Redox Reactions oxidation reduction reactions. Ch 22 sec 1 From the combustion of gasoline to the...

Redox Reactions

oxidation reduction reactions

Ch 22 sec 1From the combustion of gasoline

to the metabolism of food - oxidation is responsible

Oxidation - Loss of electronsold definition - gain of oxygen

Reduction - Gain of electronsold definition - loss of oxygen

ONE DOES NOT OCCUR WITHOUT THE OTHER

Ch 22 sec 1From the combustion of gasoline

to the metabolism of food - oxidation is responsible

Oxidation - Loss of electronsold definition - gain of oxygen

Reduction - Gain of electronsold definition - loss of oxygen

ONE DOES NOT OCCUR WITHOUT THE OTHER

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are needed to see this picture.

LEO the lion goes GERLEO the lion goes GER

Losing Electrons is OxidationGaining Electrons is ReductionExample

Mg + S ---> Mg2+ + S2-

Magnesium is oxidized (aka reducing agent)

Sulfur is reduced (aka oxidizing agent)

Losing Electrons is OxidationGaining Electrons is ReductionExample

Mg + S ---> Mg2+ + S2-

Magnesium is oxidized (aka reducing agent)

Sulfur is reduced (aka oxidizing agent)

Redox reactions are usually presented as two

components

Redox reactions are usually presented as two

componentsMg ----> Mg2+ + 2e-

S + 2e- ---> S2-

Identifying transfers of electrons is easy for ionic reactions.

What about covalent where there is not a transfer of electrons but a sharing?

Mg ----> Mg2+ + 2e-

S + 2e- ---> S2-

Identifying transfers of electrons is easy for ionic reactions.

What about covalent where there is not a transfer of electrons but a sharing?

Consider 2H2 + O2 ---> 2H2O

Consider 2H2 + O2 ---> 2H2O

Which element is reduced and oxidized? Explain

Oxygen is the electron hog. The partial gain of electrons means it is reduced and hydrogen’s partial loss means it is oxidized.

Which element is reduced and oxidized? Explain

Oxygen is the electron hog. The partial gain of electrons means it is reduced and hydrogen’s partial loss means it is oxidized.

Your Turn4Fe + 3O2 ---> 2Fe2O3

Your Turn4Fe + 3O2 ---> 2Fe2O3

Write the 2 redox reactions. Which is oxidized and with is reduced.

answerFe --> Fe3+ + 3e- oxidized (reducing agent)

O2 + 2e- ---> O2- reduced (oxidizing agent)

Corrosion occurs more rapidly in the presence of salts and acids. Why?

These are conducting solutions that make electron transfer easier.

Write the 2 redox reactions. Which is oxidized and with is reduced.

answerFe --> Fe3+ + 3e- oxidized (reducing agent)

O2 + 2e- ---> O2- reduced (oxidizing agent)

Corrosion occurs more rapidly in the presence of salts and acids. Why?

These are conducting solutions that make electron transfer easier.

Sometimes Corrosion is good.

Sometimes Corrosion is good.

Aluminum oxidizes to form tightly packed aluminum oxide particles. This is a protective covering.

Iron needs to be coated to protect it because the oxide is not tightly packed.

Aluminum oxidizes to form tightly packed aluminum oxide particles. This is a protective covering.

Iron needs to be coated to protect it because the oxide is not tightly packed.

Ch 22 sec 2Assigning Oxidation Numbers

Ch 22 sec 2Assigning Oxidation Numbers

Redox equations will be balanced using oxidation numbers

Rules1. The oxidation number of a monatomic

ion is equal to its ionic charge.2. The oxidation number of hydrogen in a

compound is +1 except in a metal hydride. Example - NaH H is -1

3. The oxidation number for oxygen is -2 unless in a peroxide H2O2 where it is -1

Redox equations will be balanced using oxidation numbers

Rules1. The oxidation number of a monatomic

ion is equal to its ionic charge.2. The oxidation number of hydrogen in a

compound is +1 except in a metal hydride. Example - NaH H is -1

3. The oxidation number for oxygen is -2 unless in a peroxide H2O2 where it is -1

more rulesmore rules

The oxidation number of an uncombined atom (elemental form) is 0.

