IB Chemistry on Redox Titration, Biological Oxygen Demand and Redox.

1. http://lawrencekok.blogspot.com
Prepared by
Lawrence Kok
Tutorial on Redox Titration, BOD and Biological
Oxygen Demand using Winkler method

2. Titration
Redox TitrationAcid Base Titration
Primary standard acids
- Potassium hydrogen phthalate
Primary standard bases
- Anhydrous sodium carbonate
10.6 g Na2CO3
Standard 0.1M Na2CO3
10.6g in 1 L
Volumetric Burette
Accurate
known conc
Unable to prepare accurate conc of NaOH/HCI due to
•Hygroscopic nature NaOH – Absorb water vapour
•HCI in vapour state – Difficult to measure amt
VolumetricBurette
Standard 0.1M KHP
20.4 g KHP20.4 g in 1L
Unknown
Conc NaOH
Unknown
Conc HCI
? ?
Standardize NaOH
using KHP
Standardize H
CI
using Na
2 CO
3
Accurate
known conc
Acid/Base Titration Redox Titration
Neutralization bet acid/base Redox bet oxidizing/reducing agent
Transfer proton/H+
from acid to base Transfer elec from reducing to oxidizing agent
Indicator for colour change No indicator needed
Acid Base Titration
- One reactant – must be standard (known conc) or capable being standardised
- Equivalent point – equal amt neutralize each other
- End point measurable/detectable by colour change (indicator), pH change /conductivity

3. Oxidizing
Agent
Reducing
Agent
MnO4
-
Fe2+
Cr2O7
2-
SO2
HNO3 I-
H2O2 H2S
CI2 SO3
2-
KIO3 Vitamin C
OCI-
/Cu2+
Oxalate/
C2O4
2-
Titration
Redox TitrationAcid Base Titration
Burette/Titrant
Oxidizing agent
?
Acid/Base Titration Redox Titration
Neutralization bet acid/base Redox bet oxidizing/reducing agent
Transfer proton/H+
from acid to base Transfer elec from reducing to oxidizing agent
Indicator for colour change No indicator needed
Redox Titration
- One reactant – must be standard (known conc) or capable being standardised
- Reaction bet Oxidizing agent/Titrant with Reducing agent/Analyte
- Titrant of known concentration
- Stoichiometrically equivalent amt titrant/titrand added
- No indicator needed. Detectable by colour change of Oxidizing/Reducing agent
Analyte/reducing agent
Titrand Redox Titration used to determine:
-Amount of copper in brass
-Amount Fe/iron in iron pill/food
-Amount H2O2 commercial peroxide solution
-Amount OCI -
/hypochlorite/CI2 in bleach
-Amount Vitamin C
-Amount Dissolve oxygen content/BOD
-Amount ethanol in beer/wine
-Amount oxalate acidAnalyte
to
be
determ
ine
?
MnO4
-
+ 5Fe2+
+ 8H+
Mn→ 2+
+ 5Fe3+
+ 4H2O Cr2O7
2-
+ 6Fe2+
+ 14H+
2Cr→ 3+
+ 6Fe3+
7H2O
Iron determination using MnO4
-
/ Cr2O7
2-
purple colourless orange green
add MnO4
-
till endpoint
↓
turn purple (excess MnO4
-
)
add Cr2O7
2-
till endpoint
↓
turn orange (excess Cr2O7
2-
)

4. Redox Titration Calculation- % Iron in iron tablet
Iron tablet contain hydrated iron (II) sulphate (FeSO4.7H2O). One tablet weighing 1.863g
crushed, dissolved in water and solution made up to total vol of 250ml. 10ml of this
solution added to 20ml of H2SO4 and titrated with 0.002M KMnO4. Average 24.5ml need
to reach end point. Cal % iron(II) sulphate in iron tablet.
1
10ml transfer
20ml acid added
1.863 g
250ml
KMnO4
M = 0.002M
V = 24.5 ml
Fe2+
M = ?
V = 30ml
MnO4- + 5Fe2+
+ 8H+
Mn→ 2+
+ 5Fe2+
+ 4H2O
M = 0.002M M = ?
V = 24.5ml
Mole ratio – 1: 5
Using mole ratio
Mole KMO4
-
= MV
= (0.002 x 0.0245)
= 4.90 x 10-5
Mole ratio (1 : 5)
• 1 mole KMO4
-
react 5 mole Fe2+
• 4.90 x 10-5
KMO4
-
react 2.45 x 10-4
Fe2+
M V = 1
M V 5
0.002 x 0.0245 = 1
Moles Fe2+
5
Moles = 2.45 x 10-4
Fe2+
Mass of (expt yield) = 1.703g
Mass of (Actual tablet) = 1.863g
% Fe in iron tablet = 1.703 x 100%
1.863
= 91.4%
Mole  Mass
Mole x RMM = Mass FeSO4
6.125 x 10-3
x 278.05 = 1.703g FeSO4
Using formula
10ml sol contain - 2.45 x 10-4
Fe2+
250ml sol contain - 250 x 2.45 x 10-4
Fe2+
10
= 6.125 x 10-3
mole Fe2+
FeSO4.7H2O  FeSO4 + 7H2O
1 mol  1 mol + 7 mol
FeSO4  Fe2+
+ SO4
2-
1 mol  1mol + 1mol
6.125 x 10-3
mol  6.125 x 10-3
mole Fe2+
1
2
3
4
Video on % Iron in iron tablet
Video on Fe2+
/KMnO4 titration calculation

5. Redox Titration Calculation- % Iron in iron tablet
One iron tablet weighing 2.00g crushed, dissolved in water/acid to convert it to Fe2+
and
solution titrated with 0.100M KMnO4. Average 27.5ml KMnO4 needed to reach end point.
Cal mass of iron and % iron in iron tablet. How equivalent point is detected ?
2
iron solution titrated
2.000 g
KMnO4
M = 0.100M
V = 27.5 ml
Fe2+
M = ?
1MnO4- + 5Fe2+
+ 8H+
Mn→ 2+
+ 5Fe2+
+ 4H2O
M = 0.100M M = ?
V = 27.5ml
Mole ratio – 1: 5
Using mole ratio
Mole KMO4
-
= MV
= (0.100 x 0.0275)
= 0.00275
Mole ratio (1 : 5)
• 1 mole KMO4
-
react 5 mole Fe2+
• 0.00275 KMO4
-
react 0.01375Fe2+
M aVa = 1
Mb Vb 5
0.100 x 0.0275 = 1
Moles Fe2+
5
Moles = 0.01375 mol Fe2+
Mass of (expt yield) = 0.7679g
Mass of (Actual tablet) = 2.000g
% Fe in iron tablet = 0.7679 x 100%
2.000
= 38.4 %
Mole  Mass
Mole x RMM = Mass Fe
0.01375 x 55.85 = 0.7679g Fe
Using formula
1
2
3
Video on % Iron in iron tablet
Video on Fe2+
/KMnO4 titration calculation
MnO4
-
– In burette is purple – Turns colourless react with Fe2+
All Fe2+
used up at equivalence point – excess KMnO4
-
turn purple
4

