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Copyright© by Houghton Mifflin Company. All rights reserved. Oxidation-Reduction Reactions (Redox)
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Page 1: Copyright© by Houghton Mifflin Company. All rights reserved. Oxidation-Reduction Reactions (Redox)

Copyright© by Houghton Mifflin Company. All rights reserved.

Oxidation-Reduction Reactions(Redox)

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What is the difference between acid/base What is the difference between acid/base reactions and redox reactions?reactions and redox reactions?

Acid/base reactions – proton transfer (p+)

Redox reactions– electron transfer (e-)

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Flow of electronsFlow of electrons

Electrons respond to differences in potential by moving from the region of high potential to the region of low potential.

-+

High Ep

Low Ep

e-

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Flow of electronsFlow of electrons

Cl

low electronegativity

high electronegativity

e-Li

Lithium loses the e- tug-of-war with chloride.

+-

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TerminologyTerminology

Cations:– positively charged ions– generally metals–NH4

+ is the exceptionAnions:

– negatively charged ions– non-metals–complex ions

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Oxidation: –When a substances loses e-

Reduction: –When a substance gains e-

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Electron Transfer and TerminologyElectron Transfer and Terminology

Lose electrons: Oxidation

Gain electrons: Reduction.

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oxidizedreduced

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Half-reactionsHalf-reactions

Ca(s) → Ca2+(aq) + 2e-

– oxidation half reaction

2H+(aq) + 2e- → H2(g)

– reduction half reaction

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Half-reactions add togetherHalf-reactions add together

Ca(s) → Ca2+(aq) + 2e-

2H+(aq) + 2e- → H2(g)

Ca(s) + 2H+ + 2e- Ca2+ + 2e- + H2(g)

Ca(s) + 2H+(aq) Ca2+

(aq) + H2(g)

+

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Ca(s) + 2H+(aq) Ca2+

(aq) + H2(g)

Ca(s) has lost two e- to 2 H+(aq) to become Ca2+

(aq). Ca(s) has been oxidized to Ca2+

(aq)

At the same time 2 electrons are gained by 2 H+(aq) to

form H2(g) . We say H+(aq) is reduced to H2(g) .

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Half-reactions add togetherHalf-reactions add together

Cu(s) → Cu2+(aq) + 2e-

Ag+(aq) + e- → Ag(s)

Cu(s) + 2Ag+(aq) + 2e-

Cu2+(aq) + 2e- + 2Ag(s)

Cu(s) + 2Ag+(aq) Cu2+

(aq) + 2Ag(s)

+ ( ) x 2

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Iron comes from iron ore which is taken out of the ground by mining.

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The pure iron is obtained by heating the ore at very high temperatures in a furnace with limestone to remove impurities.

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This heap of iron ore pellets will be used in steel production.

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1. Hot blast from Cowper stoves2. Melting zone3. Reduction zone of ferrous oxide4. Reduction zone of ferric oxide5. Pre-heating zone6. Feed of ore, limestone and coke7. Exhaust gases8. Column of ore, coke and limestone9. Removal of slag10. Tapping of molten pig iron11. Collection of waste gases

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Why is gaining electrons called reduction?Why is gaining electrons called reduction?

Reduction originally meant the loss of oxygen from a compound.–2 Fe2O3(s) + C(s) → 4 Fe(s) + 3 CO2(g)

Iron ore is reduced to metallic iron. The size of the pile gets smaller, hence the word reduction.

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Why is losing electrons called oxidation?Why is losing electrons called oxidation?

Oxidation originally meant the combination of an element with oxygen.

– 4 Fe(s) + 3 O2(g) → 2 Fe2O3(g)

– C(s) + O2(g) → CO2(g)

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It Takes Two: Oxidation-ReductionIt Takes Two: Oxidation-Reduction

In all reduction-oxidation (redox) reactions, one species is reduced at the same time as another is oxidized.

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Oxidizing Agent:– the species which causes

oxidation is called the oxidizing agent.

– substances that gains electrons

– the oxidizing agent is always reduced

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Reducing Agent:– the species which causes

reduction is called the reducing agent.

– the reducing agent is always oxidized.

– substances that give up electrons

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Cu(s) + 2 Ag+(aq) → Cu2+

(aq) + Ag(s)oxidated reduced

R.A. O.A.