For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal 0.

For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion.

The oxidation number of an uncombined atom (elemental form) is 0.

For any neutral compound, the sum of the oxidation numbers of the atoms in the compound must equal 0.

For a polyatomic ion, the sum of the oxidation numbers must equal the ionic charge of the ion.

What is the oxidation number of elements in the following

compounds?

What is the oxidation number of elements in the following

compounds?1. SO2

S is +4 and O is -2

2. K2SO4

K is +1 S is +6 O is -2

1. SO2

S is +4 and O is -2

2. K2SO4

K is +1 S is +6 O is -2

Oxidation numbers change in redox reaction. No change no

redox.

Oxidation numbers change in redox reaction. No change no

redox.An increase in the oxidation

number indicates oxidationA decrease in the oxidation

number indicates reductionExample - What is being oxidized

and what is being reduced?2AgNO3 + Cu --> Cu(NO3)2 + 2Ag

+1 +5 -2 0 +2 +5 -2 0

An increase in the oxidation number indicates oxidation

A decrease in the oxidation number indicates reduction

Example - What is being oxidized and what is being reduced?2AgNO3 + Cu --> Cu(NO3)2 + 2Ag

+1 +5 -2 0 +2 +5 -2 0

Ch 22 sec 3Classifying Reactions

Ch 22 sec 3Classifying Reactions

Either electrons are transferred or they are not.

Redox reactions include single-replacement, combination, decomposition and combustion.

Others - double replacement and acid/base reactions.

Color changes signify redox reactionsvideo example

Either electrons are transferred or they are not.

Redox reactions include single-replacement, combination, decomposition and combustion.

Others - double replacement and acid/base reactions.

Color changes signify redox reactionsvideo example

another redox clipanother redox clip



Balancing Redox Reactions

Two Methods

Balancing Redox Reactions

Two MethodsMany reactions are too complex to be

balanced by trial and error. All reactions should be balanced in this

manner. You were taught to balance atoms because for simple reactions it works. Balancing atoms may balance the equation and may not be the correct balanced equation. Atoms along with charges need to be balanced.

Many reactions are too complex to be balanced by trial and error.

All reactions should be balanced in this manner. You were taught to balance atoms because for simple reactions it works. Balancing atoms may balance the equation and may not be the correct balanced equation. Atoms along with charges need to be balanced.

oxidation-number-change method

oxidation-number-change method

balanced by comparing the increase and decrease of oxidation numbers

Example:Step 1 - Assign Oxidation Numbers +1 +6 -2 +1-2 0 +1-2+1 +3 -2

+4-2

K2Cr2O7 + H2O + S --> KOH + Cr2O3 + SO2

balanced by comparing the increase and decrease of oxidation numbers

Example:Step 1 - Assign Oxidation Numbers +1 +6 -2 +1-2 0 +1-2+1 +3 -2

+4-2


+1 +6 -2 +1-2 0 +1-2+1 +3 -2 +4-2


Step 2 - Identify which atoms are oxidized and which are reduced.

Cr is reduced and S is oxidizedOnto step 3Use a bracket line to connect the atoms

that undergo oxidation and another line to connect those that undergo reduction. Then write the oxidation-number change at the midpoint of each line.

+1 +6 -2 +1-2 0 +1-2+1 +3 -2 +4-2


Step 2 - Identify which atoms are oxidized and which are reduced.

Cr is reduced and S is oxidizedOnto step 3Use a bracket line to connect the atoms

that undergo oxidation and another line to connect those that undergo reduction. Then write the oxidation-number change at the midpoint of each line.

