Chemical Reduction of Hexavalent Chromium with Sulfur Dioxide: 18-Env-B5 Industrial & Hazardous Waste Management, December 2018, Q10

Question 10: Chemical Reduction of Hexavalent Chromium with Sulfur Dioxide (15 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.

Quantity	Value
Hexavalent chromium concentration	650 mg/L
Flow rate	35 L/min

Find. The mass rate of SO₂ required to reduce all Cr⁶⁺ to the far less toxic Cr³⁺.

Approach. Balance the reduction reaction (Cr⁶⁺→Cr³⁺ by SO₂, SO₂ oxidized to sulfate) to get the stoichiometric mole ratio, convert to a mass ratio, then apply it to the chromium mass-loading rate.

Balance the reduction reaction. Cr⁶⁺ gains 3 electrons to become Cr³⁺; SO₂ (S⁴⁺) loses 2 electrons, oxidizing to sulfate (S⁶⁺). Equating electrons transferred (3 mol e^- per mol Cr; 2 mol e^- per mol SO₂) gives the overall balanced reaction: $$2\text{H}_2\text{CrO}_4+3\text{SO}_2\longrightarrow\text{Cr}_2(\text{SO}_4)_3+2\text{H}_2\text{O}$$ i.e. $\boxed{1.5\ \text{mol SO}_2\ \text{per mol Cr}}$ (3 mol SO₂ per 2 mol Cr).
Convert to a mass ratio. Using $MW_{Cr}=52.0$ g/mol and $MW_{SO_2}=64.06$ g/mol: $$\frac{m_{SO_2}}{m_{Cr}}=1.5\times\frac{64.06}{52.0}=\boxed{1.848\ \text{g SO}_2/\text{g Cr}}$$
Chromium mass-loading rate. From the given concentration and flow: $$\dot{m}_{Cr}=650\ \text{mg/L}\times35\ \text{L/min}=22{,}750\ \text{mg/min}=22.75\ \text{g/min}$$
SO₂ required. Apply the mass ratio from Step 2 to the chromium loading rate: $$\dot{m}_{SO_2}=22.75\ \text{g/min}\times1.848=\boxed{42.0\ \text{g/min}\ (60.5\ \text{kg/day},\ 133\ \text{lb/day})}$$

Final results
Quantity	Value
Reaction	2H₂CrO₄ + 3SO₂ → Cr₂(SO₄)₃ + 2H₂O
Stoichiometric mass ratio	1.848 g SO₂/g Cr
Cr⁶⁺ loading rate	22.75 g/min
SO₂ required	42.0 g/min (60.5 kg/day)

Check: this is the stoichiometric (theoretical) minimum SO₂ demand from the balanced electron-transfer reaction; a full-scale design would apply a safety/excess factor (commonly 1.2–1.5× stoichiometric) to ensure complete reduction given imperfect mixing and competing side reactions, and would also confirm the reaction is carried out at low pH (typically 2–3), since Cr⁶⁺ reduction by SO₂ is acid-catalyzed and proceeds far more slowly near neutral pH.

Topic: Stoichiometric chemical reduction of hexavalent to trivalent chromium

Key relations: Reduction half-reaction: $\text{Cr}^{6+}+3e^-\rightarrow\text{Cr}^{3+}$; Oxidation half-reaction: $\text{S}^{4+}(\text{SO}_2)\rightarrow \text{S}^{6+}(\text{SO}_4^{2-})+2e^-$; balanced overall reaction gives 1.5 mol SO₂ per mol Cr

Why this works: equating electrons lost by the reductant (SO₂) to electrons gained by the oxidant (Cr⁶⁺) is the general method for balancing any redox reaction and directly yields the stoichiometric reagent dose without needing to guess the molecular formula of the reaction products; Cr³⁺ is targeted specifically because it is far less toxic and mobile than Cr⁶⁺, and readily precipitates as Cr(OH)₃ for subsequent removal (the natural next treatment step, alluded to in Question 2(b)'s chemical stabilization).

Common pitfall: using a 1:1 mol ratio (one SO₂ per Cr) by analogy with a simple substitution reaction, instead of deriving the true 1.5:1 ratio from the electron-transfer balance — this understates the SO₂ dose by 33%.

Source: LaGrega, Buckingham & Evans, Hazardous Waste Management, 2nd ed.; Nemerow & Dasgupta, Industrial and Hazardous Waste Treatment, 2nd ed.

Question 10 of 11: Chemical Reduction of Hexavalent Chromium with Sulfur Dioxide

Question 10: Chemical Reduction of Hexavalent Chromium with Sulfur Dioxide (15 marks)