# Determining the reduction potential of a half cell reaction at a given pH

$$E^\circ_\ce{Cu^2+|Cu} = \pu{0.34 V}$$. What will be reduction potential at $$\mathrm{pH} = 14$$ for the same couple? Given that $$K_\mathrm{sp}$$ of $$\ce{Cu(OH)2}$$ is $$10^{-19}$$.

### My Attempt

I think that this problem might be solved by using the Nernst equation

$$E = E^\circ - \frac{0.0591}{n}\log_{10} K,$$

but I don't know how do I apply that. We have the $$E^\circ$$ and the number of electrons participating (which is equal to 2), but I don't know how do I determine the equilibrium constant.

Any help or hint would be appreciated.

• Hint: try to link solubility product and $K$; start by writing the reaction of dissolution of copper(II) hydroxide. – andselisk Mar 12 '19 at 20:45
• pH is going to affect that equilibrium. – andselisk Mar 12 '19 at 20:52
• I am so sorry. That does give me the correct answer. Made a calculation error. Thanks anyways. – Tony Mar 12 '19 at 21:18
• If you think you figured it out, you can post an answer to your own question:) – andselisk Mar 12 '19 at 21:40
• The gist is that the standard potential is for a copper solution in its standard state, ie 1 molar $\ce{Cu^{2+}}$. Since the concentration of copper is so much less the reduction potential of a copper solution at pH 14 is going to be much greater. I'd guess so much greater than you'd have water hydrolysis instead of copper plating. – MaxW Mar 12 '19 at 23:35

The K you refer to relates to the overall cell reaction and should be Q, the reaction quotient. This is only equal to K when the potential difference between the two 1/2 cells has fallen to zero.

This question does not refer to a full cell, but to a 1/2 cell. The form of The Nernst Equation you must use, assuming we are at 298K, is:

$$\rm E=E^\circ+\dfrac{0.0591}{z}log\dfrac{[oxidised\,\, form]}{[reduced\,\, form]}$$

Where z is the number of moles of electrons transferred which, in this case = 2.

This becomes:

$$\rm E=E^\circ+\dfrac{0.0591}{2}log[Cu^{2+}]$$

The 1/2 cell in question is:

$$\rm Cu_{(aq)}^{2+}+2e\rightleftharpoons Cu_{(s)}\,\,\,\,\,\,\,\,\,\,E^\circ=+0.34\,V$$

At such a high pH the copper(II) ions will precipitate.

$$\rm pH=14\,\,\,so\,\,\,pOH=0$$

So $$\rm c[OH^-]=1\,\,mol/l$$

This will precipitate out the copper(II) ions as the hydroxide.

So we have:

$$\rm Cu(OH)_{2(s)}\rightleftharpoons Cu_{(aq)}^{2+}+2OH_{(aq)}^-$$

$$\rm K_{sp}=[Cu^{2+}][OH^-]^2=10^{-19}$$

$$[\ce{OH-}]$$ is 1 (pH is 14), so $$[\ce{Cu^2+}]$$ is equal to $$\pu{e-19}$$.

As you can see this is a very low concentration. Applying Le Chatelier's Principle to the 1/2 cell couple we would predict that reducing the concentration of Cu(II) will drive the position of equilibrium to the left. This would push out electrons and make the electrode potential of the 1/2 cell more negative.

Putting in the numbers into the Nernst Equation:

$$\rm E = 0.34+\dfrac{0.0591}{2}\log \pu{e-19}$$

$$\rm E = 0.34- 1.12\,\,\,\,\,V$$

$$\rm E = -0.78\,\,\,\,V$$

As predicted, it has been reduced considerably.