+1 +6 -2 +1-2 0 +1-2+1 +3 -2 +4-2


Step 4 - Make the total increase in oxidation number equal to the total decrease in oxidation number by using appropriate coefficients.


The coefficient is the number of atoms needed. 2 K2Cr2O7 + H2O + 3S --> KOH + 2Cr2O3 + 3SO2

+1 +6 -2 +1-2 0 +1-2+1 +3 -2 +4-2


Step 4 - Make the total increase in oxidation number equal to the total decrease in oxidation number by using appropriate coefficients.


The coefficient is the number of atoms needed. 2 K2Cr2O7 + H2O + 3S --> KOH + 2Cr2O3 + 3SO2

-3

+4

(4)(-3)=-12

(3)(+4)=+12

Step 5 - Finish balancing by inspection 2K2Cr2O7 + 2H2O + 3S --> 4KOH + 2Cr2O3 + 3SO2

Your Turn - Balance using the oxidation-number-change

method.

As2O3 + Cl2 + H2O --> H3AsO4 + HCl (2)(-1)=-2 +3 -2 0 +1 -2 +1 +5 -2 +1-1

As2O3 + Cl2 + H2O --> H3AsO4 + HCl (1)(+2)=+2

As2O3 + 2Cl2 +5H2O --> 2H3AsO4 + 4HCl

Step 5 - Finish balancing by inspection 2K2Cr2O7 + 2H2O + 3S --> 4KOH + 2Cr2O3 + 3SO2

Your Turn - Balance using the oxidation-number-change

method.

As2O3 + Cl2 + H2O --> H3AsO4 + HCl (2)(-1)=-2 +3 -2 0 +1 -2 +1 +5 -2 +1-1

As2O3 + Cl2 + H2O --> H3AsO4 + HCl (1)(+2)=+2

As2O3 + 2Cl2 +5H2O --> 2H3AsO4 + 4HCl

Using Half-ReactionsUsing Half-Reactions

Good for ionic reactionsTwo equations used - one shows

oxidation the other reduction.Then combined together in the last step.

Good for ionic reactionsTwo equations used - one shows

oxidation the other reduction.Then combined together in the last step.

Balance using 1/2 reactionsExample:

Balance using 1/2 reactionsExample:

KMnO4 + HCl --> MnCl2 + Cl2 + H2O + KClStep 1 - Write the unbalanced equation

in ionic form. Place oxidation numbers above.

+1 +7-2 +1 -1 +2 -1 0 +1 -2 +1 -1 K1+ + MnO4

1- + H1+ + Cl1- --> Mn2+ + 2Cl1- + Cl2 + H2O + K1+ + Cl1-

Step 2 - Write separate 1/2 reactions for the oxidation reduction process.

Cl1- --> Cl2 MnO4

1- --> Mn2+

KMnO4 + HCl --> MnCl2 + Cl2 + H2O + KClStep 1 - Write the unbalanced equation


+1 +7-2 +1 -1 +2 -1 0 +1 -2 +1 -1 K1+ + MnO4

1- + H1+ + Cl1- --> Mn2+ + 2Cl1- + Cl2 + H2O + K1+ + Cl1-

Step 2 - Write separate 1/2 reactions for the oxidation reduction process.

Cl1- --> Cl2 MnO4

1- --> Mn2+

Step 3 - Balance the atoms in each 1/2 reaction. When reactions take place in an acid solution you will need to use water and H+ to balance hydrogen and oxygen atoms.

2Cl1- --> Cl2 MnO4

1- + 8H+ --> Mn2+ + 4H2O

atoms are balanced but charges are not.

Step 4 - Add electrons to one side of each 1/2 reaction to balance the charges.

2Cl1- --> Cl2 + 2e-

MnO41- + 8H+ + 5e- --> Mn2+ + 4H2O


2Cl1- --> Cl2 MnO4

1- + 8H+ --> Mn2+ + 4H2O

atoms are balanced but charges are not.


