This document discusses oxidation-reduction (redox) reactions and their environmental applications. It defines key terms like: 1) Biochemical oxygen demand (BOD) which measures the amount of oxygen used by microorganisms to break down organic waste in water, with higher BOD indicating more pollution. 2) Parts per million (ppm) which is used to measure very dilute solutions like dissolved oxygen in water, important for sustaining aquatic life. 3) The Winkler titration method for determining dissolved oxygen concentration in water samples, which involves a series of redox reactions and titration with sodium thiosulfate.

1. 9.1
OXIDATION AND REDUCTION
By: Merinda Sautel
Alameda Int’l Jr/Sr High School
Lakewood, CO
msautel@jeffco.k12.co.us

2. Topic 9.1
Happy Days video
(2:30)

3. Example Problem
 Consider the balanced redox reaction of
potassium manganate(VII) with ammonium
iron(II) sulfate.
5Fe2+ + MnO4
- + 8H+  5Fe3+ + Mn2+ + 4H2O
notice spectator ions are left out of the equation and only
focusing what was reduce and oxidized
 In a titration to determine the concentration
of a potassium manganate(VII) solution, 28.0
cm3 of the potassium manganate(VII) reacted
completely with 25.0 cm3 of a 0.0100 mol/dm-
3 solution of iron(II) sulfate. Determine the
concentration, in g dm-3, of the potassium
manganate(VII) solution.

4. on’t want to use the previous formula because I already know how to do this and don’t want to
learn a new formula.
𝑴𝑨 =
𝒏
𝑽
… look at balanced equation…. 𝑴𝑩 =
𝒏
𝑽
𝟎. 𝟎𝟏𝟎𝟎𝑴 =
𝒏
.𝟎𝟐𝟓𝟎𝑳
…mole(n) ratio… 𝑴𝑩 =
𝒏
.𝟎𝟐𝟖𝟎𝑳
n = .000250 mole(n) ratio is 5 to 1 n = .0000500
𝑴𝑩 =
.𝟎𝟎𝟎𝟎𝟓𝟎𝟎
.𝟎𝟐𝟖𝑳
𝑴𝑩 = 𝟎. 𝟎𝟎𝟏𝟕𝟗
need answer in g dm-3, not Molarity
𝟎. 𝟎𝟎𝟏𝟕𝟗 𝐦𝐨𝐥 𝐊𝐌𝐧𝐎𝟒 𝒙
𝟏𝟓𝟖. 𝟎𝟒 𝒈 𝐊𝐌𝐧𝐎𝟒
𝟏 𝒎𝒐𝒍 𝐊𝐌𝐧𝐎𝟒
0.283 g dm -3
from the
periodic
table

6. Enviromental application of redox
chemistry
 aquatic life depends on CO2 an O2
dissolved in water
 O2 is non-polar, while H20 is polar
› therefore, solubility of oxygen in water is
very low
 it decreases with increase in temperature
 0°C is 14.6 ppm (parts per million)
 20°C is 7.6 ppm (parts per million)

7.  ppm is used for very dilute solutions
 =
𝒎𝒂𝒔𝒔 𝒐𝒇 𝒄𝒐𝒎𝒑𝒐𝒏𝒆𝒏𝒕 𝒊𝒏 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒈𝒓𝒂𝒎𝒔
𝒕𝒐𝒕𝒂𝒍 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒈𝒓𝒂𝒎𝒔
x 106
 .00419 g in 1 liter of solution (1 liter H2O = 1000 ml = 1000 g)
 .00419 g / 1000 g x 106
 = 4.19 ppm
 to sustain a healthy aquatic environment,
dissolved oxygen should be >6 ppm

8. BOD
 Biochemical oxygen demand (also
called biological oxygen demand), or
BOD for short, is used as a measure of
the quality of water. It is a measure of
the amount of oxygen used by
microorganisms to oxidise the organic
matter in the water.

9.  Any organic pollutants in river water will
be decomposed (oxidised) by
microorganisms (aerobic bacteria) in the
water and this process uses up
dissolved oxygen.
 The higher the BOD, the more organic
waste there is in water.

10.  If, for instance, sewage is released into
a river or lake this will greatly increase
the BOD – the water is more polluted.
 BOD is defined as the amount of oxygen
used by the aerobic microorganisms in
water to decompose the organic matter
in the water over a fixed period of time
(usually 5 days) at a fixed temperature
(usually 20 °C).

11.  The basic principle in measuring BOD is
to compare the initial amount of
dissolved oxygen in a sample of water
with the amount present when the
sample has been incubated for 5 days at
20 °C.
 If a dissolved oxygen concentration of 9
ppm, which after incubation for 5 days
falls to 4 ppm, the BOD is 9 − 4, or 5
ppm.

12. A typical method for determining the
amount of dissolved oxygen is the
Winkler titration method.
 1. The basic chemistry behind the Winkler
method is that manganese(II) sulfate is
added to the water and the manganese(II)
ions are oxidised
 2. Manganese(II) sulfate is converted to
manganese(II) hydroxide in the presence of
hydroxide ions.

13.  The sample is acidified with sulfuric acid to
produce manganese(IV) sulfate:
 There is no change in oxidation number in
this reaction.
 Iodide ions are oxidised to I2 by the
manganese(IV) ions:

14.  This iodine can then be titrated against a
standard sodium thiosulfate solution:
 The outcome from these equations is that
the number of moles of dissolved oxygen is
1/4 of the number of moles of sodium
thiosulfate used in the titration – or the
mass of oxygen is eight times the number
of moles of sodium thiosulfate.