# Calculate the potential of each electrode, $\ce{Mn(OH)2}/\ce{Mn}$ ox/red couple

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.