How do I understand which is the anode and which is the cathode in this problem?

We have a concentration cell with hydrogen electrodes. It's made up of two half-cells. One half cell contains [BaCl(aq)]=0.5 M, the other one has [BeOH(aq)]=0.5 M. Draw the cell diagram and calculate e.m.f at 25ºC

So I tried this:

$$\ce{Pt}(\ce{H2} \, 1~\mathrm{atm})| \ce{H+,\,Ba^{+2}} \,(0.5 M)||\,\ce{Be^{+2}}\, (0.5 M)|\ce{H2}(g,1~\mathrm{atm})|\ce{Pt}$$ I don't know how I should think to write the correct diagram. Can you please help me?

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    $\begingroup$ You should first assume one is the cathode and the other is the anode. Then you can calculate the theoretical emf - if the emf > 0, then that reaction will be spontaneous and it means you got the right cathode and right anode. If the calculated emf < 0, it means that that reaction is not spontaneous, which means that the reverse reaction is spontaneous; that happens if you exchange the cathode and anode. $\endgroup$ – orthocresol Nov 18 '15 at 19:29
  • $\begingroup$ If I apply Nernst' eq I'd get the potential of each half cell, right? How do I do that? For instance, for the first half-cell is it correct to write $E_{(1)}= E^ 0 - \dfrac{0.0591}{n}\, \log{\left( \dfrac{[0.5M]}{[1\,atm]}\right)}$, with n=2 because Barium has +2e-? $\endgroup$ – WobblyWindows Nov 18 '15 at 19:39
  • $\begingroup$ This is a personal preference but I prefer to write out the full cell equation first, then apply the Nernst equation to the whole equation - so you have $E = E^\circ - (RT/nF)\ln Q$ where $Q$ is the familiar reaction quotient. 1 atm is the standard pressure for gases, so you can leave out those terms with 1 atm. The only terms that contribute to ln Q are the concentrations. If you want, you can also do the same thing with the individual half-cells and although I'm not 100% sure myself (because I don't do it this way ), your equation looks correct. $\endgroup$ – orthocresol Nov 18 '15 at 19:42
  • $\begingroup$ Ok, but in the exercises so far I've used $\log_{10}$ instead of $\ln{(\cdot)}$. Is n=2 in this scenario? Also, if Q= 0.5/0.5=1 then $\log1=0$, correct? I feel there's something missing here. Should I be considering which half-cell is more acidic (more $\ce{H3O+}$ ions in it) in order to find out who's the anode? $\endgroup$ – WobblyWindows Nov 18 '15 at 19:47
  • $\begingroup$ The ln and log10 forms are equivalent. No, Q probably isn't 0.5/0.5. You can't be reducing $\ce{Ba^2+}$ and $\ce{Be^2+}$ simultaneously - something has to be oxidised - it's either going to be $\ce{H2O}$ in the $\ce{BaCl2}$ solution, or $\ce{OH-}$ in the $\ce{Be(OH)2}$ solution, depending on which electrode is the cathode - either way, the concentration of the oxidised species is not going to be 0.5. There is no quick and easy way to see which is the anode and which is the cathode. You have to work out the theoretical emf for one way and if it doesn't work, you do the other way. $\endgroup$ – orthocresol Nov 18 '15 at 19:55

Let's restate the problem: we have two hydrogen half-cells with other stuff in them. Hydrogen electrodes will have voltages related to their pH, because that is the definition of a hydrogen electrode. Then it goes haywire: does one half-cell contain BaCl2 and the other contain Be(OH)2 at 0.5 M - certainly not BaCl and BeOH. In fact, Be(OH)2 is amphoteric and soluble at only 2 ppm, so even estimating the pH of that cell would be a blind guess; you could guess the BaCl2 solution to be about 7.0, but might be a tiny bit different.

I suspect the question is not one that has actually been created and measured, but is rather a made-up question with no dependence on reality, to ask how to visualize two hydrogen electrodes in pH 7 (the BaCl2 solution) and pH 13.69 (pH of 0.5 M OH), which could not occur because of the low solubility and amphotericity of Be(OH)2. So the whole cell could be written:

Pt(H2,1 atm)|H+(aq,10e-7 M)||(H+(aq,10e-13.69 M)|Pt(H2,1 atm)

or, just consider both half cells to be at pH 7 and there's no overall cell voltage (my preference).

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    $\begingroup$ Whoever originally designed the question has no idea of real electrochemical cells or even basic chemistry because not only the formulae are wrong in the original question but Be(OH)2 has limited solubility as you pointed out. Since the title reads concentration cells, both cells $must$ have the same ions. The question is messed up by typos. However the idea is nice, how to identify anode and cathode $experimentally$? $\endgroup$ – M. Farooq Dec 1 '19 at 2:44

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