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1) A redox reaction is a chemical reaction in which electrons are transferred.

2) Number of electrons lost by one species equals number of electrons gained by the other species.

3) Reduction is a process in which e- are gained.

4) Oxidation is a process in which e- are lost

5) A reducing agent donates e- and is oxidized.

6) A oxidizing agent gains e- and is reduced.

WS 15-1

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Electric potential (V), EElectric potential (V), Eoo

Work that must be done to move an electric charge between specified points.

Electric potential differences are measured in volts .

Standard conditions:

– At 25oC with all ions at 1 mol/L concentrations and all gases at 1.00 atm pressure

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Standard Reduction PotentialsStandard Reduction Potentials

We cannot measure the potential of an individual half-cell!

We assign a particular cell as being our reference cell and then assign values to other electrodes on that basis. ( H2 – half cell )

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[H+] = 1.00 mol/L

H2 (g)

e-

Pt gauze

The Standard Hydrogen electrode

• Eo (H+(aq)/H2(g)) half-cell = 0.000 V

p{H2(g)} = 1.00 atm

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Electric potential (V), EElectric potential (V), Eoo

If the net potential is a positive number then the reaction is spontaneous. Products are favoured.

If the net potential is a negative number then the reaction is non-spontaneous. Reactants are favoured.

Half cell potentials are not doubled or tripled as per balancing. We are only comparing potentials.

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Only one of these two reactions is possible. Only one of these two reactions is possible. Which one?Which one?

Cu(s) + 2 Ag+(aq) → Cu2+

(aq) + 2 Ag(s)

Cu2+(aq) + 2 Ag(s) → Cu(s) + 2 Ag+

(aq)

Use data table values, electrical potential, on page 7 of your data

books. (2009)

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Compare the two half reactions that Compare the two half reactions that make up the reaction.make up the reaction.

Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+

(aq

Cu2+(aq)

+ 2e- → Cu (s)

Eo = +0.34

2Ag(s) → 2Ag+(aq)

+ 2e- Eo = -0.80

Cu2+(aq) + 2Ag(s) → Cu(s) + 2Ag+

(aq)

Eo = -0.46

Negative potential, non-spontaneous

+

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Compare the two half reactions that Compare the two half reactions that make up the reaction.make up the reaction.

Cu(s) + 2Ag+(aq)

→ Cu2+(aq) + 2Ag(s)

Cu(s) → Cu2+(aq) + 2e- Eo = -0.34

2Ag+(aq) + 2e- → 2Ag(s) Eo = +0.80

Cu(s) + 2 Ag+(aq)

→ Cu2+(aq) + 2 Ag(s)

Eo = +0.46

Positive potential, spontaneous

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ProblemProblem

Write the oxidation/reduction half reactions and the net ionic equation when zinc is placed in Ni(NO3)2 solution. Identify the O.A. and R.A. and state if the reaction is spontaneous or non-spontaneous.

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ProblemProblem

Ni(NO3)2 → Ni2+(aq) + 2NO3-

(aq)

Zn(s) + Ni2+(aq) → ?

Oxidation: Zn(s) → Zn2+(aq) + 2e- +0.76

Reduction: Ni2+(aq) + 2e- → Ni(s) - 0.26

Spectator ion

Add half reactions

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ProblemProblem

Zn(s) + Ni2+(aq) → Zn2+

(aq) + Ni(s) +0.50

R.A. O.A.Positive potential, spontaneous

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SOA

SRA

high attraction for electrons

low attraction for electrons

decreasing strength

decreasing strength

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Spontaneous shortcutSpontaneous shortcut

Locate the O.A. on the left and the R.A. on the right of the table. If the O.A. is higher up on the table than the R.A. then the reaction is spontaneous.

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ProblemProblem

Explain what happens when nickel is placed in a zinc nitrate solution.

Ni(s) + Zn2+(aq) →

R.A. O.A.

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Zn +2 (aq) is the strongest Oxidizing agent and therefore is Reduced

Zn+2 (aq) + 2e Zn (s)Ni(s) is the strongest Reducing agent andTherefore is Oxidized Ni(s) Ni +2 (aq) + 2e (must reverse)

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NET REDOX REACTIONNET REDOX REACTION

ADD REACTIONS/REMOVE ELECTRONSZn2+ (aq) + 2e Zn(s) - 0.76 VNi (s) Ni 2+ (aq) + 2e + 0.26 V (r)

Zn2+ (aq) Ni(s) Zn(s) + Ni2+ (aq) - 0.50VNON SPONTANEOUS REACTION

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On the tableOn the table

Ni(s)

Zn2+(aq)

R.A. is above the O.A.

