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Prepared by Lawrence Kok

Tutorial on Redox Titration, BOD and Biological Oxygen Demand using Winkler method

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 KHP 20.4 g in 1L

Unknown Conc NaOH

Unknown Conc HCI

? ?

Standardize NaOH

u

sing KHP

Standardize HCI

using Na2 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

Oxidizing Agent

Reducing Agent

MnO4- Fe2+

Cr2O72- SO2

HNO3 I-

H2O2 H2S

CI2 SO32-

KIO3 Vitamin C

OCI-/Cu2+ Oxalate/C2O4

2-

Titration

Redox TitrationAcid Base Titration

Burette/TitrantOxidizing 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 agentTitrand 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 acidAnal

yte

to b

e

det

erm

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 Cr2O72- till endpoint

↓ turn orange (excess Cr2O7

2-)

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 g250ml

KMnO4

M = 0.002MV = 24.5 ml

Fe2+

M = ? V = 30ml

MnO4- + 5Fe2+ + 8H+ Mn→ 2+ + 5Fe2+ + 4H2OM = 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 = 1M V 50.002 x 0.0245 = 1 Moles Fe2+ 5Moles = 2.45 x 10-4 Fe2+

Mass of (expt yield) = 1.703gMass 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 + 1mol6.125 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

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 = 0.100MV = 27.5 ml

Fe2+

M = ?

1MnO4- + 5Fe2+ + 8H+ Mn→ 2+ + 5Fe2+ + 4H2OM = 0.100M M = ? V = 27.5ml

Mole ratio – 1: 5

Using mole ratio

Mole KMO4- = MV

= (0.100 x 0.0275) = 0.00275Mole ratio (1 : 5)• 1 mole KMO4

- react 5 mole Fe2+

• 0.00275 KMO4-react 0.01375 Fe2+

M aVa = 1Mb Vb 50.100 x 0.0275 = 1 Moles Fe2+ 5Moles = 0.01375 mol Fe2+

Mass of (expt yield) = 0.7679gMass 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

1 mol 1 mol1OCI- + 2I- + 2H+ I→ 2 + 1CI- + H2O

I2 + 2S2O32- 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 = 0.0206MV = 17.3ml

I2

M = ?

2S2O32- + 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 S2O32-

10.0ml OCI-

transfer

V = 250mlM = 8.9 x 10-3 M

20ml transfer

1g KI excess added

Mole S2O32- = 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 S2O32-- 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 VM 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 = 10M = ?

titrated

Water added till 250ml

1

Using direct formula

M V (OCI+) = 1 = 1M V (S203

2-) 2 2 Moles of OCI+ = 1 0.0206 x 0.0173 2Moles of OCI-

= 1.78 x 10-4

2

3

4

5

6

Hypochlorous acid = bleachOxidizing 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-

2S2O32- + I2 ↔S4O6

2- + 2I-

1 mol 1 mol1OCI- + 2I- + 2H+ I→ 2 + CI- + H2O

I2 + 2S2O32- 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 = 0.020MV = 38.5 ml

I2

M = ?

2S2O32- + 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 S2O32-

10ml bleach transfer

1g KI excess added

Mole S2O32- = 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 S2O32-- 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

1000M = 0.0387M

titrated

1

Using direct formula

M V (OCI+) = 1 = 1M V (S203

2-) 2 2 Moles of OCI+ = 1 0.020 x 0.03865 2Moles 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 = bleachActive 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-

2S2O32- + I2 ↔S4O6

2- + 2I-

2 mol 1 mol 2Cu2+ + 4I- I→ 2 + 2CuI

I2 + 2S2O32- 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 = 0.1MV = 28.2ml

I2

M = ?

2S2O32- + 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 S2O32-

Pour into Volumetric flask

V = 250mlM = ?

25ml transfer

1g KI excess + starch added

Mole S2O32- = 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 S2O32-- react 1.41 x 10-3 I2

Mole ratio – 2: 1

2Cu2+ + 4I- I→ 2 + 2CuIMole = ? 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 VM 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 MMass 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 = 1M V (S203

2-) 2 1Moles of Cu2+ = 1 0.1 x 0.0282 1Moles of Cu2+

= 2.82 x 10-3

1

2

3

45

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

2S2O32- + I2 ↔S4O6

2- + 2I-

Click here here for copper determination

2 mol 1 mol 2Cu2+ + 4I- I→ 2 + 2CuI

I2 + 2S2O32- 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 + 2H2OStep 2 : 4I- + 2Cu2+ 2CuI + I→ 2

Step 3 : I2 + 2S2O32- 2I→ - + S4O6

2-

Average vol S2O3 2- was 28.50ml.