6. 1 mol 1 mol
1OCI-
+ 2I-
+ 2H+
I→ 2 + 1CI-
+ H2O
I2 + 2S2O3
2-
S→ 4O6
2-
+ 2I-
1 mol 2 mol
10.0ml bleach (OCI -
) diluted to total vol of 250ml. 20.0ml is
added to 1g of KI (excess) and iodine produced is titrated with
0.0206M Na2S2O3.Using starch indicator, end point was 17.3ml.
Cal molarity of OCI-
in bleach.
Redox Titration Calculation – OCI-
in Bleach
3
Na2S2O3
M = 0.0206M
V = 17.3ml
I2
M = ?
2S2O3
2-
+ I2 S→ 4O6
2-
+ 2I-
M = 0.0206 Mole = ?
V = 17.3ml V = 0.02
Mole ratio (1 : 2)
1 mole OCI-
: 1 mole I2 : 2 mole S2O3
2-
1 mole OCI-
2 mole S2O3
2-
10.0ml OCI-
transfer
V = 250ml
M = 8.9 x 10-3
M
20ml transfer
1g KI excess
added
Mole S2O3
2-
= MV
= (0.0206 x 0.0173)
= 3.56 x 10-4
Mole ratio (2 : 1)
• 2 mole S2O3
2-
react 1 mole I2
• 3.56 x 10-4
S2O3
2--
react 1.78 x 10-4
I2
Mole ratio – 2: 1
1OCI-
+ 2I-
+ 2H+
I→ 2 + 1CI-
+ H2O
1CIO-
I2
Mole = ? Mole = 1.78 x 10-4
Mole ratio – 2: 1
Mole ratio (1 : 1)
• 1 mole OCI-
1 mole I2
• 1.78 x 10-4
OCI-
1.78 x 10-4
I2
Moles of OCI-
= M x V
M x V = 1.78 x 10-4
M x 0.02 = 1.78 x 10-4
M = 1.78 x 10-4
002
M = 8.9 x 10-3
M diluted 25x
Mole bef dilution = Mole aft dilution
M1 V1 = M2V2
M1 = Ini molarity M2
= Final molarity
V1
= Initial vol V2
= Final vol
M1 V1 = M2 V2
M1 x 10 = 8.9 x 10-3
x 250
M1 = 8.9 x 10-3
x 250
10
M1 = 0.222M
Diuted 25x
V = 10
M = ?
titrated
Water added
till 250ml
1
Using direct formula
M V(OCI+
) = 1 = 1
M V(S203
2-
) 2 2
Moles of OCI+
= 1
0.0206 x 0.0173 2
Moles of OCI-
= 1.78 x 10-4
2
3
4
5
6
Hypochlorous acid = bleach
Oxidizing agent = OCI-
Iodometric titration
I2/thiosulphate/starch
↓
I -
oxidized by OA to I2
↓
I2 react with starch
(blue black colour)
↓
S2O3
2-
added to reduce I2
↓
I2 used up – blue black
disappear
2I-
+ OCI-
↔ I2 + CI-
2S2O3
2-
+ I2 ↔S4O6
2-
+ 2I-

7. 1 mol 1 mol
1OCI-
+ 2I-
+ 2H+
I→ 2 + CI-
+ H2O
I2 + 2S2O3
2-
S→ 4O6
2-
+ 2I-
1 mol 2 mol
10.0ml bleach (OCI-
) react with KI (excess), iodine produced is titrated
with 0.020M Na2S2O3.Using starch indicator, end point was 38.65 ml.
Cal molarity of OCI-
in bleach.
Redox Titration Calculation – OCI-
in Bleach
4
Na2S2O3
M = 0.020M
V = 38.5 ml
I2
M = ?
2S2O3
2-
+ I2 S→ 4O6
2-
+ 2I-
M = 0.020 Mole = ?
V = 38.55ml
Mole ratio ( 1 : 2)
1 mole OCI-
: 1 mole I2 : 2 mole S2O3
2-
1 mole OCI-
2 mole S2O3
2-
10ml bleach
transfer
1g KI excess
added
Mole S2O3
2-
= MV
= (0.020 x 0.03865)
= 7.73 x 10-4
Mole ratio (2 : 1)
• 2 mole S2O3
2-
react 1 mole I2
• 7.73 x 10-4
S2O3
2--
react 3.865 x 10-4
I2
Mole ratio – 2: 1
1OCI-
+ 2I-
+ 2H+
I→ 2 + 2CI-
+ H2O
1OCI-
I2
Mole = ? Mole = 3.865 x 10-4
Mole ratio – 1: 1
Mole ratio (1 : 1)
• 1 mole OCI-
1 mole I2
• 3.865 x 10-4
OCI-
3.865 x 10-4
I2
M x V = Moles OCI-
M x 10 = 3.865 x 10 -4
1000
M = 0.0387M
titrated
1
Using direct formula
M V(OCI+
) = 1 = 1
M V(S203
2-
) 2 2
Moles of OCI+
= 1
0.020 x 0.03865 2
Moles of OCI-
= 3.5865 x 10-4
2
3
4
5
Video on OCI-
in bleach
Sample OCI-
calculation. Click here to view
Conc OCI-
Hypochlorous acid = bleach
Active oxidizing agent = OCI-
Iodometric titration
I2/thiosulphate/starch
↓
I -
oxidized by OA to I2
↓
I2 react with starch
(blue black colour)
↓
S2O3
2-
added to reduce I2
↓
I2 used up – blue black
disappear
2I-
+ OCI-
↔ I2 + CI-
2S2O3
2-
+ I2 ↔S4O6
2-
+ 2I-

8. 2 mol 1 mol
2Cu2+
+ 4I-
I→ 2 + 2CuI
I2 + 2S2O3
2-
S→ 4O6
2-
+ 2I-
1 mol 2 mol
2.5g brass react with 10ml conc HNO3 producing Cu2+
ions. Solution made up to 250ml using water in a
volumetric flask. Pipette 25ml of solution into conical flask. Na2CO3 added to neutralize excess acid. 1g KI
(excess) and few drops of starch added to flask. Titrate with 0.1M S2O3
2-
and end point, reached when 28.2ml
added. Find molarity copper ions and % copper found in brass.
Redox Titration Calculation - % Cu in Brass
5
Na2S2O3
M = 0.1M
V = 28.2ml
I2
M = ?
2S2O3
2-
+ I2 S→ 4O6
2-
+ 2I-
M = 0.1M Mole = ?
V = 28.2ml
Mole ratio ( 1 : 1)
2 mole Cu2+
: 1 mole I2 : 2 mole S2O3
2-
2 mole Cu2+
2 mole S2O3
2-
Pour into
Volumetric flask
V = 250ml
M = ?
25ml transfer
1g KI excess + starch
added
Mole S2O3
2-
= MV
= (0.1 x 0.0282)
= 2.82 x 10-3
Mole ratio (2 : 1)
• 2 mole S2O3
2-
react 1 mole I2
• 2.82 x 10-3
S2O3
2--
react 1.41 x 10-3
I2
Mole ratio – 2: 1
2Cu2+
+ 4I-
I→ 2 + 2CuI
Mole = ? 1.41 x 10-3
I2 Mole ratio – 2: 1
Mole ratio (2 : 1)
• 2 mole Cu2+
1 mole I2
• 2.82 x 10-3
Cu2+
1.41 x 10-3
I2
Mole of Cu2+
= M x V
M x V = 2.82 x 10-3
M x 0.025 = 2.82 x 10-3
M = 2.82 x 10-3
0.025
M = 1.13 x 10-1
M
Mass Cu = Molarity Cu x RAM
Mass Cu = (0.113 x 63.5)g Cu in 1000ml
= 7.18g  Cu in 1000ml
= 1.79g  Cu in 250ml
10 ml HNO3
titrated
Water added
till 250ml
2.5g brass
% Cu in brass = mass Cu x 100%
mass brass
= 1.79 x 100%
2.5
= 71.8%
Using formulaUsing mole ratio
Using formula
M V(Cu2+
) = 2 = 1
MV(S203
2-
) 2 1
Moles of Cu2+
= 1
0.1 x 0.0282 1
Moles of Cu2+
= 2.82 x 10-3
1
2
3
4
5
6
Iodometric titration
I2/thiosulphate/starch
↓
I -
oxidized by OA to I2
↓
I2 react with starch
(blue black colour)
↓
S2O3
2-
added to reduce I2
↓
I2 used up – blue black
disappear
4I-
+ 2Cu+
↔ I2 + 2CuI
2S2O3
2-
+ I2 ↔S4O6
2-
+ 2I-
Click here here for copper determination