Non-spontaneous

WS 15-23

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DISPROPORTIONATIONDISPROPORTIONATION

redox reactions in which the oxidizing agent and the reducing agent are the same species

2 H2O2 (l) ------> 2 H2O (l) + 2 O2 (g)

O ( -1) O ( -2) O ( 0)Oxygen -------- > reduced / OAOxygen ---------------------> oxidized / RA

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Predicting redox reactionsPredicting redox reactions1) List all species present.2) Choose the strongest oxidizing and reducing

agent. Watch for acids ( H +) Also water H2O3) Write the reduction half reaction, as written in

the data book. 4) Write the oxidation half reaction, reverse the

equation in the data book.5) Balance number of electrons.6) Add the two half reactions together to form the

net ionic equation.7) Predict if reaction is spontaneous or not.

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ProblemsProblems

A mixture of bromine gas and chlorine gas is added to a solution of copper (II) sulphate and a copper strip.

Br2(g)

Cl2(g)

H20(l)

Cu2+(aq)

Cu(s)

SO42-

(aq)

RA

OACl2(g) + 2e- → 2 Cl-

(aq)

Cu(s) → Cu2+(aq) + 2e-

Cl2(g) + Cu(s) → 2 Cl-(aq) + Cu2+

(aq)

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ProblemsProblems

Lead is placed in a zinc nitrate solution.

NO3-(aq)

H20(l)

Zn2+(aq)

Pb(s)RA

OA

Non-spontaneous

OA is below RA

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ProblemsProblemsA few drops of Hg(l) are dropped into a solution

which is 1.0 M in both sulphuric acid and potasium permanganate. MnO4

-(aq)

SO42-

(aq)

H20(l)

K+(aq)

Hg(l)

H+(aq)

RA

OA

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ProblemsProblemsA few drops of Hg(l) are droped into a solution

which is 1.0 M in both sulphuric acid and potasium permanganate.

MnO4-(aq) + 8 H+

(aq) + 5e- → Mn2+(aq) + 4 H2O(l)

Hg(l) → Hg2+(aq) + 2e-

2MnO4-(aq) + 16H+

(aq) + 5Hg(l) → 2Mn2+(aq) + 8H2O(l) + 5Hg2+

(aq)

( ) x2 ( ) x5

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General RulesGeneral Rules

Metal (+) ions are oxidizing agents.Nonmetal (-) ions are reducing

agents.Metal elements are reducing agents.Nonmetal elements are oxidizing

agents.

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Building a redox table Building a redox table (method one)(method one) One can use experimental evidence to determine the

relative strengths of oxidizing and reducing agents. The greater the number of spontaneous reactions, the

stronger the oxidizing agent. This means we can rank oxidizing agents according to the

number of spontaneous reactions. By convention the strongest oxidizing agent is at the top

left in a redox table and the strongest reducing agent is at the bottom right of the table.

SOA

SRA

Reduction Table

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Problem: Make a redox tableProblem: Make a redox table

Cu(s) Mg(s) Ag(s) Zn(s)

Cu2+(aq) ____ ____ ____ ____

Mg2+(aq) ____ ____ ____ ____

Ag+(aq) ____ ____ ____ ____

Zn2+(aq) ____ ____ ____ ____

√√

√√

Virtual Lab

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REDOX TABLEREDOX TABLEMg(s) is most reactive - 3 timesZn(s) is second - 2 timesCu(s) is third - 1 timesAg(s) is most unreactive 0 timesTHEREFORE Mg oxidizes easiest and is the strongest reducing agentMg is placed in the lower right hand side then Zn(s) …Cu(s) and Ag(s) is last.Now you fill in the reduction reactions

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Activity Series - Redox TableActivity Series - Redox Table

Ag+(aq) + 1e- Ag(s)

Cu2+(aq) + 2e- Cu(s)

Zn2+(aq) + 2e- Zn(s)

Mg2+(aq) + 2e- Mg(s)

Strongest reducing agents

Weaker

Strongest oxidizing agents

Weaker

Mg is the strongest reducing agent as it oxidizes the most and is on the lower right side of the table.

