1
$\begingroup$

So the exercise goes like this:

We consider the chemical battery(i have no idea what is its actual english name since i'm translating), $\ce{Mn_{(s)}/Mn(OH)_2//Cu^{2+}/Cu_{(s)}}$ Where $\ce{Mn}$ is submerged in a saturated solution of $\ce{Mn(OH)_{2}}$ of $\ce{pH}=9.86$

first is to calculate the constant of solubility $\ce{kS}$ and $S$ this i think i can do, since we have the concentration of $\ce{OH^-}$, and we have the relation $\ce{2S=[OH^-]}$ then we can get the $S$ then the $\ce{kS}$ which is equal to $\ce{kS=[OH^-]^2*[Mn^{2+}]}$ with $\ce{Mn^{2+}}=S$

Then comes the question that I couldn't do, it says to calculate the potential of each electrode, for the electrode with the Copper element i can do, but for the electrode with the $\ce{Mn(OH)2}/\ce{Mn}$ couple I'm confused. We have the demi-reaction of Oxidation(Since the standard potential of the Copper couple is higher than the Manganese one):

$\ce{Mn +2OH^-\leftrightarrow Mn(OH)_2+2e^-}$

Now for the electrode potential:

$\ce{E=E^0+\frac{0.06}{2}log(\frac{1}{[OH^-]^2})}$

This is the one I'm not sure of at all.

Another question is to deduce the potential of the oxred couple $\ce{Mn^{2+}/Mn}$ from the previous data we have. I think this one I have to equal the potentials of the couple $\ce{Mn(OH)_2/Mn}$ and $\ce{Mn^{2+}/Mn}$ since they coexist at the same electrode(I'm not sure of this either and I need confirmation)

Note: The potential standard of $\ce{Mn(OH)_2/Mn}$ is a given.

This isn't a homework, just me trying to understand the oxred batteries better and getting stuck.

Thanks for your help!

$\endgroup$
  • 1
    $\begingroup$ @OscarLanzi That is my bad hahaha $\endgroup$ – Mario SOUPER Aug 23 at 20:18

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.