Calculate the potential of hydrogen electrode in contact with a solution of pH 10.

My attempt:

Cell reaction: $\ce{ H^+ + e^- -> \frac{1}{2}H_2 }$

I've attempted to solve this myself by using Nernst equation which yielded the answer -0.591 V. But I believe that Nernst equation is used when there are two electrodes, but my friends are saying that

"how will you make a cell from one electrode?".

So guys, please rectify the problem.

  • 4
    $\begingroup$ Easy. Choose smarter friends. Half cell potentials are measured relative to the Standard Hydrogen Electrode which is arbitrarily assigned an EMF value of 0.00 $\endgroup$ – MaxW Mar 9 '19 at 5:07
  • $\begingroup$ @MaxW, can I assume the solution to be an electrode, and thereby making a Galvanic cell from the two? $\endgroup$ – Rahul Verma Mar 9 '19 at 5:15

"how will you make a cell from one electrode?".

There is no need to, because there is nothing which prohibits you from writing a Nernst equation for a half cell.* Remember the rule no.1 in electrochemistry, whenever you mention half-cell potential, your next automatic question should be "with respect to what?". If the textbook is quiet and you are required to look up tables, it is understood that it is with reference to the hydrogen electrode under standard condition which is set at 0 V (exactly).

Now at pH 10, your standard conditions have changed and the activity or hydrogen ion concentration differs. It is no longer zero. So proceed with the Nernst equation for the half cell and work it out.

*(Yes, there is no single electrode galvanic cell, but for fun and general knowledge, people have done single electrode reduction using electrostatics (Bard and others). Faraday did it centuries ago. It was published in a very reputable magazine.)

  • $\begingroup$ Thanks. One last question: In general, does basic solution acts as an electrode? $\endgroup$ – Rahul Verma Mar 9 '19 at 7:46
  • $\begingroup$ No. Half cell is hypothetical and it cannot exist alone, but Nernst equation can be written for a half cell. $\endgroup$ – M. Farooq Mar 9 '19 at 13:28
  • $\begingroup$ Search the definition of an electrode en.wikipedia.org/wiki/Electrode $\endgroup$ – M. Farooq Mar 9 '19 at 13:32

You can't make a cell with one electrode, but there is such a thing as a cell with only one half-reaction. This paper involves the use of such a cell to generate a solvated-electron solution of magnesium in ammonia.

Suppose you have a magnesium anode, a chemically inert cathode, and an electrolyte that is a dissolved magnesium salt $\ce{MgY2}$. (You have to choose the salt with some care because you have a small, multiply charged cation in a low polarity solvent, which leads to a limited range of soluble salts. Also in practice the authors generated the magnesium salts from commercially available ammonium salts that were converted to the magnesium salts in situ.) You apply the current and the anode duly produces a reaction that is probably most accurately rendered thusly:

$\ce{Mg(s) + Y^- -> MgY^+ + 2e^-}$

(i.e. the magnesium ions are probably in the form of ion pairs, which being only singly charged would make a more stable solvated-electron solution.) The electrons go over to the cathode and pour into the solution with no reaction at all, becoming the solvated electrons!

We are unfamiliar with such a situation in water solvent because the solvated electrons in water get consumed so quickly that they in effect react at the cathode surface. Thus you are forced to include the cathodic half-reaction. But in the ammoniacal case, these electrolytically generated magnesium solutions are finding use in activating fluorinated polymer surfaces, which is driving the research.

  • $\begingroup$ The magnesium oxidation process in liquid ammonia is very interesting. Current must be flowing in the system. Did they mention what was happening at the cathode? $\endgroup$ – M. Farooq Mar 9 '19 at 15:42
  • $\begingroup$ The electrons just keep going, intermingling with the cations formed at the anode. They do not last forever (and to get them to last anyway, you need a sufficient amount of supporting electrolyte, which suggests to me the formation of ion pairs). But when they do react it's in solution, not at the cathode. $\endgroup$ – Oscar Lanzi Mar 9 '19 at 15:46

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