1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.