2Cl1- --> Cl2 + 2e-

MnO41- + 8H+ + 5e- --> Mn2+ + 4H2O

Step 5 - numbers of electrons must be equal. Multiply each reaction by a number to make electrons equal. In this case make electrons equal 10.

10Cl1- --> 5Cl2 + 10e-

2MnO41- + 16H+ + 10e- --> 2Mn2+ + 8H2O

Step 6 - Add the 1/2 reactions to show an overall equation.

10Cl1- + 2MnO41- + 16H+ + 10e- --> 5Cl2 + 10e- + 2Mn2+ + 8H2O

Remove terms that are the same on both sides.

10Cl- + 2MnO4 + 16H+ --> 5Cl2 + 2Mn2+ + 8H2O


10Cl1- --> 5Cl2 + 10e-

2MnO41- + 16H+ + 10e- --> 2Mn2+ + 8H2O


10Cl1- + 2MnO41- + 16H+ + 10e- --> 5Cl2 + 10e- + 2Mn2+ + 8H2O

Remove terms that are the same on both sides.

10Cl- + 2MnO4 + 16H+ --> 5Cl2 + 2Mn2+ + 8H2O

Step 7 - Add spectator ions and balance the equation.

From step 1 K1+ + MnO4

1- + H1+ + Cl1- -->

Mn2+ + 2Cl1- + Cl2 + H2O + K1+ + Cl1-

10Cl- + 2K+ + 2MnO4 + 16H+ + 6Cl- --> 5Cl2 + 2Mn2+ + 4Cl-+ 8H2O + 2K+ + 2Cl-

Summarize spectator and nonspectator ions. 16Cl- + 2K+ + 2MnO4 + 16H+ --> 5Cl2 + 2Mn2+ + 6Cl- + 8H2O + 2K+

The equation is now balanced for atoms and charge.Now you can rewrite it into:2KMnO4 + 16HCl --> 2MnCl2 + 5Cl2 + 8H2O + 2KCl

Step 7 - Add spectator ions and balance the equation.

From step 1 K1+ + MnO4

1- + H1+ + Cl1- -->

Mn2+ + 2Cl1- + Cl2 + H2O + K1+ + Cl1-

10Cl- + 2K+ + 2MnO4 + 16H+ + 6Cl- --> 5Cl2 + 2Mn2+ + 4Cl-+ 8H2O + 2K+ + 2Cl-

Summarize spectator and nonspectator ions. 16Cl- + 2K+ + 2MnO4 + 16H+ --> 5Cl2 + 2Mn2+ + 6Cl- + 8H2O + 2K+

The equation is now balanced for atoms and charge.Now you can rewrite it into:2KMnO4 + 16HCl --> 2MnCl2 + 5Cl2 + 8H2O + 2KCl

Balance using 1/2 reaction method

Your Turn:

Balance using 1/2 reaction method

Your Turn:S + HNO3 --> SO2 + NO + H2OStep 1 - Write the unbalanced equation


0 +1 +5 -2 +4-2 +2-2 +1 -2

S + H+ + NO3- --> SO2 + NO + H2O

Identify what is oxidized and what is reduced.

S + HNO3 --> SO2 + NO + H2OStep 1 - Write the unbalanced equation


0 +1 +5 -2 +4-2 +2-2 +1 -2

S + H+ + NO3- --> SO2 + NO + H2O

Identify what is oxidized and what is reduced.

Step 2 - Write separate half-reactions for the oxidation and reduction process.

S --> SO2

NO3- --> NO


2H2O + S --> SO2 + 4H+

4H+ + NO3- --> NO + 2H2O

note: charges are not balanced

Step 2 - Write separate half-reactions for the oxidation and reduction process.