Redox Titration Calculation - % Cu in Brass

6

Na2S2O3

M = 0.1MV = 28.5ml

I2

M = ?V = 100ml

2S2O32- + 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 S2O32-

V = 100mlM = ?

1g KI excess/starch added

Mole S2O32- = 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 S2O32-- react 1.41 x 10-3 I2

Mole ratio – 2: 1

2Cu2+ + 4I- I→ 2 + 2CuIMole = ? 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 RAMMass 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 = 1M V (S203

2-) 2 1Moles of Cu2+ = 1 0.1 x 0.0285 1Mole of Cu 2+ = 2.85 x 10-3

M V (Cu2+) = 2 = 1M V (S203

2-) 2 1M x 0.100 = 1 0.1 x 0.0285 1Conc Cu2+

= 2.85 x 10-2

1

2

3

45

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

2S2O32- + I2 ↔S4O6

2- + 2I-

Click here here for copper determination expt

Cal Amt S2O32-

Cal Conc/Mole/Mass CuCal % Cu by mass in brassCal % error (Lit value = 44.2 % Cu)

% error = Expt value x 100% Lit value= (44.2 – 39.7) x 100% 44.2= 10.2%

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 = 0.040MV = 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 = 2Mb Vb 50.04 x 0.03562 = 2 Mole C2O4

2- 5Mol C2O4

2- = 3.55 x 10 -3

Mass of (expt yield) = 0.476 gMass 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 C2O42- used up at equivalence point – excess KMnO4

- turn purple

?

Oxidizing Agent

Reducing Agent

MnO4- Fe2+

Cr2O72- SO2

HNO3 I-

H2O2 H2S

CI2 SO32-

KIO3 Vitamin C

CIO-/Cu2+ Oxalate/C2O4

2-

MnO4– reduced to Mn2+

C2O42- oxidized to CO2

(+7) ON decrease ↓ (+2)

(+3) ON increase ↑ (+4)

4

M V (KIO3) = 1M V (C6H8O6) 30.002 x 0.0255 = 1Mole C6H8O6 3Mole 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 molKIO3 + 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 = 0.002MV = 25.5ml

Vit CM = ?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 = 25mlM = ?

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 10-4Mole 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+

Cr2O72- SO2

HNO3 I-

H2O2 H2S

CI2 SO32-

KIO3 Vitamin C

CIO-/Cu2+ Oxalate/C2O4

2-

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 = 0.02114MV = 28.64 ml

H2O2

M = ?

2MnO4- + 5H2O2

+ 6H+ 2Mn→ 2+ + 5O2 + 8H2OM = 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 = 2M V 50.02114 x 0.02864 = 2 Mole H2O2

5Mol H2O2

= 1.5135 x 10 -3

Using formula

1

2

?

Oxidizing Agent

Reducing Agent

MnO4- Fe2+

Cr2O72- SO2

HNO3 I-

H2O2 H2S

CI2 SO32-

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 VM 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 = 0.06052 M (Diluted sample)Original sample = 0.06052 x 10

= 0.6052 M

Conc H2O2 = 0.6052MRMM H2O2 = 34Mass H2O2 = 0.6052 x 34 = 20.60g in 1000 ml = 2.06g in 100ml = 2.06%

3

Stronger oxidizing agent reduce weaker oxidizing agent

Cr2O72- 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+ + 11H2OCalculate % by mass of ethanol. Explain how end point is determined?

10

Cr2O72-

M = 0.055MV = 9.25 ml

C2H5 OHM = ?

2Cr2O72- + 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 = 2M V 30.055 x 0.0925 = 2 MV 3Mol 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 RAMMass = 7.63 x 10-3 x 46Mass = 0.351 g

3

Click here practical breath analyzer using dichromate

AlcoholC2H5OH

Ethanoic acidCH3COOH

Cr2O72- – In burette is orange– Turns green react with C2H5OH

All C2H5OH used up at equivalence point – excess Cr2O72- turn orange

oxidized

Dichromate Cr2O7

2-Chromate Cr3+

reduced

Oxidizing Agent

Reducing Agent

MnO4- Fe2+

Cr2O72- SO2

HNO3 I-

H2O2 H2S

CI2 SO32-

KIO3 Vitamin C

CIO-/Cu2+ Ethanol/C2H4OH

?

