21-Mat-A3 Structure and Characterization of Materials · Dec-10-Met-A3 2018
Question 7 of 7: Electrometallurgy (20 marks)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2018 — 10-Met-A3, Metal Extraction Processes. Three hours, closed book, one approved calculator (Casio or Sharp). Seven problems of 20 marks each; the rubric asks for any five, and only the first five in the answer book are marked. All seven are solved here, because this set is a study resource rather than an exam script.
Note on the exam title. The printed exam header reads 10-Met-A3, Metal Extraction Processes. The content is extractive metallurgy — mineral processing, mass and heat balances, pyrometallurgical roasting, copper and aluminum production, and electrometallurgy — and is answered as such.
Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:
F. Habashi, Textbook of Pyrometallurgy — roasting, smelting, mass and heat balances.
T. Rosenqvist, Principles of Extractive Metallurgy, 2nd ed. — ironmaking, non-ferrous smelting, electrometallurgy.
B. A. Wills and J. Finch, Wills' Mineral Processing Technology, 8th ed. — pulp density, comminution, froth flotation.
D. R. Gaskell, Introduction to the Thermodynamics of Materials — heat capacities, reaction enthalpies, electrochemistry.
W. G. Davenport et al., Extractive Metallurgy of Copper, 5th ed. — smelting/converting/refining and SX-EW flowsheets.
ASM Handbook, Vol. 2 (Properties and Selection: Nonferrous Alloys); CIM (Canadian Institute of Mining) practice literature for Canadian smelter/refinery practice (e.g. the historic Cominco smelter at Trail, BC, and Rio Tinto's Kitimat, BC aluminum smelter).
Given. A $\mathrm{Zn}$/$\mathrm{Cu^{2+}}$ galvanic (Daniell-type) cell with standard reduction potentials for the two half-reactions, and (part d) non-standard concentrations of $\mathrm{Cu^{2+}}$ and $\mathrm{Zn^{2+}}$.
Given data
Quantity
Symbol
Value
Reduction potential, $\mathrm{Zn^{2+}/Zn}$
$E^{\circ}_{Zn}$
−0.76 V
Reduction potential, $\mathrm{Cu^{2+}/Cu}$
$E^{\circ}_{Cu}$
0.34 V
Electrons transferred
$n$
2
Temperature
$T$
298.15 K
Non-standard $[\mathrm{Cu^{2+}}]$ (part d)
—
0.5 M
Non-standard $[\mathrm{Zn^{2+}}]$ (part d)
—
1.5 M
Find. (a) $E^{\circ}_{cell}$, (b) $\Delta G^{\circ}$, (c) the equilibrium constant $K$, and (d) the non-standard cell potential $E$ by the Nernst equation.
Approach. Identify the cathode (the more positive/more easily reduced half-reaction, copper) and anode (zinc, oxidized), get $E^{\circ}_{cell}$ from their difference, then chain the standard thermodynamic relations $\Delta G^{\circ}=-nFE^{\circ}_{cell}$ and $\Delta G^{\circ}=-RT\ln K$ to reach $K$, and finally apply the Nernst equation for the stated non-standard concentrations.
Part (a) — standard cell potential. In the cell reaction as written, zinc is oxidized (anode) and $\mathrm{Cu^{2+}}$ is reduced (cathode), so
$$E^{\circ}_{cell}=E^{\circ}_{cathode}-E^{\circ}_{anode}=E^{\circ}_{Cu}-E^{\circ}_{Zn}=0.34-(-0.76)$$
$$\boxed{E^{\circ}_{cell}=1.10\ \text{V}}$$
A positive $E^{\circ}_{cell}$ confirms the reaction proceeds spontaneously left to right as written, which is exactly what a galvanic (rather than electrolytic) cell requires.
Part (b) — standard free energy. With $n=2$ electrons transferred per mole of reaction and Faraday's constant $F=96{,}485\ \text{C/mol}$:
$$\Delta G^{\circ}=-nFE^{\circ}_{cell}=-2\times96{,}485\times1.10=-212{,}267\ \text{J/mol}$$
$$\boxed{\Delta G^{\circ}=-212.3\ \text{kJ/mol}}$$
The large negative $\Delta G^{\circ}$ reflects the wide 1.10 V separation between the two half-cell potentials — zinc and copper sit far apart on the electrochemical series, which is exactly why this pairing is the textbook galvanic cell.
Part (c) — equilibrium constant. Equating the two expressions for $\Delta G^{\circ}$ gives $\ln K=nFE^{\circ}_{cell}/(RT)$:
$$\ln K=\frac{nFE^{\circ}_{cell}}{RT}=\frac{2\times96{,}485\times1.10}{8.314\times298.15}=\frac{212{,}267}{2478.8}$$
$$\ln K=85.6$$
Converting to base 10 (a number this large is only meaningfully reported as an order of magnitude):
$$\log_{10}K=\frac{\ln K}{\ln10}=\frac{85.6}{2.303}$$
$$\boxed{K\approx10^{37.2}}$$
A $K$ of this size means the reaction runs essentially to completion — consistent with a cell potential as large as 1.10 V, since $K$ depends exponentially on $E^{\circ}_{cell}$.
Part (d) — non-standard cell potential (Nernst equation). For the reaction $\mathrm{Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu}$ (solids at unit activity), the reaction quotient is $Q=[\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}]$. At $[\mathrm{Cu^{2+}}]=0.5$ M and $[\mathrm{Zn^{2+}}]=1.5$ M:
$$Q=\frac{1.5}{0.5}=3.0$$
$$E=E^{\circ}_{cell}-\frac{RT}{nF}\ln Q=1.10-\frac{8.314\times298.15}{2\times96{,}485}\ln(3.0)=1.10-(0.01285)(1.099)$$
$$\boxed{E\approx1.086\ \text{V}}$$
Raising $[\mathrm{Zn^{2+}}]$ and lowering $[\mathrm{Cu^{2+}}]$ relative to standard conditions both push the reaction back toward its reactants (Le Châtelier), so $E$ drops slightly below $E^{\circ}_{cell}$ — but only by 14 mV, because the $RT/nF$ prefactor is small and the shift in $Q$ from 1 to 3 is modest.