9. 2 mol 1 mol
2Cu2+
+ 4I-
I→ 2 + 2CuI
I2 + 2S2O3
2-
S→ 4O6
2-
+ 2I-
1 mol 2 mol
Brass is a copper alloy. Analysis carried out to determine copper.
Iodometric titration was performed.
Step 1 : Cu + 2HNO3 + 2H+
Cu→ 2+
+ 2NO2 + 2H2O
Step 2 : 4I-
+ 2Cu2+
2CuI + I→ 2
Step 3 : I2 + 2S2O3
2-
2I→ -
+ S4O6
2-
Average vol S2O3 2- was 28.50ml.
Redox Titration Calculation - % Cu in Brass
6
Na2S2O3
M = 0.1M
V = 28.5ml
I2
M = ?
V = 100ml
2S2O3
2-
+ I2 → S4O6
2-
+ 2I-
M = 0.1M Mole = ?
V = 28.5ml
Mole ratio ( 1 : 1)
2 mole Cu2+
: 1 mole I2 : 2 mole S2O3
2-
2 mole Cu2+
2 mole S2O3
2-
V = 100ml
M = ?
1g KI excess/starch
added
Mole S2O3
2-
= MV
= (0.1 x 0.0285)
= 2.85 x 10-3
Mole ratio (2 : 1)
• 2 mole S2O3
2-
react 1 mole I2
• 2.85 x 10-3
S2O3
2--
react 1.41 x 10-3
I2
Mole ratio – 2: 1
2Cu2+
+ 4I-
I→ 2 + 2CuI
Mole = ? 1.41 x 10-3
I2
Mole ratio – 2: 1
Mole ratio (2 : 1)
• 2 mole Cu2+
1 mole I2
• 2.82 x 10-3
Cu2+
1.41 x 10-3
I2
Mass Cu = Mole Cu x RAM
Mass Cu = (2.85 x 10-3
x 63.5) g Cu
= 0.181 g
titrated
HNO3 and water
added till 100ml
0.456g
brass
% Cu in brass = mass Cu x 100%
mass brass
= 0.181 x 100%
0.468
= 39.7%
Using formulaUsing mole ratio
Using formula
M V(Cu2+
) = 2 = 1
MV(S203
2-
) 2 1
Moles of Cu2+
= 1
0.1 x 0.0285 1
Mole of Cu 2+
= 2.85 x 10-3
M V(Cu2+
) = 2 = 1
MV(S203
2-
) 2 1
M x 0.100 = 1
0.1 x 0.0285 1
2+ -2
1
2
3
4
5
6
Iodometric titration
I2/thiosulphate/starch
↓
I -
oxidized by OA to I2
↓
I2 react with starch
(blue black colour)
↓
S2O3
2-
added to reduce I2
↓
I2 used up – blue black
disappear
4I-
+ 2Cu+
↔ I2 + 2CuI
2S2O3
2-
+ I2 ↔S4O6
2-
+ 2I-
Click here here for copper determination expt
Cal Amt S2O3
2-
Cal Conc/Mole/Mass Cu
Cal % Cu by mass in brass
Cal % error (Lit value = 44.2 % Cu)
% error = Expt value x 100%
Lit value
= (44.2 – 39.7) x 100%
44.2
= 10.2%

10. Redox Titration Calculation- % purity of oxalate ion
Purity of sodium oxalate Na2C2O4 is determine by redox titration with standard 0.040M KMnO4. 35.62 ml KMnO4 needed to
reach end point. Cal % w/w Na2C2O4 in sample. How equivalent point is detected ?
7
oxalate solution titrated
0.5116 g
KMnO4
M = 0.040M
V = 35.62 ml
C2O4
2-
M = ?
2MnO4- + 5C2O4
2-
+ 16H+
2Mn→ 2+
+ 10CO2 + 8H2O
M = 0.040M M = ?
V = 35.62 ml
Mole ratio – 2: 5
Using mole ratio
Mole KMO4
-
= MV
= (0.040 x 0.03562)
= 1.42 x 10-3
Mole ratio (2 : 5)
• 2 mol KMO4
-
react 5 mol C2O4
2-
• 1.42 x 10-3
KMO4
-
react 3.55 x 10-3
C2O4
2-
M aVa = 2
Mb Vb 5
0.04 x 0.03562 = 2
Mole C2O4
2-
5
Mol C2O4
2-
= 3.55 x 10-3
Mass of (expt yield) = 0.476 g
Mass of (Actual tablet) = 0.5116 g
% w/w in Na2C2O4 = 0.476 x 100 %
0.5116
= 93 %
Mole  Mass
Mole x RMM = Mass Na2C2O4
3.55 x 10-3
x 134 = 0.476 g Fe
Using formula
1
2
3
MnO4
-
– In burette is purple – Turns colourless react with C2O4
2-
All C2O4
2-
used up at equivalence point – excess KMnO4
-
turn purple
?
Oxidizing
Agent
Reducing
Agent
MnO4
-
Fe2+
Cr2O7
2-
SO2
HNO3 I-
H2O2 H2S
CI2 SO3
2-
KIO3 Vitamin C
CIO-
/Cu2+
Oxalate/
C2O4
2-
MnO4
–
reduced to Mn2+
C2O4
2-
oxidized to CO2
(+7) ON decrease ↓ (+2)
(+3) ON increase ↑ (+4)
4

11. M V(KIO3) = 1
MV (C6H8O6) 3
0.002 x 0.0255 = 1
Mole C6H8O6 3
Mole C6H8O6 = 1.53 x 10-4
Mole C6H8O6 = M x V
M x V = 1.53 x 10-4
M x 0.025 = 1.53 x 10-4
M = 1.53 x 10-4
0025
M = 6.12 x 10-3
M
1 mol 3 mol
KIO3 + 5KI + 6H+
3I→ 2 + 6K+
+ 3H2O
3C6H8O6 + 3I2 3C→ 6H6O6 + 6I-
+ 6H+
3 mol 3 mol
Iodometric titration was performed on Vit C, (C6H8O6). 25ml Vit C is titrated with 0.002M KIO3
from burette, using excess KI and starch. Average vol KIO3 is 25.5ml. Cal molarity of Vit C.
Redox Titration Calculation – Vitamin C quantification
8
KIO3
M = 0.002M
V = 25.5ml
Vit C
M = ?
V = 25ml
KIO3 + 5KI + 6H+
3I→ 2 + 3H2O + 6K=
M = 0.002M Mole = ?
V = 25.5ml
Mole ratio (1 :3)
1 mol KIO3 : 3 mol I2 : 3 mol C6H8O6
1 mol KIO3 3 mol C6H8O6
V = 25ml
M = ?
25ml transfer
1g KI excess + starch
added
Mole KIO3 = MV
= (0.002 x 0.0255)
= 5.10 x 10-5
Mole ratio (1 : 3)
• 1 mole KIO3 produce 3 mole I2
• 5.10 x 10-5
KIO3 produce 1.53 x 10-4
I2
Mole ratio – 1: 3
3C6H8O6 + 3I2 3C→ 6H6O6 + 6I-
+ 6H+
Mole = ? 1.53 x 10-4
Mole ratio – 3: 3
Mole ratio (1 : 3)
• 1 mol KIO3 react 3 mol C6H8O6
• 5.10 x 10-5
KIO3 react 1.53 x 10-4
C6H8O6
Mole C6H8O6 = M x V
M x V = 1.53 x 10-4
M x 0.025 = 1.53 x 10-4
M = 1.53 x 10-4
0025
M = 6.12 x 10-3
M
titrated
Using mole ratio Using formula
Using formula
Vitamin C
1
2
3
4
Click here here to view sample Vitamin C expt
?
Oxidizing
Agent
Reducing
Agent
MnO4
-
Fe2+
Cr2O7
2-
SO2
HNO3 I-
H2O2 H2S
CI2 SO3
2-
KIO3 Vitamin C
CIO-
/Cu2+
Oxalate/
C2O4
2-

