - マーチン Aug 2 '17 at 11:42
  • $\begingroup$ @Martin Idk how that website works. I made a mistake. $\endgroup$ – ParaH2 Aug 2 '17 at 11:48
  • $\begingroup$ @Hexacoordinate-C It's no problem. When you are reviewing suggested edits, you have the option to approve, approve and then edit (you chose that option), reject, reject and then edit yourself. If you are undecided you can also skip. In this case you should have just rejected the edit, this would have saved you some time. More importantly, thanks for helping out :D $\endgroup$ – Martin - マーチン Aug 2 '17 at 11:53

The SHE (Standard Hydrogen Electrode) is a standard equipment for measuring the relative electrode potential, as the process of measuring the actual potential is quite difficult. It consists of a platinum foil dipped in 1.0M $\ce{HCl}$ in the presence of $\ce{H2}$ gas at 1 bar of pressure.

Now when the SHE is connected to another half-cell, either oxidation or reduction may occur in the SHE.

Suppose we have to find the electrode potential of a half cell represented by "A" Now, on setting up the cell and connecting a voltmeter we get the standard potential(Estandard) of the cell i.e.
Estandard = EA - EH2

Now as our our aim is to just find EA, we take EH2 to be equal to 0
Estandard = EA - 0
Estandard = EA
The value of Estandard is taken from the voltmeter.
It should be noted that the electrode potential calculated using this method is relative(to the electrode potential of hydrogen). It isn't the absolute value. Also, the potential for the SHE is taken to be 0 for the simplification of the process.

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    $\begingroup$ The SHE also has hydrogen gas in it so that either half reaction can occur: $$\ce{2H+ +2 e- -> H2}$$ $$\ce{H2 -> 2H+ + 2e-}$$ $\endgroup$ – Ben Norris Apr 11 '15 at 12:09

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