1. Half Reaction A half reaction is either the oxidation or reduction reaction component of a redox reaction. A half reaction is obtained by considering the change in oxidation states of individual substances involved in the redox reaction. Step 1: Write the skeletons of the oxidation and reduction half-reactions. Step 2: Balance all elements other than H and O. Step 3: Balance the oxygen atoms by adding H 2 O molecules where needed. Step 4: Balance the hydrogen atoms by adding H + ions where needed.

2. Half Reaction - continued Step 5: Balance the charge by adding electrons, e -. Step 6: If the number of electrons lost in the oxidation half- reaction is not equal to the number of electrons gained in the reduction half-reaction, multiply one or both of the half- reactions by a number that will make the number of electrons gained equal to the number of electrons lost. Step 7: Add the 2 half-reactions as if they were mathematical equations. The electrons will always cancel. If the same formulas are found on opposite sides of the half-reactions, you can cancel them. If the same formulas are found on the same side of both half-reactions, combine them. Step 8: Check to make sure that the atoms and the charges balance.

3. Half Reaction - continued Electron donor: CHCl 3 (Chloroform) Electron acceptor: O 2 (1/4O 2 + H + + e - → 1/2H 2 O) Step 1: C → CO 2 ; N → NO 3 ; Cl x → Cl - CHCl 3 → CO 2 + Cl - Step 2: CHCl 3 → 1CO 2 + 3Cl - Step 3: CHCl 3 + 2H 2 O → CO 2 + 3Cl - Step 4: CHCl 3 + 2H 2 O → CO 2 + 3Cl - + 5H + Step 5: CHCl 3 + 2H 2 O → CO 2 + 3Cl - + 5H + + 2e - Step 6: Done Step 7: CHCl 3 + 2H 2 O → CO 2 + 3Cl - + 5H + + 2e - + 1/4O 2 + H + + e - → 1/2H 2 O × 2 Step 8: CHCl 3 + 1/2O 2 + H 2 O → CO 2 + 3Cl - + 3H +

4. Half Reactions CH 2 Cl 2 (Dichloromethane or methylene chloride) CH 2 Cl 2 + 2H 2 O → CO 2 + 2Cl - + 6H + + 4 e - 1/2O 2 + 2H + + 2e - → H 2 O  2 CH 2 Cl 2 + O 2 → CO 2 + 2H + + 2Cl - COD for 100 mg/L of CH 2 Cl 2 2 mol  16 g/mol  2/(1  12+2  2+2  35.45)g  100 mg/L = 73.6 mg/L 4

5. Theoretical Oxygen Demand Determine the ThOD for glycine (CH 2 (NH 2 )COOH) using the following assumptions 1. In the first step, the organic carbon and nitrogen are converted to carbon dioxide (CO 2 ) and ammonia (NH 3 ), respectively. 2. In the second and third steps, the ammonia is oxidized sequentially to nitrite and nitrate. 3. The ThOD is the sum of the oxygen required for all three steps. 5

6. Solution 1. Write balanced reaction for the carbonaceous oxygen demand. CH 2 (NH 2 )COOH + 1.5O 2  NH 3 + 2CO 2 + H 2 0 2. Write balanced reactions for the nitrogenous oxygen demand. NH 3 + 1.50 2  HNO 2 + H 2 0 HNO 2 + 0.5O 2  HNO 3 NH 3 + 2O 2  HNO 3 + H 2 O 3. Determine the ThOD. ThOD = (1.5 + 2) mol O 2 /mol glycine = 3.5 mol O 2 /mol glycine  32 g/mol O 2  43 g/mol glycine = 2.6 g O 2 /g glycine 6

7. Example A groundwater contains the following: 150 mg/L ethylene glycol; 100 mg/L phenol; 40 mg/L sulfide (S 2- ); 125 mg/L ethylene diamine hydrate (ethylene diamine is essentially nonbiodegradable). (a) Compute the COD and TOC. (b) Compute the BOD5 if the k10 is 0.2/day (BODult = 0.92 COD). (c) After treatment, the BOD5 is 25 mg/L. Estimate the COD (k10 = 0.1/day) 7

8. Solution (a) COD: Ethylene glycol: C 2 H 6 O 2 + 2.5 O 2  2 CO 2 + 3 H 2 O COD = 2.5  32/62  150 mg/L=194 mg/L Phenol: C 6 H 6 O + 7 O 2  6 CO 2 + 3 H 2 O COD = 7  32/94  100 mg/L = 238 mg/L Ethylene diamine hydrate: C 2 H 10 N 2 O + 2.5 O 2  2 CO 2 + 2 H 2 O + 2 NH 3 COD = 2.5  32/78  125 mg/L = 128 mg/L Sulfide: S 2- + 2 O 2  SO 4 2- COD = 2  32/32  40 mg/L = 80 mg/L Total COD: 194 + 238 + 128 + 80 = 640 mg/L 8

9. Solution - continued (a) TOC: Ethylene glycol: Phenol: Ethylene diamine hydrate: Total TOC: 58 + 77 + 39 = 174 mg/L 9

10. (b) Ultimate BOD: BOD ult = 0.92  COD = 0.92  (194 + 238 + 80) mg/L = 471 mg/L BOD 5 = 471 mg/L  0.9 = 424 mg/L (c) BOD ult in the effluent: COD eff = 128 mg/L + 40 mg/L + residual byproducts 10 Solution - continued