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Redox Table BuildingRedox Table Building (method two)(method two)

The spontaneity rule is used to order the oxidizing agents to produce a redox table.

Consider the following redox equations which represent spontaneous reactions from an experiment. From this evidence construct a redox table.

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Redox Table BuildingRedox Table Building3 Equations given3 Equations given

all are spontaneous reactionsall are spontaneous reactions

Co(s) + Pd2+(aq) → Co2+

(aq) + Pd(s)

Pd(s) + Pt2+(aq) → Pd2+

(aq) + Pt(s)

Mg(s) + Co2+(aq) → Mg2+

(aq) + Co(s)

Work with one equation at a time.

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Redox Table BuildingRedox Table Building

Co(s) + Pd2+(aq) → Co2+

(aq) + Pd(s)

Pd2+(aq) + 2 e- → Pd(s) reduced/stays as is

Co2+(aq) + 2 e- → Co(s) oxidized/reverse

OA is above RAReverse the oxidation reaction to

compare Pd(s) with Co(s)

RA OA

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NOTE POSITION OF THE REDUCING NOTE POSITION OF THE REDUCING AGENTSAGENTS

Pt(s) is above Pd(s) in position

Pt(s)

Pd(s) stronger reducing agent

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Redox Table BuildingRedox Table Building

Pd(s) + Pt2+(aq) → Pd2+

(aq) + Pt(s)

Pt2+(aq) + 2 e- → Pt(s)

Pd2+(aq) + 2 e- → Pd(s)

OA is above RAspontaneous reaction

OARA

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NOTE POSITION OF THE REDUCINGNOTE POSITION OF THE REDUCING AGENTS AGENTS

Pt(s) is above Pd(s) in position

Pt (s) Pd (s)

Stronger reducing agent

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Redox Table BuildingRedox Table Building

Mg(s) + Co2+(aq) → Mg2+

(aq) + Co(s)

Co2+(aq) + 2 e- → Co(s) ( stays as is)

Mg2+(aq) + 2 e-→ Mg(s) (reverse)

OA is above RAspontaneous reaction

OARA

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MAKING THE TABLEMAKING THE TABLE

USING THE 4 REDUCING REAGENTS Mg(s) Pt(s) Co(s) Pd(s) Place into the correct order using the

previous information you collecteNote the position with respect to each

other….Make the complete ½ reactions MAKE THE REDOX TABLE

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Redox Table BuildingRedox Table Building

Pt2+(aq) + 2 e- → Pt(s)

Pd2+(aq) + 2 e- → Pd(s)

Co2+(aq) + 2 e- → Co(s)

Mg2+(aq) + 2 e-→ Mg(s)

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Oxidation StatesOxidation States

Some reactions are not adequately explained with redox theories.

Chemists have developed a method of electron bookkeeping to describe the redox of molecules and complex ions.

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Oxidation StatesOxidation StatesOxidation state:

– apparent net charge that an atom would have if electron pairs belonged entirely to the more electronegative atom

Oxidation number:– a positive or negative number

assigned to a combined atom according to a set of arbitrary numbers.

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Assigning Oxidation NumbersAssigning Oxidation Numbers

1) Oxidation numbers for all uncombined elements (elemental/standard) = 0

• K(s) = 0 N2(g) = 0 S8(s) = 0

2) Oxidation number for all simple ions is equal to the charge of the ion.

• Br1-(aq) = -1 Fe3+

(aq) = +3

3) Oxidation for oxygen in a compound = -2 (except for peroxides = -1)

H2O(l) H2O2(l)-2 -1

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Assigning Oxidation NumbersAssigning Oxidation Numbers

4) Hydrogen in compounds = +1 H2O(l) (except hydrides = -1) NaH(s)

5) Sum of oxidation numbers in a compound is = 0

• H2O(l) → (2 x +1) + (1 x -2) = 0

6) Sum of oxidation numbers in a complex ion = charge of ion.

• NH4+

(aq) → (4 x +1) + (1 x -3) = +1

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ExampleExample

What is the oxidation number for carbon in CO3

2-(aq) ?

CO# + 3 OO# = -2 ? + 3 (-2) = -2 ? + -6 = -2 ? = +4

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ExampleExample

What is the oxidation number for carbon in C6H12O6 ?