4

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 microbesLots 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 microbes (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 goodBOD ↓ - 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

Water Quality

Clean Lightly polluted

Moderate polluted

Severelypolluted

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/bloomBlock 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

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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 BODIodometric titration

Biological Oxygen DemandRedox titration (Winkler Method) measure dissolve O2

BOD index

1 mol 2 mol2Mn2+ + 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

Severelypolluted

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 mol2Mn(OH)2 + O2 2MnO(OH)→ 2

2MnO(OH)2 + 8H+ + 6I- → 2I3- + 6H2O

2 mol 2 mol

2I3- + 4S2O3 2- → 6I- + 2S4O6 2-

2 mol 4 mol

Redox titration Winkler Method

DO bottle

Mn2+ salt

1g KI excessalkaline/OH-

shake

White ppt Mn(OH)2

Conc H2SO4

White ppt dissolve in acid

Na2S2O3

M = 0.05MV = 12.5ml

titrated S2O32-

1O2 + 4S2O32- 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 : S2O32-

(1 : 4)1 mol O2 : 4 mol I2 : 4 mol S2O3

2-

1 mol O2 4 mol S2O32-

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 S2O32-

1 mol 2 mol2Mn2+ + 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 excessalkaline/OH-

shake

White ppt Mn(OH)2

Conc H2SO4

White ppt dissolve in acid

Na2S2O3

M = 0.05MV = 12.5ml

titrated S2O32-

1O2 + 4S2O32- 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 : S2O32-

(1 : 4)1 mol O2 : 4 mol I2 : 4 mol S2O3

2-

1 mol O2 4 mol S2O32-

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 S2O32- used is

12.50ml. Calculate dissolved oxygen in g/dm3.

1

Mole S2O32- = 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 S2O32-

2 3 Mole O2 = 1.56 x 10-4 molMass O2 = Mole O2 x RAMMass 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

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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 KHPClick 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)

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

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 electronIncrease ON ↑

Gain electronDecrease ON ↓

Reducing Agent Oxidation

2Br - Br→ 2 + 2e-

Loss electronIncrease ON ↑

Oxidizing Agent Reduction

CI2 + 2e 2CI→ - Gain electron

Decrease ON ↓

Reducing Agent Oxidation

(NH3) -N3- N→ + 3e- Loss electronIncrease ON ↑

Oxidizing Agent Reduction

(CuO) Cu2+ + 2e Cu→Gain electronDecrease ON ↓

Reducing Agent Oxidation

Zn Zn→ 2+ + 2e- Loss electronIncrease ON ↑

Oxidizing Agent Reduction

2H+ + 2e H→ 2

Gain electronDecrease ON ↓

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 rxnlose electron

Zn + 2H+ H→ 2 + Zn2+

Zn Zn→ 2+ + 2e 2H+ + 2e H→ 2

(0) ON increase ↑ (+2)

Zn Zn→ 2+ + 2e2H+ + 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+ + 2elose electron

(0) ON increase ↑ (+2)

Reduction half eqn

Cu2+ + 2e Cu→

(+2) ON decrease ↓ (0)

gain electron

Zn Zn→ 2+ + 2eCu2+ + 2e Cu→Cu2+ + Zn Zn→ 2+ + Cu

Half equations

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+ + 2e2H+ + 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+ + 2elose electron

(0) ON increase ↑ (+2)

Reduction half eqn

Cu2+ + 2e Cu→

(+2) ON decrease ↓ (0)

gain electron Zn Zn→ 2+ + 2eCu2+ + 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++ SO42- Zn→ 2+ + SO4

2- + Cu

Complete full eqn

Complete ionic/redox eqn

spectator ions

Zn + Cu2+ Zn→ 2+ + Cu

Half equations Half equationsZn + Cu2+ Zn→ 2+ + Cu

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+ + 2ePb2+ + 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

+ 2eCI2

+ 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

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 electronIncrease ON ↑

Gain electronDecrease ON ↓

Complete full eqnOxidation 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 agentMnO4

- + 8H+ + 5Fe2+ Mn→ 2+ + 5Fe3+ 4H2O

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 electronIncrease ON ↑

Gain electronDecrease ON ↓

Complete full eqnOxidation 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→ 42- + 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-

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

+ 8H2OOxidizing agent ↓ Reduction

Reducing agent ↓ Oxidation

Oxidizing Agent Reduction

MnO4- + 5e Mn→ 2+

Reducing Agent Oxidation

H2O2 O→ 2

+ 2e- Loss electronIncrease ON ↑

Gain electronDecrease 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-

Cr2O72- + 3NO2

- + 8H+ 2Cr→ 3+ + 3NO3- + 4H2O

Cr2O72- 2Cr→ 3+

Constructing Half and complete redox equations

(+6) (+3)Cr red - ON ↓

(+3) NO2- oxi – ON ↑ (+5)