12. 25ml of undiluted H2O2 is transfer to 250ml volumetric flask. (Diluted 10x ). 25ml diluted sample was titrated with
standard 0.02114M KMnO4. 28.64 ml KMnO4 needed to reach end point. Cal conc in M H2O2 sample. Assuming
density is 1g/ml, calculate % H2O2 by weight. (Theoretical value H2O2 = 3%)
9
25ml pipette solution
KMnO4
M = 0.02114M
V = 28.64 ml
H2O2
M = ?
2MnO4
-
+ 5H2O2 + 6H+
2Mn→ 2+
+ 5O2 + 8H2O
M = 0.02114M M = ?
V = 28.64 ml
Mole ratio – 2: 5
Using mole ratio
Mole KMO4
-
= MV
= (0.02114 x 0.02864)
= 6.054 x 10-4
Mole ratio (2 : 5)
• 2 mol KMO4
-
react 5 mol H2O2
• 6.054 x 10-4
KMO4
-
react 1.513 x 10-3
H2O2
M V = 2
M V 5
0.02114 x 0.02864 = 2
Mole H2O2 5
Mol H2O2 = 1.5135 x 10-3
Using formula
1
2
?
Oxidizing
Agent
Reducing
Agent
MnO4
-
Fe2+
Cr2O7
2-
SO2
HNO3 I-
H2O2 H2S
CI2 SO3
2-
KIO3 Vitamin C
CIO-
/Cu2+
Oxalate/
C2O4
2-
MnO4
–
reduced to Mn2+
H2O2 oxidized to O2
(+7) ON decrease ↓ (+2)
(-1) ON increase ↑ (0)
Redox Titration H2O2 Calculation
Pour into
Volumetric flask
25 ml H2O2
Water added
till 250ml
Mol H2O2 = M x V
M x V = 1.513 x 10-3
M x 0.025 = 1.513 x 10-3
M = 1.513 x 10-3
0.025
M = 0.06052M (Diluted sample)
Original sample = 0.06052 x 10
= 0.6052 M
Conc H2O2 = 0.6052M
RMM H2O2 = 34
Mass H2O2 = 0.6052 x 34
= 20.60g in 1000 ml
= 2.06g in 100ml
= 2.06%
3
Stronger oxidizing agent
reduce weaker oxidizing agent

13. Cr2O7
2-
reduced to Cr3+
C2H5OH oxidized CH3COOH3
% C2H5OH by mass = mass C2H5OH x 100%
mass blood
= 0.351 x 100%
10.0
= 3.51 %
Alcohol in blood can be determined by redox titration with K2Cr2O7
3C2H5OH + 2Cr2O7
2-
+ 16H+
→ 3CH3COOH3 + 4Cr 3+
+ 11H2O
Calculate % by mass of ethanol. Explain how end point is determined?
10
Cr2O7
2-
M = 0.055M
V = 9.25 ml
C2H5 OH
M = ?
2Cr2O7
2-
+ 3C2H5OH + 16H+
3CH→ 3COOH3 + 4Cr 3+
+ 11H2O
M = 0.0550 M = ?
V = 9.25ml
Mole ratio – 3: 2
Using mole ratio
Mole Cr2O7
-2-
= MV
= (0.055 x 0.00925)
= 5.08 x 10-4
Mole ratio (2 : 3)
• 2 mol Cr2O7
2-
react 3 mol C2H5OH
• 5.08 x 10-4
Cr2O7
2-
react 7.63 x 10-3
C2H5OH
M V = 2
M V 3
0.055 x 0.0925 = 2
MV 3
Mol C2H5OH = 7.63 x 10-3
Using formula
1
2
(+7) ON decrease ↓ (+3)
(-2) ON increase ↑ (0)
Redox Titration Alcohol Calculation C2H2OH
10g of blood sample
Mass C2H5OH = Mol x RAM
Mass = 7.63 x 10-3
x 46
Mass = 0.351 g
3
Click here practical breath analyzer using dichromate
Alcohol
C2H5OH
Ethanoic acid
CH3COOH
Cr2O7
2-
– In burette is orange– Turns green react with C2H5OH
All C2H5OH used up at equivalence point – excess Cr2O7
2-
turn orange
oxidized
Dichromate
Cr2O7
2-
Chromate
Cr3+
reduced
Oxidizing
Agent
Reducing
Agent
MnO4
-
Fe2+
Cr2O7
2-
SO2
HNO3 I-
H2O2 H2S
CI2 SO3
2-
KIO3 Vitamin C
CIO-
/Cu2+
Ethanol/
C2H4OH
?
4

14. Biological Oxygen Demand
Measure amt dissolve oxygen needed by aerobic organism to break down
•Organic matter in water sample over 5 day period
•BOD polluted water – Amt dissolve oxygen need for biological decomposition
•Measure amt O2 used for biochemical decomposition of organic matter
•Measure amt O2 used to oxidize organic to produce energy for microbes
Lots of organic decomposition (uses O2)
↓
Dissolve oxygen Low used up↓
↓
Biological Oxygen Demand High ↑
↓
Level Pollution is HIGH ↑
↓
Aquatic life die /Toxic
Low Dissolve Oxygen, signify high O2 demand from microb
(organic waste contamination)
Breakdown organic matter in water consumes
oxygen by aerobic micro-organisms.
BOD High ↑
Dissolve O2 Low (O↓ 2 used up)
Level Pollution HIGH ↑
Organic waste decomposition ↑
Aquatic life die/Toxic
BOD Low ↓
Dissolve O2 High ↑ (O2 high)
Level Pollution LOW ↓
Organic waste decomposition ↓
Aquatic life thrive
BOD ↑ – No good
BOD ↓ - Good
Dissolve oxygen Level -
•Indicator of clean water
•Level of pollution
BOD ↓
Dissolve
Oxygen ↑
Click here carolina Winkler method BOD
Click here dissolve oxygen video

15. Water
Quality
Clean Lightly
polluted
Moderate
polluted
Severely
polluted
Dissolve O2,
mg/ml
DO > 6.5 4.5 – 6.5 2.0 – 4.5 < 2.0
BOD, mg/ml < 3 3 – 4.9 5 – 15 > 15
Explosive growth algae/bloom
Block sunlight for photosynthesis
Eutrophication on BOD
Excessive use fertilisers like phosphates/nitrates
Wash into river/water
Eutrophication
Explosive growth algae/bloom
↓
When die - organic decomposition
by bacteria
↓
Uses up dissolve oxygen
↓
BOD demand HIGH ↑
↓
Water polluted
Algae bloom
Dissolve oxygen Low ↓
BOD High ↑
Nutrient leach
Biological Oxygen Demand
Redox titration (Winkler Method)
measure dissolve O2
BOD index
Click here on Winkler titration method
Iodometric titration
I2/thiosulphate/starch
↓
Mn2+
oxidized by O2 to Mn4+
↓
Mn4+
oxidized I-
to I2
I2 react with starch
(blue black colour)
↓
S2O3
2-
added to reduce I2
↓
I2 used up – blue black
disappear
Measure BOD
Iodometric titration

