Galvanic cell involving a hydrogen electrode and copper electrode in sodium hydroxide

Question

A galvanic cell consists of a standard hydrogen electrode and a copper electrode. Suppose that the copper electrode is immersed in a solution that is $0.100\ \mathrm{M}$ in $\ce{NaOH}$ and that is saturated with $\ce{Cu(OH)2}$. Find cell potential

$\begin{alignat}{2}\ce{2 H+ (aq) -> H2 (g)}\quad E_0 = 0.00\ \mathrm{V}\\ \ce{Cu^2+ (aq) + 2 e- -> Cu (s)}\quad E_0 = 0.34\ \mathrm{V}\end{alignat}$

$K_\text{sp} = 1.6 \times 10^{-19}$ for $\ce{Cu(OH)2}$

I started by doing writing off the Nernst equation and realized I do not have $\ce{Cu^2+}$ concentration, and since I’m given $K_\text{sp}$ of copper, I assumed I should use it to find copper (after doing ice table for $\ce{Cu(OH)2 -> Cu^2+ + 2 OH-}$). But then I stumbled on $0.100\ \mathrm{M}$ of $\ce{NaOH}$, what should I do with it? and what does saturated mean in the given context of the problem?

Answer 1

In this case, "saturated" means that there is enough dissolved copper (II) hydroxide such that no more will dissolve, i.e. the equilibrium represented by the $K_\mathrm{sp}$, $\ce{Cu(OH)2_{\ (s)}<=>Cu^2+_{(aq)} + 2OH-_{(aq)}}$ holds true. So first, let's figure out the concentration of copper (II) hydroxide that will dissolve in 0.1 M NaOH (making the assumption, as mentioned in the comments, that $[\ce{OH^-}] = \mathrm{C_\ce{NaOH}})$:

• The $K_\mathrm{sp}$ is really really small, so we can actually ignore its contribution to the $[\ce{OH^-}]$, but for fun, you can do the whole ICE table thing and end up with: $$\begin{align} [\ce{OH^-}] &= .1 + 2x \\ [\ce{Cu^2+}] &= x \\ K_\mathrm{sp} &= [\ce{OH^-}]^2[\ce{Cu^2+}] \\ 1.6\times 10^{-19} &= (.1 + 2x)^2(x) \\ x = [\ce{Cu^2+}] &= 1.6 \times 10^{-17}\ \mathrm{M} \end{align} $$

Next, using the Nernst equation, we can calculate the copper half-cell reduction potential (Assuming T = 298.15 K):

• $$\begin{align} E &= E^\circ - \frac{RT}{zF}\ln\frac{1}{[\ce{Cu^2+}]} \\ &= 0.34\ \mathrm{V} - \frac{RT}{2F}\ln\frac{1}{1.6 \times 10^{-17}} \\ &= -0.16\ \mathrm{V} \end{align} $$

Finally, the whole cell potential:

• First, negate the calculated reduction potential to get the oxidation potential: $$\begin{align} E_\mathrm{ox} &= -E_\mathrm{red} \\ &= 0.16\ \mathrm{V} \end{align} $$

• Then, because the SHE's reduction potential is defined as 0 V, the whole cell potential is the same as the oxidation half-cell potential:

  $$\begin{align} E_\mathrm{ox} &= E_{(\ce{H})\mathrm{red}} + E_{(\ce{Cu})\mathrm{ox}} \\ &= 0\ \mathrm{V} + 0.16\ \mathrm{V} \\ &= 0.16\ \mathrm{V} \end{align} $$