Question 2 of 6: Ten hydrometallurgical/electrochemical term pairs

Problem 2 — Ten hydrometallurgical/electrochemical term pairs (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.

a) Hydrolysis vs. Hydration. Hydration is the purely electrostatic association of intact water dipoles around a dissolved ion (a solvation sheath) — no bond is broken. Hydrolysis is a chemical reaction in which the dissolved ion breaks an O-H bond of a coordinated water molecule, releasing H⁺ and forming a hydroxo-complex, e.g. $\text{M}^{2+}+\text{H}_2\text{O}\rightleftharpoons \text{M(OH)}^{+}+\text{H}^{+}$ — exactly the step whose equilibrium constant $K_1$ is derived from the speciation diagram in Problem 4. Difference: hydration is physical solvation; hydrolysis is a proton-transfer reaction that changes the ion's own speciation and lowers pH.

b) Speciation diagram vs. Distribution diagram. A speciation diagram plots the concentration (often log-molar) of every dissolved AND solid species of an element against pH or Eh at one fixed total loading — the four Eh-pH diagrams of Problem 3 are speciation diagrams. A distribution (fractional) diagram instead plots each species' FRACTION of total dissolved metal (fractions sum to 1 at every pH), discarding the absolute-concentration axis — the mol% plot of Problem 4 is a distribution diagram. Difference: a speciation diagram carries absolute-concentration/solid-stability information; a distribution diagram is normalized and shows only relative proportions, which is exactly why the crossover pH's in Problem 4 (where two fractions are equal) fix equilibrium-constant RATIOS but not absolute concentrations on their own.

c) Cementation vs. Precipitation. Cementation is a specific electrochemical (redox) recovery method: a more active metal is oxidized and dissolves while simultaneously reducing a less active dissolved metal ion onto its surface as solid metal, e.g. Problem 3(d)'s $\text{Fe(s)}+2\text{Au(CN)}_2^-\rightarrow \text{Fe}^{2+}+2\text{Au(s)}+4\text{CN}^-$. Precipitation is the broader term for forming any insoluble solid by exceeding its solubility product — no metal-to-metal electron transfer is required, e.g. Problem 5's $\text{Mn}^{2+}+\text{S}^{2-}\rightarrow \text{MnS(s)}$. Difference: cementation is always a redox reaction between two metals; precipitation is a general solubility-driven phenomenon that need not involve any oxidation-state change.

d) Anode vs. Cathode. The anode is the electrode at which OXIDATION occurs (electrons leave into the external circuit). The cathode is the electrode at which REDUCTION occurs (electrons arrive and are consumed, e.g. $\text{Zn}^{2+}+2e^-\rightarrow \text{Zn(s)}$ in Problem 6b). Difference: this assignment is identical in both galvanic and electrolytic cells; what differs (part e) is the terminal polarity.

e) Galvanic cell vs. Electrolytic cell. A galvanic (voltaic) cell converts a spontaneous reaction ($\Delta G \lt 0$) directly into electrical energy, generating its own EMF; the cathode is the positive terminal. An electrolytic cell — every electrowinning cell in Problem 6b, and every cementation-free electrolytic recovery in Problem 3(iii) — consumes external electrical energy to force a non-spontaneous reaction ($\Delta G>0$); the anode is connected to the positive terminal, the cathode to the negative. Difference: the sign of $\Delta G$ and the terminal polarity are reversed, even though oxidation still occurs at the anode and reduction at the cathode in both.

f) Solubility vs. Dissolution. Solubility is an equilibrium (thermodynamic) property — the maximum concentration that can remain dissolved at equilibrium, at given T/pH (the $K_{sp}$-controlled curves of Problem 4b). Dissolution is the kinetic PROCESS of a solid entering solution over time, controlled by surface area, agitation and driving force. Difference: solubility is a fixed equilibrium endpoint; dissolution is the time-dependent approach to that endpoint, independently fast or slow.

g) Saturation, Supersaturation, and Undersaturation. A solution is saturated when its dissolved concentration equals exactly the equilibrium solubility limit at that temperature/pH — stable equilibrium with any solid present. It is undersaturated when the dissolved concentration sits BELOW that limit — more solid could dissolve, and no precipitate can form (this is the Problem 5(a) Mn²⁺/H₂S case: $Q_{\text{Mn}} \lt K_{sp}$, undersaturated, no MnS). It is supersaturated when the dissolved concentration sits ABOVE that limit — a metastable state that spontaneously precipitates solid given a nucleation site (Problem 5(a)'s Cu²⁺/H₂S case, $Q_{\text{Cu}}\gg K_{sp}$, and Problem 5(c) once the pH is raised). Difference: saturation is the equilibrium boundary itself; undersaturation and supersaturation are the two metastable/stable sides of it, and $Q$ vs. $K_{sp}$ (the reaction quotient test used throughout Problem 5) is exactly how the three states are distinguished quantitatively.

h) Filtrate vs. Raffinate. The filtrate is the liquid that has passed through a filter medium, physically separated from suspended solids (Problem 1's S/L-separation product). The raffinate is the liquid phase LEFT BEHIND after a liquid/liquid solvent-extraction step removes the desired solute into the organic phase — the depleted aqueous phase, often recycled to leaching. Difference: filtrate results from solid/liquid separation; raffinate results from liquid/liquid separation and specifically denotes the extracted-FROM phase.

i) Ion exchange vs. Solvent extraction. Ion exchange (IX) uses a solid resin bearing fixed, exchangeable ionic groups; target ions swap with the resin's mobile counter-ions as solution percolates through a packed bed. Solvent extraction (SX) uses a water-immiscible organic liquid whose extractant selectively complexes the target ion and partitions it into the organic phase, mechanically separated in a settler. Difference: IX is a solid-phase process suited to dilute solutions with no liquid/liquid separation step; SX is a liquid/liquid process, generally preferred at higher concentrations, requiring mixer-settler equipment.

j) Electrowinning vs. Electrorefining. Electrowinning (EW) — Problem 6a/6b — deposits metal by electrolysis FROM A PURIFIED LEACH SOLUTION onto an inert or starter-sheet cathode; the anode is typically inert (Pb-alloy, Ti), evolving O2 (as assumed for the Zn cell of Problem 6b). Electrorefining (ER) purifies an already-metallic, impure anode by dissolving it electrolytically and re-depositing pure metal at the cathode, leaving insoluble impurities as anode slimes (defined in Problem 6a). Difference: EW's anode is inert and its feed is a solution; ER's anode IS the impure metal being purified and dissolves as the cell runs.