1. Redox Titration Reactions and The Winkler Method
Topic 9.1
Happy Days video (2:30)

2. Redox titration reactions
in the titration lab so far this year…
recall if you know 2 out of 3 in one solution:
M = # moles /volume(L)
𝑴= 𝒏 𝑽
you can always solve for the unknown
then look at the balanced equation
determine the number of moles reacted with in the other solution
then solve for its unknown
this is actually called volumetric analysis

3. 𝟏 𝒗𝑨 (nA) = 𝟏 𝒗𝑩 (nB) and hence… 𝟏 𝒗𝑨 (VA x MA) = 𝟏 𝒗𝑩 (VB x MB)
from this, one can derived the following formula to solve for molarity in one step
𝟏 𝒗𝑨 (nA) = 𝟏 𝒗𝑩 (nB) and hence…
𝟏 𝒗𝑨 (VA x MA) = 𝟏 𝒗𝑩 (VB x MB)
VA = volume of reactant A
MA = molarity of reactant A
VB = volume of reactant B
MB = molarity of reactant B
va and vb are the coefficients in the balanced equation

4. 5Fe2+ + MnO4- + 8H+  5Fe3+ + Mn2+ + 4H2O
Example Problem
Consider the balanced redox reaction of potassium manganate(VII) with ammonium iron(II) sulfate.
5Fe2+ + MnO H+  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 mol dm-3 solution of iron(II) sulfate. Determine the concentration, in g dm-3, of the potassium manganate(VII) solution.

5. need answer in g dm-3, not Molarity from the periodic table
I don’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 = mole(n) ratio is 5 to n =
𝑴𝑩= .𝟎𝟎𝟎𝟎𝟓𝟎𝟎 .𝟎𝟐𝟖𝑳
𝑴𝑩= 𝟎.𝟎𝟎𝟏𝟕𝟗 mol dm-3
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
g in 1 liter of solution (1 liter H2O = 1000 ml = 1000 g)
g / 1000 g x 106
= 4.19 ppm or
= 𝒎𝒂𝒔𝒔 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒆 𝒊𝒏 𝒎𝒈 𝒗𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝒔𝒐𝒍𝒖𝒕𝒊𝒐𝒏 𝒊𝒏 𝒅𝒎𝟑
.00419g would be mg / 1 dm3
to sustain a healthy aquatic environment, dissolved oxygen should be >6 ppm

8. The Winkler method
can be used to measure the degree of pollution in a water sample
saturate a sample of water with oxygen
incubate for 5 days to allow microorganisms to oxidize the organic matter
use an iodine/thiosulfate redox titration to measure amount of remaining dissolved oxygen
the measurement is biochemical oxygen demand (BOD)
defined as the amount of O2 required by aerobic biological organisms to break down organic material present in a given water sample for 5 days at a constant temp (20°C)
measured in ppm or mg/L (same thing)

9. when organic matter is discharged into water, it provides a source of food for bacteria
when bacteria multiply and their uptake of oxygen is greater than what is replaced by photosynthesis, the body of water becomes depleted of oxygen
element
Aerobic conditions
Anaerobic conditions C
CO2
Methane (CH4) H
H2O
Hydrogen sulfide (H2S) N
NO3-
Ammonia (NH3) S
SO42-
H2S P
PO43-
phosphine (PH3)

10. Worked example measuring BOD using the Winkler method
An iodine/thiosulfate redox titration is carried out to measure the dissolved oxygen present in a water sample:
A 50 cm3 water sample is taken and saturated with oxygen for five days
The following reactions (don’t need to know) are related to the Winkler method
2 Mn2+ (aq) + 1O2 (aq) + 4OH– → 2 MnO(OH)2(s)   
1MnO(OH)2(s) + 2 I–(aq) + 4H+ → Mn2+(aq) + 1I2(aq) + 3H2O 
2 S2O32-(aq) + 1I2 (aq) → S4O62-(aq) + 2 I– (aq)
It was found that 5.25 cm3 of a mol dm-3 solution of sodium thiosulfate (Na2S2O3) was required to react with the iodine produced.

11. Comment on the BOD value obtained.
Determine the concentration of dissolved oxygen in ppm in the sample of water
calculate # moles of S2O32- and work backwards to moles of O2
2.6 x 10-5 moles S2O32- / 4 = 6.56 x 10-6 moles O2
calculate molarity of O2
1.31 x 10-4 mol dm-3
ppm is 4.19
Deduce the BOD, in ppm, of the water sample assuming the maximum solubility of oxygen in water is 9.00 ppm at 293K.
9.00 – 4.19 = 4.81 so BOD is 4.81 ppm
Comment on the BOD value obtained.
Results show reasonable water quality
untreated sewage has a BOD range of ppm.