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 H2O molecules

where needed. Step 4: Balance the hydrogen atoms by adding H+ ions

where needed.

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.

Half Reaction - continued

Electron donor: CHCl3 (Chloroform) Electron acceptor: O2 (1/4O2 + H+ + e- → 1/2H2O)Step 1: C → CO2; N → NO3; Clx → Cl-

CHCl3 → CO2 + Cl- Step 2: CHCl3 → 1CO2 + 3Cl- Step 3: CHCl3 + 2H2O → CO2 + 3Cl-

Step 4: CHCl3 + 2H2O → CO2 + 3Cl- + 5H+ Step 5: CHCl3 + 2H2O → CO2 + 3Cl- + 5H+ + 2e-

Step 6: DoneStep 7: CHCl3 + 2H2O → CO2 + 3Cl- + 5H+ + 2e- + 1/4O2 + H+ + e- → 1/2H2O × 2Step 8: CHCl3 + 1/2O2 + H2O → CO2 + 3Cl- + 3H+

Half Reactions

CH2Cl2 (Dichloromethane or methylene chloride)

CH2Cl2 + 2H2O → CO2 + 2Cl- + 6H+ + 4 e-

1/2O2 + 2H+ + 2e- → H2O 2

CH2Cl2 + O2 → CO2 + 2H+ + 2Cl-

COD for 100 mg/L of CH2Cl2

2 mol 16 g/mol 2/(112+22+235.45)g 100 mg/L = 73.6 mg/L

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Theoretical Oxygen Demand

Determine the ThOD for glycine (CH2(NH2)COOH) using the following assumptions

1. In the first step, the organic carbon and nitrogen are converted to carbon dioxide (CO2) and ammonia (NH3), 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.

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Solution

1. Write balanced reaction for the carbonaceous oxygen demand.

CH2(NH2)COOH + 1.5O2 NH3 + 2CO2 + H2 0 2. Write balanced reactions for the nitrogenous

oxygen demand. NH3 + 1.502 HNO2 + H20 HNO2 + 0.5O2 HNO3

NH3 + 2O2 HNO3 + H2O3. Determine the ThOD.

ThOD = (1.5 + 2) mol O2/mol glycine = 3.5 mol O2/mol glycine 32 g/mol O2

43 g/mol glycine= 2.6 g O2/g glycine

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Example

A groundwater contains the following: 150 mg/L ethylene glycol; 100 mg/L phenol; 40 mg/L sulfide (S2-); 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)

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Solution

(a) COD: Ethylene glycol: C2H6O2 + 2.5 O2 2 CO2 + 3 H2O

COD = 2.532/62 150 mg/L=194 mg/L Phenol: C6H6O + 7 O2 6 CO2 + 3 H2O

COD = 732/94 100 mg/L = 238 mg/LEthylene diamine hydrate: C2H10N2O + 2.5 O2 2 CO2 + 2 H2O + 2 NH3

COD = 2.532/78 125 mg/L = 128 mg/LSulfide: S2- + 2 O2 SO4

2-

COD = 232/32 40 mg/L = 80 mg/LTotal COD: 194 + 238 + 128 + 80 = 640 mg/L

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Solution - continued

(a) TOC:Ethylene glycol:

Phenol:

Ethylene diamine hydrate:

Total TOC: 58 + 77 + 39 = 174 mg/L

mg/L 58mg/L 15062

122TOC

mg/L 77mg/L 10094

126TOC

mg/L 93mg/L 12578

122TOC

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(b) Ultimate BOD: BODult = 0.92 COD = 0.92 (194 + 238 + 80) mg/L = 471

mg/L

BOD5 = 471 mg/L 0.9 = 424 mg/L

(c) BODult in the effluent:

CODeff = 128 mg/L + 40 mg/L + residual byproducts

9.0101BOD

BOD 2.05

ult

5

L/mg 5.36101

L/mg 251.05

mg/L 40mg/L 39.70.92

mg/L 36.5

0.92

BODCOD ult

10

Solution - continued