Lead recovery from battery scrap by chloride leaching and electrowinning: 21-Mat-B2 Pyrometallurgy, May 2015, Q4

Problem 4 — Lead recovery from battery scrap by chloride leaching and electrowinning (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.

1. Reactions for each step.

(a) Leaching – chloride-complexation-assisted metathesis converts the insoluble sulphate to a soluble chloro-complex: $$\text{PbSO}_4(\text{s})+4\text{Cl}^-(\text{aq})\rightarrow \text{PbCl}_4^{2-}(\text{aq})+\text{SO}_4^{2-}(\text{aq}).$$ (Written more simply as $\text{PbSO}_4(\text{s})+2\text{Cl}^-\rightarrow \text{PbCl}_2(\text{aq})+\text{SO}_4^{2-}$ if chloro-complexation is not carried further.)

(b) Purification – lime and HCl together neutralize and precipitate the sulphate as gypsum while regenerating chloride for recycle: $$\text{Ca(OH)}_2(\text{s})+2\text{HCl}(\text{aq})+\text{SO}_4^{2-}(\text{aq})\rightarrow \text{CaSO}_4\!\cdot\!2\text{H}_2\text{O}(\text{s})\downarrow\ +\ 2\text{Cl}^-(\text{aq}).$$ Ca(OH)2 neutralizes the HCl's acidity (keeping pH neutral) while the liberated Ca2+ precipitates the sulphate as gypsum, and the chloride is regenerated in solution for reuse.

(c) Electrowinning – cathodic Pb deposition paired with anodic oxygen evolution (per the question's own statement that "metallic lead (and oxygen)" are produced): $$\text{Cathode: } \text{PbCl}_4^{2-}+2e^-\rightarrow \text{Pb(s)}+4\text{Cl}^-,\qquad \text{Anode: } \text{H}_2\text{O}\rightarrow \tfrac12\text{O}_2(\text{g})+2\text{H}^++2e^-.$$ Overall: $$\text{PbCl}_4^{2-}+\text{H}_2\text{O}\rightarrow \text{Pb(s)}+4\text{Cl}^-+2\text{H}^++\tfrac12\text{O}_2(\text{g}).$$

2. Flowsheet.

Fig. 3 — lead-from-battery-scrap flowsheet: neutral NaCl-brine leach, lime/HCl purification (gypsum by-product), filtration, and electrowinning (Pb cathode, O2 anode), with the regenerated chloride recycled back to leaching.

3-4. Electrowinning potentials (Given/Find/Approach for the calc parts).

Given. $E^\circ(\text{Pb}^{2+}/\text{Pb})=-0.13$ V; $E^\circ(\tfrac12\text{Cl}_2/\text{Cl}^-)=+1.36$ V; pH = 7 (does not enter either half-reaction directly, since neither involves H⁺); assumed "reasonable" concentrations, stated explicitly per the question's own instruction: $[\text{Pb}^{2+}]=0.1$ M (typical electrowinning bath concentration) and $[\text{Cl}^-]=5$ M (a near-saturated NaCl brine, consistent with "brine solution" in part (a)).

Find. (3) the minimum cathode potential to deposit Pb; (4) the maximum anode potential before Cl2 gas evolves.

Approach. Apply the Nernst equation to each of the two GIVEN half-reactions separately, at 25°C ($\frac{RT}{F}\ln 10=0.0592$ V).

(3) Minimum electrowinning (cathode) potential. Pb deposits once the cathode is driven to (or below) its Nernst-corrected equilibrium potential: $$E_{\text{Pb}^{2+}/\text{Pb}}=E^\circ+\dfrac{0.0592}{2}\log[\text{Pb}^{2+}]=-0.13+\dfrac{0.0592}{2}\log(0.1)=-0.13-0.030=\boxed{-0.160\ \text{V vs. SHE}}.$$ Any cathode potential at or more negative than $-0.160$ V will electrowin lead.
(4) Maximum electrowinning (anode) potential to avoid Cl2. Cl2 gas begins to evolve once the anode reaches the Nernst-corrected $E(\text{Cl}_2/\text{Cl}^-)$ at 1 atm Cl2: $$E_{\text{Cl}_2/\text{Cl}^-}=E^\circ-0.0592\log[\text{Cl}^-]=1.36-0.0592\log(5)=1.36-0.041=\boxed{+1.32\ \text{V vs. SHE}}.$$ Keeping the anode BELOW this value (e.g. near the O2-evolution potential of ≈0.81 V at pH 7, well below 1.32 V) is why the process normally produces O2, not Cl2, at the anode — a substantial safety margin that a concentrated brine narrows slightly (concentrated Cl^- LOWERS the Cl2 threshold from 1.36 V toward 1.32 V) but does not eliminate.

