Topic 9.1 - Oxidation and reduction Flashcards
Oxidation
Gains oxygen
Loses electrons
Increase in oxidation number
Reduction
Loses oxygen (addition of hydrogen)
Gains electrons
Decrease in oxidation number
Oxidising agent
The oxidising agent oxidises the other reactant and is often reduced
Reducing agent
The reducing agent reduces the other reactant and is often oxidised
Half equations
Fe²⁺ + MnO₄⁻ -> Fe³⁺ + Mn²⁺
Step 1) Assign oxidation states
Fe²⁺ = +2 Mn = +7 O₄ = -8
Step 2) Deduce who is being oxidised/reduced
Oxidised = iron Reduced = Mn
Step 3) State half equations
Fe²⁺ -> Fe³⁺ e⁻
MnO₄ + 5e⁻ -> Mn²⁺
Step 4) Balance equations so electrons are equal
5Fe²⁺ -> 5Fe³⁺ 5e⁻
MnO₄ + 5e⁻ -> Mn²⁺
Step 5) Write redox equation
5Fe²⁺ + MnO₄ -> 5Fe³⁺ + Mn²⁺
Step 6) Check total charge on reactants/products (incl O₂)
reactant = 9 products = 17
Step 7) Add H⁺ and H₂O to sides needing them
5Fe²⁺ + MnO₄ +8H⁺ -> 5Fe³⁺ + Mn²⁺ + 4H₂O
Reactive series
More reactive will displace less reactive.
Zn (s) + Cu²⁺ (aq) -> Zn²⁺ (aq) + Cu (s)
Chlorine and ozone purposes
Chlorine is a good antibacterial and antiseptic as Cl₂ (chlorine), NaOCl (sodium hypochlorite), and CaO(Cl)₂ (calcium hypochlorite) can be added to water to form HOCl (hypochlorous acid) which kills pathogens.
Ozone (O₃) kills pathogens.
Chlorine -> Kills viruses and is cheap to produce, but leaves a residual taste and unpleasant odour, may form toxic side products (ie CHCl₃ (trichloromethane/chloroform)) which are often carcinogenic.
Ozone -> doesn’t kill viruses, expensive, but leaves no residual taste or odour and forms less toxic side products than chlorine reactions do
The Winkler method
Used to measure the amount of dissolved oxygen in water
Oxygen’s solubility is temperature dependent: as temp increases, solubility decreases.
Oxygen is non-polar and water is polar
BOD
Amount of oxygen used by aerobic bacteria in water to decompose organic matter over a period of 5 days
Biochemical oxygen demand is the degree of organic pollution in a sample water.
Pure water is <1 ppm
Concentration (ppm)
Mass of component in solution (mg) / volume of sol (dm³)
BOD equation
BOD = final conc O₂ - initial conc O₂
Working out moles of oxygen in Winkler methods
If one mole of sodium thiosulphate is titrated, what would the number of moles of dissolved oxygen be?
2Mn(OH)₂ + O₂ -> 2MnO(OH)₂
MnO(OH)₂ + 4H⁺ -> Mn⁴⁺ + 3H₂O
Mn⁴⁺ + 2I⁻ -> Mn²⁺ + I₂
2S₂O₃²⁻ + I₂ -> S₄O₆²⁻ + 2I⁻
Thiosulfate = S₂O₃²⁻
2Mn(OH)₂ + O₂ -> 2MnO(OH)₂
MnO(OH)₂ + 4H⁺ -> Mn⁴⁺ + 3H₂O
Mn⁴⁺ + 2I⁻ -> Mn²⁺ + I₂
2S₂O₃²⁻ + I₂ -> S₄O₆²⁻ + 2I⁻
1 mole 2S₂O₃²⁻- reacts with 0.5 moles I₂
- 5 mole I₂ formed by 0.5 mol Mn⁴⁺
- 5 mole Mn⁴⁺ formed by 0.5 mol MnO(OH)₂
- 5 mole MnO(OH)₂ formed by 0.25 moles O₂
Calculating the concentration of dissolved oxygen in water in ppm
100cm³ water used, formed iodine titrated with 5 x 10⁻³ mol dm⁻³ conc. 16.00 cm³ required
2Mn(OH)₂ + O₂ -> 2MnO(OH)₂
MnO(OH)₂ + 4H⁺ -> Mn⁴⁺ + 3H₂O
Mn⁴⁺ + 2I⁻ -> Mn²⁺ + I₂
2S₂O₃²⁻ + I₂ -> S₄O₆²⁻ + 2I⁻
Another sample of the same water was incubated at 20°C for 5 days. Winkler method found to be 2.20 mg dm⁻³ dissolved oxygen.
Find BOD, is it polluted?
Step 1:
Calculate thiosulfate moles:
moles = volume (dm3) x conc moles = 0.016 x 5x10⁻³ moles = 8.00 x10⁻⁵ mol
Step 2: calculate iodine moles moles is a 2:1 ratio so moles = 8x10⁻⁵/2 moles = 4.00 x10-5
Step 3:
Calculate Mn⁴⁺ moles
moles is a 1:1
moles = 4x10⁻⁵
Step 4: MnO(OH)₂ to Mn⁴⁺ is 1:1 MnO(OH)₂ to O₂ is 2:1 moles O₂ = 4x10-5/2 moles = 2x10⁻⁵
Step 5: Work out mass of O₂ mass = moles x Mr mass = 2x10-5 x 32.00 mass = 6.4x10-3 g dm⁻³ mass = 6.4 mg dm⁻³ 6.40 ppm
Step 6:
Find BOD:
6.40 - 2.20 = 4.20
4.20 is <5 so it’s unpolluted
Typical BOD values
Pure water = <1 ppm
Untreated sewage = 350ppm
Brewery effluent = 500ppm
Abattoir water = 3000ppm
Typical oxidised elements and their products (C, H, N, S, P)
Element - aerobic product - anaerobic product
Carbon - CO₂ - CH₄
Hydrogen - H₂O - CH₄/NH₃/H₂S/PH₃
Nitrogen - NO₃⁻ - NH₃
Sulfur - SO₄²⁻ - H₂S
Phosphorus - PO₄³⁻ - PH₃
