Copper Half-Cell Potential and Electroplating: 04-BS-11, May 2013, Q5

Question 5: Copper Half-Cell Potential and Electroplating (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) Mass Cu $=40$ g dissolved as Cu$^{2+}$ in $V=2$ L; $E^0=+0.337$ V; $n=2$; $T=298$ K (room temperature). (b) Deposit thickness $=500\ \mu\text{m}$; cathode diameter $=2.5$ cm (circular disk); C.D. $=400$ A/ft$^2$; $\rho_{Cu}=8.96$ g/cm$^3$; $M_{Cu}=63.54$ g/mol; $F=96{,}500$ C/mol.

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

Fig. Q5(b) — electroplating cell: Cu anode, CuSO₄ electrolyte, circular cathode (Ø 2.5 cm) plated at 400 ASF.

Approach

Part (a) is a direct application of the Nernst equation using the molar concentration of Cu$^{2+}$ formed from the dissolved mass. Part (b) first sets the current from the recommended current density and cathode area, then finds the plating time from Faraday's law using the mass of copper needed for the specified deposit thickness.

(a) Concentration of Cu²⁺. $$n_{Cu}=\frac{40}{63.54}=0.6295\ \text{mol}\ \Rightarrow\ [\text{Cu}^{2+}]=\frac{0.6295}{2}=0.3148\ \text{mol/L}.$$
Nernst equation. $$E=E^0+\frac{RT}{nF}\ln[\text{Cu}^{2+}]=0.337+\frac{(8.314)(298)}{(2)(96{,}500)}\ln(0.3148)=\boxed{0.322\ \text{V}}.$$
(b) Cathode area and current. Diameter $2.5$ cm $\Rightarrow$ $A=\tfrac{\pi}{4}(2.5)^2=4.909\ \text{cm}^2=5.284\times10^{-3}\ \text{ft}^2$ (1 ft$^2$ = 929.0 cm$^2$). At 400 A/ft$^2$: $$I=(400)(5.284\times10^{-3})=\boxed{2.11\ \text{A}}.$$
Mass of copper to deposit. Volume $=A\times$thickness $=4.909\times0.0500=0.2454\ \text{cm}^3$: $$m=\rho_{Cu}V=8.96\times0.2454=\boxed{2.199\ \text{g}}.$$
Faraday's law ⇒ plating time. Moles Cu $=2.199/63.54=0.03461$ mol; charge $Q=nFn_{Cu}=2\times96{,}500\times0.03461=6680$ C: $$t=\frac{Q}{I}=\frac{6680}{2.11}=\boxed{3161\ \text{s}}=52.7\ \text{min}.$$

Quantity	Result
(a) Electrode potential, $E$	0.322 V
(b) Cathode area	4.909 cm²
(b) Current, $I$	2.11 A
(b) Plating time	3161 s = 52.7 min

Concept & Theory

Topic. Electrochemical half-cell potentials (Nernst equation) and Faraday's laws of electrolysis.

Key relations. $E=E^0+\frac{RT}{nF}\ln[\text{M}^{n+}]$; $I=\text{C.D.}\times A$; $m=\dfrac{ItM}{nF}$ (Faraday's law).

Why this works. The Nernst equation corrects the standard potential for the actual ion activity present, here fixed by how much copper was dissolved; in electroplating, the recommended current density sets the deposition rate (grain quality/adhesion), so the current follows directly from the plated area, and Faraday's law converts the resulting charge passed into the mass (and hence thickness) of metal deposited.

Common pitfall. Using natural log versus $\log_{10}$ inconsistently in the Nernst equation, or forgetting to convert the current density's ft$^2$ basis to the cathode's cm$^2$ area (or vice versa) before multiplying.

Source. Callister & Rethwisch, Materials Science and Engineering, 9th ed., Ch. 17 (corrosion and electrochemistry).

Question 5 of 8: Copper Half-Cell Potential and Electroplating

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

Approach