S --> SO2

NO3- --> NO


2H2O + S --> SO2 + 4H+

4H+ + NO3- --> NO + 2H2O

note: charges are not balanced


2H2O + S --> SO2 + 4H+ + 4e-

4H+ + NO3- + 3e- --> NO + 2H2O


6H2O + 3S --> 3SO2 + 12H+ + 12e-

16H+ + 4NO3- + 12e- --> 4NO + 8H2O


2H2O + S --> SO2 + 4H+ + 4e-

4H+ + NO3- + 3e- --> NO + 2H2O


6H2O + 3S --> 3SO2 + 12H+ + 12e-

16H+ + 4NO3- + 12e- --> 4NO + 8H2O


6H2O + 3S + 16H+ + 4NO3- + 12e- -->

3SO2 + 12H+ + 12e- + 4NO + 8H2ORemove terms that are the same on

both sides.3S + 4H+ + 4NO3

- --> 3SO2 + 4NO + 2H2OStep 7 - Add spectator ions and balance the

equation. This reaction does not have spectator ions. Therefore final answer is:

3S + 4HNO3 --> 3SO2 + 4NO + 2H2O


6H2O + 3S + 16H+ + 4NO3- + 12e- -->

3SO2 + 12H+ + 12e- + 4NO + 8H2ORemove terms that are the same on

both sides.3S + 4H+ + 4NO3

- --> 3SO2 + 4NO + 2H2OStep 7 - Add spectator ions and balance the

equation. This reaction does not have spectator ions. Therefore final answer is:

3S + 4HNO3 --> 3SO2 + 4NO + 2H2O

Electrochem IntroElectrochem Intro

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Ch 23 sec 1Electrochemical Cells

Ch 23 sec 1Electrochemical Cells

When a strip of zinc metal is placed in CuSO4. Electrons are transferred from Zn to Cu.Zn(s) + Cu2+

(aq) --> Zn2+(aq) + Cu(s)

The flow of electrons is an electric current.Any conversion between electrical and

chemical energy is an electrochemical process. The device that converts between the two is an electrochemical cell.

When a strip of zinc metal is placed in CuSO4. Electrons are transferred from Zn to Cu.Zn(s) + Cu2+

(aq) --> Zn2+(aq) + Cu(s)

The flow of electrons is an electric current.Any conversion between electrical and

chemical energy is an electrochemical process. The device that converts between the two is an electrochemical cell.

Voltaic CellsVoltaic Cells

Convert chemical energy into electrical energy

Half-cell is part of the cell where oxidation or reduction takes place.

A half-cell consists of a metal rod or strip immersed in a solution of its ions.

Convert chemical energy into electrical energy

Half-cell is part of the cell where oxidation or reduction takes place.

A half-cell consists of a metal rod or strip immersed in a solution of its ions.

zinc-copper reactionzinc-copper reaction

One 1/2 cell has a zinc rod immersed in zinc sulfate

One 1/2 cell has a copper rod immersed in copper (II) sulfate

Cells are separated by a salt bridgea tube containing a strong electrolyte often K2SO4

A wire carries the electrons in the external circuit from the zinc rod to the copper rod.

The driving force is the spontaneous redox reaction

One 1/2 cell has a zinc rod immersed in zinc sulfate

One 1/2 cell has a copper rod immersed in copper (II) sulfate

Cells are separated by a salt bridgea tube containing a strong electrolyte often K2SO4

A wire carries the electrons in the external circuit from the zinc rod to the copper rod.

The driving force is the spontaneous redox reaction

Electrode - a conductor in a circuit that carries electrons to or from a substance.

Anode - electrode at which oxidation occurs. Electrons are produced at the anode so it it labeled the negative electrode.

Cathode - electrode at which reduction takes place. Electrons are consumed at the electrode. The cathode is labeled the positive electrode.

Neither electrode is really charged. The moving electrons balance any charge that might build up as oxidation and reduction occur.

Electrode - a conductor in a circuit that carries electrons to or from a substance.

Anode - electrode at which oxidation occurs. Electrons are produced at the anode so it it labeled the negative electrode.

Cathode - electrode at which reduction takes place. Electrons are consumed at the electrode. The cathode is labeled the positive electrode.