Cr2O72- + 3NO2

- + 8H+ 2Cr→ 3+ + 3NO3- + 4H2O

Oxidizing agent ↓ Reduction

Reducing agent ↓ Oxidation

Oxidizing Agent Reduction

Cr2O72- + 6e- 2Cr→ 3+

Reducing Agent Oxidation

NO2- NO→ 3

- + 2e- Loss electronIncrease ON ↑

Gain electronDecrease ON ↓

Complete full eqnOxidation half eqnReduction half eqn

1. Balance # O - add H2O

2. Balance # H add H+

3. Balance # charges - add electrons

4. Balance # electron transferx 3x 1

Cr2O72- + 14H+ + 6e- 2Cr→ 3+ + 7H2O

3NO2-+ 3H2O 3NO→ 3

- + 6H+ + 6e-+

Cr2O72- + 3NO2

- + 8H+ 2Cr→ 3+ + 3NO3- + 4H2O

Cr2O72- 2Cr→ 3+ + 7H2O

Cr2O72- + 14H+ 2Cr→ 3+ + 7H2O

Cr2O72- + 14H+ + 6e- 2Cr→ 3+ + 7H2O

Cr2O72- + 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-

Cr2O72- + 6Fe2+ + 14H+ 2Cr→ 3+ + 6Fe3+ + 7H2O

Cr2O72- 2Cr→ 3+

Constructing Half and complete redox equations

(+6) (+3)Cr red - ON ↓

(+2) Fe2+ oxi – ON ↑ (+3)

Cr2O72- + 6Fe2+ + 14H+ 2Cr→ 3+ + 6Fe3+ + 7H2O

Oxidizing agent ↓ Reduction

Reducing agent ↓ Oxidation

Oxidizing Agent Reduction

Cr2O72- + 6e- 2Cr→ 3+

Reducing Agent Oxidation

Fe2+ Fe→ 3+ + e- Loss electronIncrease ON ↑

Gain electronDecrease ON ↓

Complete full eqnOxidation half eqnReduction half eqn

1. Balance # O - add H2O

2. Balance # H add H+

3. Balance # charges - add electrons

4. Balance # electron transferx 6x 1

Cr2O72- + 14H+ + 6e- 2Cr→ 3+ + 7H2O

6Fe2+ 6Fe→ 3+ + 6e-+

Cr2O72- 2Cr→ 3+ + 7H2O

Cr2O72- + 14H+ 2Cr→ 3+ + 7H2O

Cr2O72- + 14H+ + 6e- 2Cr→ 3+ + 7H2O

Cr2O72- + 14H+ + 6e- 2Cr→ 3+ + 7H2O

Cr2O72- + 6Fe2+ + 14H+ 2Cr→ 3+ + 6Fe3+ 7H2O

Fe2+ Fe→ 3+

Fe2+ Fe→ 3+ + e

6Fe2+ 6Fe→ 3+ + 6e

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 electronDecrease ON ↓

Complete full eqnOxidation half eqnReduction half eqn

1. Balance # O - add H2O

2. Balance # H add H+

3. Balance # charges - add electrons

4. Balance # electron transferx 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

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 electronIncrease ON ↑

Gain electronDecrease ON ↓

Complete full eqnOxidation half eqnReduction half eqn

1. Balance # O - add H2O

2. Balance # H add H+

3. Balance # charges - add electrons

4. Balance # electron transferx 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

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 electronIncrease ON ↑

Gain electronDecrease ON ↓

Complete full eqnOxidation half eqnReduction half eqn

1. Balance # O - add H2O

2. Balance # H add H+

3. Balance # charges - add electrons

4. Balance # electron transferx 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-

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 electronIncrease ON ↑

Gain electronDecrease ON ↓

Complete full eqnOxidation half eqnReduction half eqn

1. Balance # O - add H2O

2. Balance # H add H+

3. Balance # charges - add electrons

4. Balance # electron transferx 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

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 electronIncrease ON ↑

Gain electronDecrease ON ↓

Complete full eqnOxidation 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→ 42- + 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→ 42- + 4H+ + 2e-

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+ + H22Fe2+ + 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+

Cr2O72- 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

Cr2O72- + 3NO2

- + 8H+ 2Cr→ 3+ + 3NO3- + 4H2O

1 2

3

oxidizing agent

oxidizing agent

oxidizing agent

reducing agent

reducing agent

reducing agent

Acknowledgements

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Prepared by Lawrence Kok

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