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I have come across many online articles and books where I see the following:

Gaseous electrode

Gases such as $\ce{H2}$ and $\ce{Cl2}$ can act as anode and as cathode by loosing electrons ans by accepting electrons, respectively. For example,

$\ce{Pt,H2 | H+}$ (acts as oxidation potential)
or \begin{align} \ce{0.5 H2 (g)(\pu{1 atm}) &-> H+ (aq) + e-} \\ E_\mathrm{cell} &= E_\mathrm{cell}^\circ -\frac{0.059}{1}\log{[\ce{H+}]} \\ &= 0 + 0.059\mathrm{pH} \end{align}

Aren't the two species in different phases? If so, aren't they supposed to be separated by a '|' and not by a comma?

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    $\begingroup$ What actually reacts is hydrogen adsorbed on platinum, which I wouldn't really call a different phase. You might argue that if we interpret it like that, then the gaseous hydrogen is missing, but everyone knows it is there anyway, so why bother. $\endgroup$ – Ivan Neretin Oct 20 '17 at 5:42
  • $\begingroup$ (Without bothering to look at the textbook definition) I think that the "|" denotes the two different sides of the equation (in an abbreviated manner, since Pt would technically belong on both sides, being effectively a structure-providing catalyst). $\endgroup$ – TAR86 Oct 20 '17 at 5:45
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Since $\ce{Pt}$ is not a reactant and only a catalyst, one only wants to write $\ce{H_2}$ on the reactant side. When it is specified, since it is not involved in the reaction, the | is not warranted since Pt doesn't show up in the reaction as the comment by TAR86 pointed out.

Generally, the electrode material is not entered into the shorhand notation since the material of the electrode doesn't change the thermodynamic potential. In this case, the choice of $\ce{Pt}$ is critical and that is why it is specified. The potential of the electrode would be affected by the details of the measurement device with any other electrode material that is a worse catalyst.

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    $\begingroup$ My question isn't why platinum is included. Rather it is about the semantics regarding why there is a comma after platinum. $\endgroup$ – Vishal Subramanyam Rajesh Oct 20 '17 at 10:57
  • $\begingroup$ I edited the answer. It is since the Pt is not showing up on either side of the reaction. $\endgroup$ – Burak Ulgut Oct 20 '17 at 14:10

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