Question 5 of 7: Copper Half-Cell Potential; Electroplating Current and Time

Question 5: Copper Half-Cell Potential; Electroplating Current and Time (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. (a) $m_{Cu}=40$ g, $V=2$ L, $M_{Cu}=63.54$ g/mol (page-1 table), $E^0=+0.337$ V, $n=2$. (b) coating thickness $=500\,\mu$m $=0.05$ cm, cathode diameter $=2.5$ cm, current density $=400$ A/ft$^2$, $\rho_{Cu}=8.96$ g/cm$^3$, Faraday’s constant $F=96{,}500$ C/mol.

Find. (a) Electrode potential $E$. (b) Plating current $I$ and time $t$.

Approach

(a) is a direct application of the Nernst equation, using the given formula sheet's form $E=E^0+(0.0592/n)\log_{10}[\text{ion}]$ at room temperature. (b) combines a geometric/Faraday’s-law calculation (mass of copper needed for the specified thickness $\rightarrow$ moles $\rightarrow$ charge) with the specified current density (which fixes the current from the cathode area alone); time then follows from charge $=$ current $\times$ time.

1. (a) Molar concentration of Cu$^{2+}$. $$n_{mol}=\frac{40}{63.54}=0.6295\ \text{mol}\ \Rightarrow\ [\text{Cu}^{2+}]=\frac{0.6295}{2} =0.3148\ \text{mol/L}$$
2. (a) Nernst equation. $$E=E^0+\frac{0.0592}{n}\log_{10}[\text{Cu}^{2+}]=0.337+\frac{0.0592}{2}\log_{10}(0.3148)$$ $$=0.337+0.0296(-0.5019)=0.337-0.0149$$ $$\boxed{E\approx0.322\ \text{V}}$$ (slightly below $E^0$, as expected: the Cu$^{2+}$ concentration here is below the 1 mol/L standard-state reference, and the Nernst correction is small since $\log_{10}(0.31)$ is only mildly negative).
3. (b) Cathode area and plating current. One face of the 2.5 cm-diameter disc: $$A=\frac{\pi}{4}(2.5)^2=4.909\ \text{cm}^2=4.909/929.0=5.284\times10^{-3}\ \text{ft}^2$$ $$I=(\text{C.D.})\times A=400(5.284\times10^{-3})$$ $$\boxed{I\approx2.11\ \text{A}}$$
4. (b) Mass of copper to be deposited. $$\text{Volume}=A\times\text{thickness}=4.909\ \text{cm}^2\times0.05\ \text{cm}=0.2454\ \text{cm}^3$$ $$m=\text{Volume}\times\rho_{Cu}=0.2454(8.96)=2.199\ \text{g}$$
5. (b) Plating time from Faraday’s law. Using the given formula $w=ItM/(nF)$, solved for $t$: $$t=\frac{w\,n\,F}{I\,M}=\frac{2.199(2)(96{,}500)}{2.113(63.54)}=\frac{424{,}400}{134.3}$$ $$\boxed{t\approx3161\ \text{s}\approx52.7\ \text{min}\ (0.878\ \text{hr})}$$