Neither electrode is really charged. The moving electrons balance any charge that might build up as oxidation and reduction occur.

The electrochemical process in a zinc-copper voltaic cell. These steps occur at

the same time.

The electrochemical process in a zinc-copper voltaic cell. These steps occur at

the same time.

1. Electrons are produced at the zinc rod. Zn --> Zn2+ + 2e- (anode)

2. The electrons leave the the zinc anode and travel through the external circuit to the copper rod. If a bulb is in the circuit it will light.

3. Electrons enter the copper rod and interact with copper ions in solution.Cu2+ + 2e- --> Cu

1. Electrons are produced at the zinc rod. Zn --> Zn2+ + 2e- (anode)

2. The electrons leave the the zinc anode and travel through the external circuit to the copper rod. If a bulb is in the circuit it will light.

3. Electrons enter the copper rod and interact with copper ions in solution.Cu2+ + 2e- --> Cu

To complete the circuit, both positive and negative ions move through the aqueous solutions via the salt bridge.

Summary:Zn --> Zn2+ + 2e-

Cu2+ + 2e- --> CuZn(s) + Cu2+

(aq) --> Zn2+(aq) + Cu(s)

To complete the circuit, both positive and negative ions move through the aqueous solutions via the salt bridge.

Summary:Zn --> Zn2+ + 2e-

Cu2+ + 2e- --> CuZn(s) + Cu2+

(aq) --> Zn2+(aq) + Cu(s)

Dry Cells (batteries)Dry Cells (batteries)

A voltaic cell in which the electrolyte is a paste.

A zinc container is filled with a thick , moist electrolyte paste of manganese (IV) oxide, (MnO2), ZnCl2, NH4Cl and water.

A graphite rod is embedded in the paste.The zinc container is the anode and the

graphite rod is the cathode.

A voltaic cell in which the electrolyte is a paste.

A zinc container is filled with a thick , moist electrolyte paste of manganese (IV) oxide, (MnO2), ZnCl2, NH4Cl and water.

A graphite rod is embedded in the paste.The zinc container is the anode and the

graphite rod is the cathode.

The thick paste and its surrounding paper liner prevent the contents of the cell from freely mixing.

Dry Cell Reactions:Zn --> Zn2+ + 2e-

2MnO2 + 2NH4+ + 2e- --> Mn2O3 + 2NH3 +

H2OThe graphite serves as a conductor

even though it is the cathode. The Mn is actually reduced.

Alkaline battery - same as dry cell but the paste is KOH to prevent buildup of ammonia gas.

The thick paste and its surrounding paper liner prevent the contents of the cell from freely mixing.

Dry Cell Reactions:Zn --> Zn2+ + 2e-

2MnO2 + 2NH4+ + 2e- --> Mn2O3 + 2NH3 +

H2OThe graphite serves as a conductor

even though it is the cathode. The Mn is actually reduced.

Alkaline battery - same as dry cell but the paste is KOH to prevent buildup of ammonia gas.

Lead Storage BatteriesLead Storage Batteries

A battery is actually a group of cells connected together.

A 12 volt car battery consists of 6 cells.One set of grids packed with spongy

lead (anode).The other grid is (cathode) is packed

with PbO2.The grids are immersed in concentrated

sulfuric acid.

A battery is actually a group of cells connected together.

A 12 volt car battery consists of 6 cells.One set of grids packed with spongy

lead (anode).The other grid is (cathode) is packed

with PbO2.The grids are immersed in concentrated

sulfuric acid.

The reactions:Pb + SO4

2- --> PbSO4 + 2e- (oxidation)

PbO2 + 4H+ + SO42- + 2e- --> PbSO4 +

2H2O (reduction)

OverallPb + PbO2 + 2H2SO4 --> 2PbSO4 +

2H2O

The sulfate slowly builds up and concentration of acid decreases.

The car’s generator reverses the reaction.

