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

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?

• This is the way a homework question should be written, nicely done! – user15489 May 30 '15 at 7:05
• ehm,thank you,would appreciate if you help me to solve it though haha:) – Jx1 May 30 '15 at 7:08
• How is the $K_{sp}$ related to the concentration of the ions? – LDC3 May 30 '15 at 18:13
• uh, it's like regular K?? the part im stuck with is, what do I do with the 0.100??? because if i solve it as regular K( eg: do ice table, take square root of the Ksp) i dont get the right answer – Jx1 May 30 '15 at 20:36
• Do consider the fact that you have the concentration of NaOH. If you assume near complete disassociation of NaOH, you have the concentration of hydroxide ions in solution. Did you try plugging it in to the expression for solubility product? – getafix 20 mins ago – getafix May 31 '15 at 0:34

## 1 Answer

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}