6 CO# + 12 HO# + 6 OO# = 0 6 (?) + 12 (+1) + 6 (-2) = 0 6 (?) + 12 + -12 = 0 ? = 0

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If you have 2 unknowns?If you have 2 unknowns?

First ionize the substance in water. Then work out the two resulting ions separately.

Example: CuSO4(aq)

CuSO4(s) → Cu2+(aq) + SO4

2-(aq)

Simple ion

+2

Solve as a complex ion

4(-2) + 1(x) = -2

x = +6

+6 -2

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Assign oxidation numbers to chlorine in each of the Assign oxidation numbers to chlorine in each of the

following chemicals.following chemicals.

HCl(aq)

Cl2(g)

NaClO (s)

Cl-(aq)

HClO3(aq)

ClO3(aq)

-10+1-1+5+6

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Who cares about oxidation numbers?Who cares about oxidation numbers?

Determining oxidation numbers allows us to predict electron transfer.

If there is an increase in oxidation number then oxidation occurs.

If there is a decrease in oxidation number then reduction occurs.

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ProblemProblem

Determine the oxidation numbers for all atoms and ions in the following redox equation and indicate which substance is undergoing oxidation and reduction.

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ProblemProblem

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)

+1

-4

0 -2

+4

-2

+1

C is oxidized

O is reduced

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Identifying Redox Reactions….Which of these are Redox reactions?

1) N2O4(g) → 2NO2(g)

2) Cl2(g) + 2NaBr(aq) → 2 NaCl(aq) + Br2(l)

3) PbCl2(aq)+ K2SO4(aq) → 2KCl(aq) + PbSO4(aq)

4) 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

5) 2K(s) + 2H2O(l) → 2KOH(aq) + H2(g)

+4 -2 +4 -2

0 +1 -1 +1 -1 0

NO

YES

+2 -1 +1 +6 -2 +1 -1 +2 +6 -2

NO

+1 -2 +1 +1 +6 -2 +1 +6 -2 +1 -2

NO

0 +1 -2 +1 -2+1 0

YES

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Ion electron methodIon electron method

Under Acidic conditionsUnder Acidic conditions

1. Identify oxidized and reduced species

Write the half reaction for each.

2. Balance the half rxn separately except H & O’s.

Balance: Oxygen by adding H2O

Balance: Hydrogen by adding H+

Balance: Charge by adding e -

3. Multiply each half reaction by a coefficient.

Must have the same # of e- in both half-rxn.

4. Add the half-rxn together, the e - should cancel.

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Balancing Half ReactionsBalancing Half Reactions

MnO4¯ → Mn2+

MnO4¯ → Mn2+ + 4 H2O

8 H+ + MnO4¯ → Mn2+ + 4 H2O

5 e¯ + 8 H+ + MnO4¯ → Mn2+ + 4 H2O

Note All elements balance

Note The charge is balanced / LHS 2 + = RHS 2 +

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Balancing Half ReactionsBalancing Half Reactions

NO3¯ → HNO2

NO3¯ → HNO2 + H2O

3 H+ + NO3¯ → HNO2 + H2O

2 e¯ + 3 H+ + NO3¯ → HNO2 + H2O

NOTE : CHARGE IS BALANCED / LHS = RHS ( 0)

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Example: Acidic ConditionsExample: Acidic ConditionsI- + S2O8

-2 I2 + S2O42-

Oxidation: I- I2

Reduction:S2O8-2 S2O4

2-

Bal. chemical and e- : 2 I- I2 + 2 e-

Bal. chemical O and H : 8e- + 8H+ + S2O8-2 S2O4

2- + 4H2O

Mult 1st rxn by 4: 8I- 4 I2 + 8e-

Add rxn 1 & 2: 8I- 4 I2 + 8e-

8e- + 8H+ + S2O8-2 S2O4

2- + 4H2O

8I- + 8H+ + S2O8-2 4 I2 + S2O4

2- + 4H2O

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Example: Acidic ConditionsExample: Acidic ConditionsNO3

- + Bi NO2 + Bi3+

Half Rxn (oxid): Bi Bi3+

Half Rxn (red): NO3- NO2

Bal. chemical and e- : Bi Bi3+ + 3 e-

Bal. chemical O and H : 1e- + 2H+ + NO3- NO2

+ H2OMult 2nd rxn by 3: 3e- + 6H+ + 3NO3

- 3NO2 + 3H2O

Add rxn 1 & 2: Bi Bi3+ + 3 e-

3 e- + 6 H+ + 3 NO3- 3NO2 + 3H2O

Bi + 6 H+ + 3 NO3- Bi3+

+ 3 NO2 + 3 H2O

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1. Procedure identical to that under acidic conditions1. Procedure identical to that under acidic conditionsBalance the half rxn separately except H & O’s.