16. Biological Oxygen Demand
Redox titration (Winkler Method)
measure dissolve O2
BOD index
1 mol 2 mol
2Mn2+
+ O2 + 4OH-
2MnO→ 2 + 2H2O
2MnO2 + 4I-
+ 4H+
4I→ 2 + 2Mn2-
+ 4H2O
4I2 + 4S2O3
2-
→ 4I-
+ 2S4O6
2-
Click here on Winkler titration method
Water Quality Clean Lightly
polluted
Moderately
polluted
Severely
polluted
Dissolve O2,
mg/ml
DO > 6.5 4.5 – 6.5 2.0 – 4.5 < 2.0
BOD, mg/ml < 3 3 – 4.9 5 – 15 > 15
Dissolve O2 reacts with alkaline manganese (Mn2+
) to form (Mn4+
)
4Mn2+
+ 4OH- 2Mn(OH)→ 2
1 mol 2 mol
2Mn(OH)2 + O2 2MnO(OH)→ 2
2MnO(OH)2 + 8H+
+ 6I-
→ 2I3
-
+ 6H2O
2 mol 2 mol
2I -
+ 4S O 2-
→ 6I-
+ 2S O 2-
Redox titration Winkler Method
DO bottle
Mn2+
salt
1g KI excess
alkaline/OH-
shake
White ppt Mn(OH)2
Conc
H2SO4
White ppt dissolve in acid
Na2S2O3
M = 0.05M
V = 12.5ml
titrated S2O3
2-
1O2 + 4S2O3
2-
products→
M = ? M = 0.05M
V = 12.5ml
I-
oxidized to I2 by Mn2+
O2
M = ?
V = 500ml
2 mol 4 mol
4 mol 4 mol
Mole ratio O2 : S2O3
2-
(1 : 4)
1 mol O2 : 4 mol I2 : 4 mol S2O3
2-
1 mol O2 4 mol S2O3
2-
Brown I2 sol form
Starch added
Iodometric titration
I2/thiosulphate/starch
↓
Mn2+
oxidized by O2 to Mn4+
↓
Mn4+
oxidized I-
to I2
I2 react with starch
(blue black colour)
↓
S2O3
2-
added to reduce I2
↓
I2 used up – blue black
disappear
Water sample
added
1 mol O2 : 4 mol S2O3
2-

17. 1 mol 2 mol
2Mn2+
+ O2 + 4OH-
2MnO→ 2 + 2H2O
2MnO2 + 4I-
+ 4H+
4I→ 2 + 2Mn2-
+ 4H2O
4I2 + 4S2O3
2-
→ 4I-
+ 2S4O6
2-
Dissolve O2 reacts with alkaline manganese (Mn2+) to form (Mn4+)
Redox titration Winkler Method
DO bottle
Mn2+
salt
1g KI excess
alkaline/OH-
shake
White ppt Mn(OH)2
Conc
H2SO4
White ppt dissolve in acid
Na2S2O3
M = 0.05M
V = 12.5ml
titrated S2O3
2-
1O2 + 4S2O3
2-
product→
M = ? M = 0.05M
V = 12.5ml
I-
oxidized to I2 by Mn2+
O2
M = ?
V = 500ml
2 mol 4 mol
4 mol 4 mol
Mole ratio O2 : S2O3
2-
(1 : 4)
1 mol O2 : 4 mol I2 : 4 mol S2O3
2-
1 mol O2 4 mol S2O3
2-
Brown I2 sol
form
Starch added
Iodometric titration
I2/thiosulphate/starch
↓
Mn2+
oxidized by O2 to Mn4+
↓
Mn4+
oxidized I-
to I2
I2 react with starch
(blue black colour)
↓
S2O3
2-
added to reduce I2
↓
I2 used up – blue black
disappear
Water sample
added
500ml water tested for dissolve oxygen by adding Mn2+
in alkaline solution, followed by addition
of KI and acid. I2 produced is reduced by titrating with 0.05M S2O3
2-
. Average vol S2O3
2-
used is
12.50ml. Calculate dissolved oxygen in g/dm3
.
1
Mole S2O3
2-
= MV
= (0.05 x 0.0125)
= 6.25 x 10-4
Mole ratio (1 : 4)
• 1 mole O2 react 4 mole S2O3
2-
? 6.25 x 10-4
S2O2
2-
6.25 x 10-4
= 1.56 x 10-4
4
1 mol O2 : 4 mol S2O3
2-
2 3
Mole O2 = 1.56 x 10-4
mol
Mass O2 = Mole O2 x RAM
Mass O2 = (1.56 x 10-4
x 32.0)g
= (5.00 x 10-3
)g in 500ml
= 0.01 g in 1000ml
= 0.01g/dm3
4
Click here on Winkler titration methodClick here on Winkler titration method

18. Titration for IA (DCP) assessment
Acid Base Titration
Standardization HCI
with primary std Na2CO3
Click here for expt 4.2
Standardization NaOH
with primary std KHP
Click here or here for expt`
Titration bet NaOH
with std HCI
Click here for expt 4.2a
Titration bet HCI
with std NaOH
Click here for expt 4.2a
Determining water crystallization
in hydrated Na2CO3 with std HCI
Click here for expt 4.4
Standardization KMnO4
with std ammonium
iron(II) sulphate
Click here for expt 4.5
Iron (II) determination
with std KMnO4
Click here for expt 4.6
Hypochlorite (OCI-
) in bleach
with iodine/thiosulphate
Click here for expt 4.8
Determining ethanoic acid
in vinegar with std NaOH
Click here for expt 4.3
Copper(II) determination
in brass
with iodine/thiosulphate
Click here or here for expt`
Click here for more expt
Standardization KI/I2
with std KIO3
Click here for expt 4.7
Click here for more expt
Determining acetylsalicylic acid
in aspirin with std NaOH
Click here or here for expt`
Click here for more expt
Vit C determination with
iodine/thiosulphate
Click here or here for expt
Click here more detail expt
Standardization Expt Acid/Base Titration Expt
Standardization Expt
Redox Titration Expt
Redox Titration
Standardization KI/I2
with std
sodium thiosulphate
Click here for expt 4.7 Iodine/thiosulphate (iodometric titration)

19. CI2 + 2KBr-
2KCI→ + Br2
3CuO+ 2NH3 3H→ 2O+ 3Cu+ N2
Redox (Oxidation and Reduction)
(+7) (+2)Mn red - ON ↓
(+2) Fe oxi – ON ↑ (+3)
MnO4
-
+ Fe2+
+ 8H+
Mn→ 2+
+ Fe3+
4H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing
Agent
Reducing
Agent
MnO4
-
Fe2+
Reduction Oxidation
Oxidizing
Agent
Reducing
Agent
CI2 Br-
Reduction Oxidation
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
(0) CI red – ON ↓ (-1)
(-1) Br-
oxi – ON ↑ (0)
Oxidizing
Agent
Reducing
Agent
CuO NH3
Reduction Oxidation
Reducing agent
↓
Oxidation
(-3) NH3 oxi – ON ↑ (0)
Oxidizing agent
↓
Reduction (+2) Cu red – ON ↓ (0)
2HCI + Zn H→ 2 + ZnCI2
(0) Zn oxi – ON ↑ (+2)Reducing agent
↓
Oxidation
Oxidizing agent
↓
Reduction
(+1) H red – ON ↓ (0)
Oxidizing
Agent
Reducing
Agent
HCI Zn
Reduction Oxidation