Quantity	Result
Minimum electrowinning (cathode) potential, $[\text{Pb}^{2+}]=0.1$ M	$\boxed{-0.160\ \text{V vs. SHE}}$
Maximum electrowinning (anode) potential to avoid Cl2, $[\text{Cl}^-]=5$ M	$\boxed{+1.32\ \text{V vs. SHE}}$

5. Why electrowinning may work poorly under very acidic conditions, and the needed process controls. At very low pH, the competing hydrogen-evolution half-reaction $\text{H}^++e^-\rightarrow\tfrac12\text{H}_2$ ($E=-0.0592\times\text{pH}$) becomes progressively LESS negative (rises toward 0 V) as pH falls, moving closer to (and potentially above) the $-0.160$ V needed to deposit Pb. Once $E(\text{H}^+/\text{H}_2)$ exceeds $E(\text{Pb}^{2+}/\text{Pb})$, hydrogen co-evolves at the cathode, wasting current on gas production instead of metal deposition and lowering the process's current efficiency — exactly the same competing-reaction mechanism analyzed quantitatively for nickel in Problem 5. Very acidic conditions can also destabilize the chloro-lead complex and encourage local re-precipitation of solid PbCl2 if chloride activity shifts. Process controls: (i) buffer/control the electrolyte pH near neutral, using the SAME lime addition already used in purification (electrowinning itself generates H⁺ at the anode via O2 evolution, so the bath acidifies over time unless continuously neutralized); (ii) maintain an adequately high, well-controlled $[\text{Pb}^{2+}]$ to keep the Pb deposition potential comfortably above the H2-evolution line; (iii) monitor and control chloride concentration to preserve the safety margin against Cl2 evolution identified in part 4; (iv) consider a diaphragm/membrane cell to isolate anolyte/catholyte pH swings.

Check. Two "reasonable value" assumptions are load-bearing, per the question's own instruction to state them: $[\text{Pb}^{2+}]=0.1$ M (typical industrial electrowinning bath concentration) and $[\text{Cl}^-]=5$ M (near-saturated brine, consistent with the leaching step's own brine description). Neither the Pb2+/Pb nor the Cl2/Cl- half-reaction involves H⁺, so the stated pH = 7 is process CONTEXT (the operating point of the whole bath) rather than a term that enters either Nernst calculation directly.

Concept & Theory

Topic. Electrowinning thermodynamics: cathode deposition potential vs. anode gas-evolution potential, and current-efficiency loss from a competing reaction.

Key relations. Nernst equation $E=E^\circ+\frac{0.0592}{n}\log Q$; current efficiency $=$ (charge to desired product)/(total charge passed).

Why this works. Each electrode's actual operating potential must sit between its own thermodynamic threshold and that of the next-easiest competing reaction at that electrode; the "window" between the desired reaction and the next competing one (Pb vs. H2 at the cathode, O2 vs. Cl2 at the anode) is exactly what a real electrowinning cell must be designed and controlled to stay inside.

Common pitfall. Forgetting that pH does not appear in a Nernst equation unless H⁺/OH^- is a REACTANT OR PRODUCT of that specific half-reaction; here neither given half-reaction involves H⁺, so pH = 7 is context, not a term to substitute.

Source. Habashi, A Textbook of Hydrometallurgy, Ch. 9 (electrowinning); Free, Hydrometallurgy, Ch. 6 (electrochemistry of aqueous processing).

Question 4 of 6: Lead recovery from battery scrap by chloride leaching and electrowinning

Problem 4 — Lead recovery from battery scrap by chloride leaching and electrowinning (20 marks)