Balance Oxygen by adding H2O

Balance Hydrogen by H+ (ADD OH - TO BOTH SIDES )

Balance charge by e- ( NEUTRALIZE H + ) result is OH - AND H + => H2O (water)

2. Mult each half rxn such that both half- rxn have same 2. Mult each half rxn such that both half- rxn have same number of electrons – number of electrons – MUST CANCEL OUT e MUST CANCEL OUT e --

3. Add the half-rxn together, the e3. Add the half-rxn together, the e-- should cancel. should cancel.

Redox Reactions - Ion electron method.Redox Reactions - Ion electron method.Under Basic conditions/ Under Basic conditions/ NOT PART OF PROGRAMNOT PART OF PROGRAM

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Example: Basic ConditionsExample: Basic Conditions

H2O2 (aq) + Cr2O7-2

(aq ) Cr 3+ (aq) + O2 (g)

red: 6e- + 14H+ + Cr2O7-2

(aq) 2Cr3+ + 7 H2O

oxid: (H2O2 (aq) O2 + 2H+ + 2e-) x 3

8 H+ + 3H2O2 + Cr2O72- 2Cr+3 + 3O2

+ 7H2O

add: 8 OH- 8 OH-

3H2O2 + Cr2O72 - + 8H2O 2Cr+3 + 3O2 + 7H2O + 8OH-

3H2O2 + Cr2O72 - + H2O 2Cr+3 + 3O2 + 8OH-

+3+6 0-1

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Redox StoichiometryRedox StoichiometryCan be used to predict or analyze a chemical

reaction.A method of reacting a solution with a known

concentration with a solution of unknown concentration.

Common oxidizing agents in redox reactions– MnO4

-(aq) → Mn2+

(aq)

– purple colorless– Cr2O7

2-(aq) → Cr3+

(aq)

– orange green

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Redox StoichiometryRedox Stoichiometry

In a titration experiment all of the Br-

(aq) ions in an acidic solution were oxidized to Br2(l) by a 0.0200 M KMnO4(aq) solution. The volume of Br-

(aq) solution was 25.0 mL and the volume of KMnO4(aq) was 15.0 mL. Calculate the concentration of Br-

(aq) ions in solution.

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MAKE HALF REACTIONSMAKE HALF REACTIONS

KMnO4 K+ MnO4- H+ Br- ions

FIND SOA AND SRA / MAKE ½ RXNMnO4

- + H+ is the SOA

Br- is the SRA

Make half reactions / data book P. 11

continue next slide -- >

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We need a balanced chemical equation to do any We need a balanced chemical equation to do any stoichiometry.stoichiometry.

MnO4-(aq) + 8 H+

(aq) + 5e- ↔ Mn2+(aq) + 4 H2O(l)

2Br–(aq) ↔ Br2(l) + 2e-

( ) x2

( ) x 5

2 MnO4-(aq) + 16 H+

(aq) + 10 Br-(aq) ↔ 2 Mn2+

(aq) + 8 H2O(l) +5 Br 2(l)

0.0200 mol/L

0.015 L

0.0003 mol

0.025 L

0.0015 mol

C = n / V

c = 0.0600 M

U 10 Br- = x K 2 MnO4 - 0.0003 mole

Br- = 0.0015 mole

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BreathalyzerBreathalyzer

The Breathalyzer device contains: – A system to sample the breath of the suspect – Two glass vials containing the chemical

reaction mixture – A system of photocells connected to a meter to

measure the color change associated with the chemical reaction

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BreathalyzerBreathalyzer

To measure alcohol, a suspect breathes into the device. The breath sample is bubbled in one vial through a mixture of sulfuric acid, potassium dichromate, silver nitrate and water. The principle of the measurement is based on the following chemical reaction:

8H+ + Cr2O72- + 3C2H5OH → 2Cr3+ + 3C2H4O + 7H2O

yellow blue

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The sulfuric acid removes the alcohol from the air into a liquid solution.