20. CI2 + 2KBr-
2KCI→ + Br2
3CuO+ 2NH3 3H→ 2O+ 3Cu+N2
Redox (Oxidation and Reduction)
(+7) (+2)Mn red - ON ↓
(+2) Fe oxi – ON ↑ (+3)
MnO4
-
+ 8H+
+ Fe2+
Mn→ 2+
+ Fe3+
4H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
MnO4
-
+ 5e Mn→ 2+
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
(0) CI red – ON ↓ (-1)
(-1) Br -
oxi – ON ↑ (0)
Reducing agent
↓
Oxidation
(-3) NH3 oxi – ON ↑ (0)
Oxidizing agent
↓
Reduction (+2) Cu red – ON ↓ (0)
2HCI + Zn H→ 2 + ZnCI2
(0) Zn oxi – ON ↑ (+2)Reducing agent
↓
Oxidation
Oxidizing agent
↓
Reduction
(+1) H red – ON ↓ (0)
Reducing Agent Oxidation
Fe 2+
Fe→ 2+
+ e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Reducing Agent Oxidation
2Br -
Br→ 2 + 2e-
Loss electron
Increase ON ↑
Oxidizing Agent Reduction
CI2 + 2e 2CI→ - Gain electron
Decrease ON ↓
Reducing Agent Oxidation
(NH3) -N3-
N→ + 3e-
Loss electron
Increase ON ↑
Oxidizing Agent Reduction
(CuO) Cu2+
+ 2e Cu→
Gain electron
Decrease ON ↓
Reducing Agent Oxidation
Zn Zn→ 2+
+ 2e-
Loss electron
Increase ON ↑
Oxidizing Agent Reduction
2H+
+ 2e H→ 2
Gain electron
Decrease ON ↓

21. Redox (Oxidation and Reduction)
Half equations
Oxidation rxn
Oxidation half eqn Reduction half eqn
Loss electron ↓
Reduction rxn
Loss hydrogen ↓ Gain oxygen ↑ Gain ON ↑ Gain electron ↑ Gain hydrogen ↑ Loss oxygen ↓ Loss ON ↓
Oxidizing AgentReducing Agent
Oxidation rxn Reduction rxn
lose electron
Zn + 2H+
H→ 2 + Zn2+
Zn Zn→ 2+
+ 2e 2H+
+ 2e H→ 2
(0) ON increase ↑ (+2)
Zn Zn→ 2+
+ 2e
2H+
+ 2e H→ 2
2H+
+ Zn Zn→ 2+
+ H2
lose electron gain electron
(+1) ON decrease ↓ (0)
Complete full eqn
Zn + Cu2+
Zn→ 2+
+ CuOxidation half eqn
Zn Zn→ 2+
+ 2e
lose electron
(0) ON increase ↑ (+2)
Reduction half eqn
Cu2+
+ 2e Cu→
(+2) ON decrease ↓ (0)
gain electron
Zn Zn→ 2+
+ 2e
Cu2+
+ 2e Cu→
Cu2+
+ Zn Zn→ 2+
+ Cu
Half equations

22. Redox (Oxidation and Reduction)
Half equations
Oxidation half eqn Reduction half eqn
Zn Zn→ 2+
+ 2e 2H+
+ 2e H→ 2
(0) ON increase ↑ (+2)
Zn Zn→ 2+
+ 2e
2H+
+ 2e H→ 2
2H+
+ Zn Zn→ 2+
+ H2
lose electron gain electron
(+1) ON decrease ↓ (0)
Complete full eqn
Oxidation half eqn
Zn Zn→ 2+
+ 2e
lose electron
(0) ON increase ↑ (+2)
Reduction half eqn
Cu2+
+ 2e Cu→
(+2) ON decrease ↓ (0)
gain electron
Zn Zn→ 2+
+ 2e
Cu2+
+ 2e Cu→
Cu2+
+ Zn Zn→ 2+
+ Cu
Half equations
Zn + 2HCI H→ 2 + ZnCI2
Zn + 2H+
+ 2CI-
H→ 2 + Zn2+
+ 2CI -
Complete ionic/redox eqn
Zn + 2H+
H→ 2 + Zn2+
spectator ionsspectator ions
Zn + 2H+
H→ 2 + Zn2+
Zn + CuSO4 ZnSO→ 4 + Cu
Zn + Cu2+
+ SO4
2-
Zn→ 2+
+ SO4
2-
+ Cu
Complete full eqn
Complete ionic/redox eqn
spectator ions
Zn + Cu2+
Zn→ 2+
+ Cu
Half equations Half equations
Zn + Cu2+
Zn→ 2+
+ Cu

23. Redox (Oxidation and Reduction)
Half equations
Oxidation half eqn Reduction half eqn
Mg Mg→ 2+
+ 2e Pb2+
+ 2e Pb→
(0) ON increase ↑ (+2)
Mg Mg→ 2+
+ 2e
Pb2+
+ 2e Pb→
Pb2+
+ Mg Mg→ 2+
+ Pb
lose electron gain electron
(+2) ON decrease ↓ (0)
Complete full eqn
Oxidation half eqn
2Br-
Br→ 2 + 2e
lose electron
(-1) ON increase ↑ (0)
Reduction half eqn
CI2 + 2e 2CI→ -
(0) ON decrease ↓ (-1)
gain electron
2Br-
Br→ 2 + 2e
CI2 + 2e 2CI→ -
CI2 + 2Br-
2CI→ -
+ Br2
Half equations
Mg + PbO Pb→ + MgO
Mg + Pb2+
+ O2-
Pb→ + Mg2+
+ O 2-
Complete ionic/redox eqn
spectator ionsspectator ions
Mg + Pb2+
Pb→ + Mg2+
2KBr + CI2 Br→ 2 + 2KCI
2K+
+ 2Br-
+ CI2 Br→ 2 + 2K+
+ 2CI -
Complete full eqn
Complete ionic/redox eqn
spectator ions
2Br-
+ CI2 Br→ 2 + 2CI-
Half equations Half equations
Mg + Pb2+
Pb→ + Mg2+
2Br-
+ CI2 Br→ 2 + 2CI-
lose electron

24. MnO4
-
+ 8H+
+ 5Fe2+
Mn→ 2+
+ 5Fe3+
+ 4H2O
Constructing Half and complete redox equation
(+7) (+2)Mn red - ON ↓
(+2) Fe oxi – ON ↑ (+3)
MnO4
-
+ 5Fe2+
+ 8H+
Mn→ 2+
+ 5Fe3+
+ 4H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
MnO4
-
+ 5e Mn→ 2+
Reducing Agent Oxidation
Fe 2+
Fe→ 2+
+ e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O -add H2O
2. Balance # H add H+
3. Balance # charges -add electrons
4. Balance # electron transfer
MnO4
-
Mn→ 2+
MnO4
-
Mn→ 2+
+ 4H2O
MnO4
-
+ 8H+
Mn→ 2+
+ 4H2O
MnO4
-
+ 8H+
+ 5e- Mn→ 2+
+ 4H2O
Fe2+
Fe→ 3+
Fe2+
Fe→ 3+
+ e-
5Fe2+
5Fe→ 3+
+ 5e-
MnO4
-
+ 8H+
+ 5e- Mn→ 2+
+ 4H2O
x 5x 1
MnO4
-
+ 8H+
+ 5e-
Mn→ 2+
+ 4H2O
5Fe2+
5Fe→ 3+
+ 5e-
+
MnO4
-
- In acidic medium
- Strong oxidizing agent
MnO4
-
+ 8H+
+ 5Fe2+
Mn→ 2+
+ 5Fe3+
4H2O

25. 2MnO4
-
+ 5SO2+ 2H2O 2Mn→ 2+
+ 5SO4
2-
+ 4H+
Constructing Half and complete redox equation
(+7) (+2)Mn red - ON ↓
(+4) SO2 oxi – ON ↑ (+6)
2MnO4
-
+ 5SO2 + 2H2O 2Mn→ 2+
+ 5SO4
2-
+ 4H+
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
MnO4
-
+ 5e Mn→ 2+
Reducing Agent Oxidation
SO2 SO→ 4
2-
+ 2e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
MnO4
-
Mn→ 2+
MnO4
-
Mn→ 2+
+ 4H2O
MnO4
-
+ 8H+
Mn→ 2+
+ 4H2O
MnO4
-
+ 8H+
+ 5e- Mn→ 2+
+ 4H2O
SO2 SO→ 4
2-
2MnO4
-
+ 16H+
+ 10e- 2Mn→ 2+
+ 8H2O
x 5x 2
2MnO4
-
+ 16H+
+ 10e-
2Mn→ 2+
+ 8H2O
5SO2 + 10H2O 5SO→ 4
2-
+ 20H+
+ 10e-
+
2MnO4
-
+ 5SO2 + 2H2O 2Mn→ 2+
+ 5SO4
2-
4H+
SO2 + 2H2O SO→ 4
2-
SO2 + 2H2O SO→ 4
2-
+ 4H+
SO2 + 2H2O SO→ 4
2-
+ 4H+
+ 2e-
5SO2 + 10H2O 5SO→ 4
2-
+ 20H+
+ 10e-

26. 2MnO4
-
+ 5H2O2 + 6H+
2Mn→ 2+
+ 5O2 + 8H2O
Constructing Half and complete redox equations
(+7) (+2)Mn red - ON ↓
(-1) H2O2 oxi – ON ↑ (0)
2MnO4
-
+ 5H2O2 + 6H+
2Mn→ 2+
+ 5O2 + 8H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
MnO4
-
+ 5e Mn→ 2+
Reducing Agent Oxidation
H2O2 O→ 2 + 2e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
MnO4
-
Mn→ 2+
MnO4
-
Mn→ 2+
+ 4H2O
MnO4
-
+ 8H+
Mn→ 2+
+ 4H2O
MnO4
-
+ 8H+
+ 5e- Mn→ 2+
+ 4H2O
2MnO4
-
+ 16H+
+ 10e- 2Mn→ 2+
+ 8H2O
x 5x 2
2MnO4
-
+ 16H+
+ 10e-
2Mn→ 2+
+ 8H2O
5H2O2 5O→ 2 + 10H+
+ 10e-
+
2MnO4
-
+ 5H2O2 + 6H+
2Mn→ 2+
+ 5O2 + 8H2O
H2O2 O→ 2
H2O2 O→ 2 + 2H+
H2O2 O→ 2 + 2H+
+ 2e-
5H2O2 5O→ 2 + 10H+
+ 10e-

27. Cr2O7
2-
+ 3NO2
-
+ 8H+
2Cr→ 3+
+ 3NO3
-
+ 4H2O
Cr2O7
2-
2Cr→ 3+
Constructing Half and complete redox equations
(+6) (+3)Cr red - ON ↓
(+3) NO2
-
oxi – ON ↑ (+5)
Cr2O7
2-
+ 3NO2
-
+ 8H+
2Cr→ 3+
+ 3NO3
-
+ 4H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
Cr2O7
2-
+ 6e- 2Cr→ 3+
Reducing Agent Oxidation
NO2
-
NO→ 3
-
+ 2e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
x 3x 1
Cr2O7
2-
+ 14H+
+ 6e-
2Cr→ 3+
+ 7H2O
3NO2
-
+ 3H2O 3NO→ 3
-
+ 6H+
+ 6e-
+
Cr2O7
2-
+ 3NO2
-
+ 8H+
2Cr→ 3+
+ 3NO3
-
+ 4H2O
Cr2O7
2-
2Cr→ 3+
+ 7H2O
Cr2O7
2-
+ 14H+
2Cr→ 3+
+ 7H2O
Cr2O7
2-
+ 14H+
+ 6e-
2Cr→ 3+
+ 7H2O
Cr2O7
2-
+ 14H+
+ 6e-
2Cr→ 3+
+ 7H2O
NO2
-
NO→ 3
-
NO2
-
+ H2O NO→ 3
-
NO2
-
+ H2O NO→ 3
-
+ 2H+
NO2
-
+ H2O NO→ 3
-
+ 2H+
+ 2e-
3NO2
-
+ 3H2O 3NO→ 3
-
+ 6H+
+ 6e-

28. Cr2O7
2-
+ 6Fe2+
+ 14H+
2Cr→ 3+
+ 6Fe3+
+ 7H2O
Cr2O7
2-
2Cr→ 3+
Constructing Half and complete redox equations
(+6) (+3)Cr red - ON ↓
(+2) Fe2+
oxi – ON ↑ (+3)
Cr2O7
2-
+ 6Fe2+
+ 14H+
2Cr→ 3+
+ 6Fe3+
+ 7H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
Cr2O7
2-
+ 6e- 2Cr→ 3+
Reducing Agent Oxidation
Fe2+
Fe→ 3+
+ e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
x 6x 1
Cr2O7
2-
+ 14H+
+ 6e-
2Cr→ 3+
+ 7H2O
6Fe2+
6Fe→ 3+
+ 6e-
+
Cr2O7
2-
2Cr→ 3+
+ 7H2O
Cr2O7
2-
+ 14H+
2Cr→ 3+
+ 7H2O
Cr2O7
2-
+ 14H+
+ 6e-
2Cr→ 3+
+ 7H2O
Cr2O7
2-
+ 14H+
+ 6e-
2Cr→ 3+
+ 7H2O
Cr2O7
2-
+ 6Fe2+
+ 14H+
2Cr→ 3+
+ 6Fe3+
7H2O
Fe2+
Fe→ 3+
Fe2+
Fe→ 3+
+ e
6Fe2+
6Fe→ 3+
+ 6e

29. Constructing Half and complete redox equations
(+5) (-1)CIO3
-
red - ON ↓
(-1) I-
oxi – ON ↑ (0)
CIO3
-
+ 6I-
+ 6H+
3I→ 2 + CI-
+ 3H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
CIO3
-
+ 6e- CI→ -
Reducing Agent Oxidation
2I-
I→ 2 + 2e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
x 3x 1
CIO3
-
+ 6H+
+ 6e-
CI→ -
+ 3H2O
6I-
3I→ 2 + 6e-
+
CIO3
-
+ 6I-
+ 6H+
3I→ 2 + CI-
+ 3H2O
CIO3
-
CI→ -
CIO3
-
CI→ -
+ 3H2O
CIO3
-
+ 6H+
CI→ -
+ 3H2O
CIO3
-
+ 6H+
+ 6e- CI→ -
+ 3H2O
CIO3
-
+ 6H+
+ 6e- CI→ -
+ 3H2O
2I-
I→ 2
2I-
I→ 2 + 2e-
6I-
3I→ 2 + 6e-
CIO3
-
+ 6H+
+ 6I-
3I→ 2 + 3H2O