The alcohol reacts with potassium dichromate to produce: – chromium sulfate – potassium sulfate – acetic acid – water

The silver nitrate is a catalyst, a substance that makes a reaction go faster without participating in it. The sulfuric acid, in addition to removing the alcohol from the air, also might provide the acidic condition needed for this reaction.

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During this reaction, the reddish-orange dichromate ion changes color to the green chromium ion when it reacts with the alcohol; the degree of the color change is directly related to the level of alcohol in the expelled air. To determine the amount of alcohol in that air, the reacted mixture is compared to a vial of unreacted mixture in the photocell system, which produces an electric current that causes the needle in the meter to move from its resting place. The operator then rotates a knob to bring the needle back to the resting place and reads the level of alcohol from the knob -- the more the operator must turn the knob to return it to rest, the greater the level of alcohol.

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yellow blue

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Bleaching AgentsBleaching Agents

Bleaching agents are compounds which are used to remove color from substances such as textiles. In earlier times textiles were bleached by exposure to the sun and air. Today most commercial bleaches are oxidizing agents, such as sodium hypochlorite (NaOCl) or hydrogen peroxide (H2O2) which are quite effective in "decolorizing" substances via oxidation.

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Bleaching AgentsBleaching Agents

The action of these bleaches can be illustrated in the following simplified way:

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Bleaching AgentsBleaching AgentsThe decolorizing action of bleaches is due in

part to their ability to remove electrons which are activated by visible light to produce the various colors. The hypochlorite ion (OCl-), found in many commercial preparations, is reduced to chloride ions and hydroxide ions forming a basic solution as it accepts electrons from the colored material as shown below.

OCl- + 2e- + HOH → Cl- + 2 OH-

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Bleaching AgentsBleaching Agents

Bleaches are often combined with "optical brighteners". These compounds are quite different from bleaches. They are capable of absorbing wavelengths of ultraviolet light invisible to the human eye, and converting these wavelengths to blue or blue-green light. The blue or blue-green light is then reflected by the substance making the fabric appear much "whiter and brighter" as more visible light is seen by the eye.

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PhotosynthesisPhotosynthesis

An example of naturally-occuring biological oxidation-reduction reactions is the process of photosynthesis. It is a very complex process carried out by green plants, blue-green algae, and certain bacteria. These organisms are able to harness the energy contained in sunlight, and via a series of oxidation-reduction reactions, produce oxygen and sugar. The overall equation for the photosynthetic process may be expressed as:

6 CO2(g) + 6 H2O(l) → C6H12O 6(aq) + 6 O2(g)

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PhotosynthesisPhotosynthesis

The equation is the net result of two processes. One process involves the splitting of water. This process is really an oxidative process that requires light, and is often referred to as the "light reaction". This reaction may be written as:

12 H2O(l) → 6 O2(g) + 24 H+(aq) + 24e-

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PhotosynthesisPhotosynthesis

Think of the light reaction, as a process by which organisms capture and store radiant energy as they produce oxygen gas. This energy is stored in the form of chemical bonds of compounds such as NADPH and ATP.

The energy contained in both NADPH and ATP is then used to reduce carbon dioxide to glucose. This reaction, shown below, does not require light, and it is often referred to as the dark reaction.

6 CO2 + 24 H+ + 24 e- → C6H12O6 + 6 H2O

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PhotosynthesisPhotosynthesis

The chemical bonds present in glucose also contain a considerable amount of potential energy. This stored energy is released whenever glucose is broken down to drive cellular processes.

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PhotosynthesisPhotosynthesisIn simplest terms, the process of

photosynthessis can be viewed as one-half of the carbon cycle. In this half, energy from the sun is captured and transformed into nutrients which can be utilized by higher organisms in the food chain. The release of this energy during the metabolic re-conversion of glucose to water and carbon dioxide represents the second half of the carbon cycle and it may be referred to as "oxidative processes".

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Cellular Respiration

C6H12O6(aq) + 2 O2(g) → Energy + CO2(g) + 2 H2O(g)

+1

0

0 -2

+4

-2

+1

C is oxidized

O is reduced

-2