30. Constructing Half and complete redox equations
(+5) (+2)NO3
-
red - ON ↓
(0) Cu oxi – ON ↑ (+2)
2NO3
-
+ 3Cu + 8H+
3Cu→ 2+
+ 2NO+ 4H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
NO3
-
+ 3e- NO→
Reducing Agent Oxidation
Cu Cu→ 2+
+ 2e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
x 3x 2
2NO3
-
+ 8H+
+ 6e-
2NO→ + 4H2O
3Cu 3Cu→ 2+
+ 6e-
+
2NO3
-
+ 3Cu + 8H+
3Cu→ 2+
+ 2NO+ 4H2O
NO3
-
NO→
NO3
-
NO + 2H→ 2O
NO3
-
+ 4H+
NO + 2H→ 2O
NO3
-
+ 4H+
+ 3e- NO + 2H→ 2O
2NO3
-
+ 8H+
+ 6e- 2NO + 4H→ 2O
Cu Cu→ 2+
Cu Cu→ 2+
+ 2e-
3Cu 3Cu→ 2+
+ 6e-
2NO3
-
+ 8H+
+ 3Cu 3Cu→ 2+
+2NO+ 4H2O

31. HNO3 +3Fe2+
+ 3H+
3Fe→ 3+
+ NO+ 2H2O
Constructing Half and complete redox equations
(+5) (+2)HNO3 red - ON ↓
(+2) Fe oxi – ON ↑ (+3)
HNO3 + 3Fe2+
+ 3H+
3Fe→ 3+
+ NO+ 2H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
HNO3 + 3e- NO→
Reducing Agent Oxidation
Fe 2+
Fe→ 3+
+ e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
x 3x 1
HNO3 + 3H+
+ 3e-
NO→ + 2H2O
3Fe2+
3Fe→ 3+
+ 3e-
+
HNO3 NO + 2H→ 2O
HNO3+ 3H+
NO + 2H→ 2O
HNO3 + 3H+
+ 3e- NO + 2H→ 2O
HNO3 + 3H+
+ 3e- NO + 2H→ 2O
Fe2+
Fe→ 3+
HNO3 + 3Fe2+
+ 3H+
3Fe→ 3+
+ NO+ 2H2O
HNO3 NO→
Fe2+
Fe→ 3+
+ e-
3Fe2+
3Fe→ 3+
+ 3e-

32. H2O2 + 2Fe2+
+2H+
2Fe→ 3+
+ 2H2O
Constructing Half and complete redox equations
(-1) (-2)H2O3 red - ON ↓
(+2) Fe oxi – ON ↑ (+3)
H2O2 + 2Fe2+
+ 2H+
2Fe→ 3+
+ 2H2O
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
H2O3 + e- H→ 2O
Reducing Agent Oxidation
Fe 2+
Fe→ 3+
+ e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
x 2x 1
H2O2 + 2H+
+ 2e-
2H→ 2O
2Fe2+
2Fe→ 3+
+ 2e-
+
Fe2+
Fe→ 3+
Fe2+
Fe→ 3+
+ e-
2Fe2+
2Fe→ 3+
+ 2e-
H2O2 + 2Fe2+
+ 2H+
2Fe→ 3+
+ 2H2O
H2O2 H→ 2O
H2O2 2H→ 2O
H2O2 + 2H+
2H→ 2O
H2O2 + 2H+
+ 2e- 2H→ 2O
H2O2 + 2H+
+ 2e- 2H→ 2O

33. CI2 + SO2 + 2H2O 2CI→ -
+ SO4
2-
+ 4H+
Constructing Half and complete redox equations
(0) (-1)CI2 red - ON ↓
(+4) SO2 oxi – ON ↑ (+6)
CI2 + SO2 + 2H2O 2CI→ -
+ SO4
2-
+ 4H+
Oxidizing agent
↓
Reduction
Reducing agent
↓
Oxidation
Oxidizing Agent Reduction
CI2 + 2e 2CI→ -
Reducing Agent Oxidation
SO2 SO→ 4
2-
+ 2e-
Loss electron
Increase ON ↑
Gain electron
Decrease ON ↓
Complete full eqn
Oxidation half eqnReduction half eqn
1. Balance # O - add H2O
2. Balance # H add H+
3. Balance # charges - add electrons
4. Balance # electron transfer
SO2 SO→ 4
2-
x 1x 1
CI2 + 2e-
2CI→ -
SO2 + 2H2O SO→ 4
2-
+ 4H+
+ 2e-
+
SO2 + 2H2O SO→ 4
2-
SO2 + 2H2O SO→ 4
2-
+ 4H+
SO2 + 2H2O SO→ 4
2-
+ 4H+
+ 2e-
CI2 + SO2 + 2H2O 2CI→ -
+ SO4
2-
+ 4H+
CI2 2CI→ -
CI2 + 2e- 2CI→ -
CI2 + 2e- 2CI→ -
SO2 + 2H2O SO→ 4
2-
+ 4H+
+ 2e-

34. Sn2+
+ 2Fe3+
Sn→ 4+
+ 2Fe2+
2Fe2+
+ CI2 2Fe→ 3+
+ 2CI-
Ca + 2H+
Ca→ 2+
+ H2
IB Redox Questions
Deduce half eqn of oxidation and reduction for the following
Ca + 2H+
Ca→ 2+
+ H2
2Fe2+
+ CI2 2Fe→ 3+
+ 2CI-
Sn2+
+ 2Fe3+
Sn→ 4+
+ 2Fe2+
0 +1 +2 0
Ca Ca→ 2+
+ 2e
2H+
+ 2e H→ 2
oxidation
reduction
+2 0 +3 -1
2Fe2+
Fe→ 3+
+ 2e
CI2 + 2e 2CI→ -
oxidation
reduction
+2 +3 +4 +2
Sn2+
Sn→ 4+
+ 2e
2Fe3+
+ 2e 2Fe→ 2+
Substances acting as oxidizing and reducing agent
2MnO4
-
+ 5H2O2 + 6H+
2Mn→ 2+
+ 5O2 + 8H2O
H2O2 + 2Fe2+
+ 2H+
2Fe→ 3+
+ 2H2O
H2O2 + 2I-
+ 2H+
I→ 2 + 2H2O
Oxidizing Agent Reducing Agent
MnO4
-
Fe2+
Cr2O7
2- SO2
HNO3 I-
H2O2 H2S
CI2 SO3
2-
Acidified H2O2 act as oxidizing agent
- Oxidizes Fe2+
to Fe3+
- Oxidizes I-
to I2
Acidified MnO4
-
act as more powerful oxidizing agent
-Oxidizes weaker oxidizing agent
H2O2 to H2O and O2
- H2O2 act as reducing agent
Identify oxidizing and reducing agent for following rxn.
5As2O3 + 2MnO4
-
+ 16H+
2Mn→ 2+
+ 5As2O5 + 8H2O 2NO3
-
+ 3Cu + 8H+
3Cu→ 2+
+ 2NO+ 4H2O
Cr2O7
2-
+ 3NO2
-
+ 8H+
2Cr→ 3+
+ 3NO3
-
+ 4H2O
1 2
3
oxidizing
agent
oxidizing
agent
oxidizing
agent
reducing
agent
reducing
agent
reducing
agent

35. Acknowledgements
Thanks to source of pictures and video used in this presentation
Thanks to Creative Commons for excellent contribution on licenses
http://creativecommons.org/licenses/
Prepared by Lawrence Kok
Check out more video tutorials from my site and hope you enjoy this tutorial
http://lawrencekok.blogspot.com