The reactions:Pb + SO4

2- --> PbSO4 + 2e- (oxidation)

PbO2 + 4H+ + SO42- + 2e- --> PbSO4 +

2H2O (reduction)

OverallPb + PbO2 + 2H2SO4 --> 2PbSO4 +

2H2O

The sulfate slowly builds up and concentration of acid decreases.

The car’s generator reverses the reaction.

Fuel CellsFuel Cells

Cells with renewable electrodes.A voltaic cell in which a fuel substance

undergoes oxidation and from which electrical energy is continuously obtained.

Do not have to be rechargedemits no air pollutants, operates

quieter, more cost effective than an electrical generator.

Cells with renewable electrodes.A voltaic cell in which a fuel substance

undergoes oxidation and from which electrical energy is continuously obtained.

Do not have to be rechargedemits no air pollutants, operates

quieter, more cost effective than an electrical generator.

The hydrogen-oxygen fuel cellThe hydrogen-oxygen fuel cell

3 compartments separated by 2 carbon electrodes.

Oxygen (the oxidizer) is fed into the cathode compartment.

Hydrogen (the fuel) is fed into the anode compartment.

The gases diffuse slowly.

3 compartments separated by 2 carbon electrodes.

Oxygen (the oxidizer) is fed into the cathode compartment.

Hydrogen (the fuel) is fed into the anode compartment.

The gases diffuse slowly.

The electrolyte in the center is a hot, concentrated solution of KOH.

Electrons enter from the oxidation 1/2 reaction at the anode then pass through an external circuit to enter the reduction 1/2 reaction at the cathode.

Summary:2H2 + 4OH- --> 4H2O + 4e- (anode)O2 + 2H2O + 4e- -->4OH- (cathode)

The overall reaction is:2H2 + O2 --> 2H2O

The electrolyte in the center is a hot, concentrated solution of KOH.

Electrons enter from the oxidation 1/2 reaction at the anode then pass through an external circuit to enter the reduction 1/2 reaction at the cathode.

Summary:2H2 + 4OH- --> 4H2O + 4e- (anode)O2 + 2H2O + 4e- -->4OH- (cathode)

The overall reaction is:2H2 + O2 --> 2H2O

Ch 23 sec 2Cell Potentials

Ch 23 sec 2Cell Potentials

Electrical Potential - cells ability to produce an electric current. Measured in Volts.

The electrical potential results from a competition for electrons.

E0 = reduction potentialCell Potential

E0 = E0red - E0

Oxid

Electrical Potential - cells ability to produce an electric current. Measured in Volts.

The electrical potential results from a competition for electrons.

E0 = reduction potentialCell Potential

E0 = E0red - E0

Oxid

Standard Cell Potentialmeasured potential when the ion

concentrations in the half-cells are 1M, gases are at 101 kPa and temperature is 25oC

Half-cell potentials cannot be measured. So, the 1/2 cells are compared to assigning a hydrogen cell 0.00 V where reduction of Hydrogen occurs.

Standard Cell Potentialmeasured potential when the ion

concentrations in the half-cells are 1M, gases are at 101 kPa and temperature is 25oC

Half-cell potentials cannot be measured. So, the 1/2 cells are compared to assigning a hydrogen cell 0.00 V where reduction of Hydrogen occurs.

Reduction takes place at the cathodeOxidation takes place at the anodeCalculating 1/2 cell potential examples:

2 examples

Zinc and Hydrogen half cells.A voltmeter reads +0.76 V.The zinc is oxidized (anode)Hydrogen Ions are reduced (cathode)oxidation Zn --> Zn2+ + 2e-

reduction 2H+ + 2e- --> H2

cell reaction Zn + 2H+ --> Zn2+ + H2

Reduction takes place at the cathodeOxidation takes place at the anodeCalculating 1/2 cell potential examples:

2 examples

Zinc and Hydrogen half cells.A voltmeter reads +0.76 V.The zinc is oxidized (anode)Hydrogen Ions are reduced (cathode)oxidation Zn --> Zn2+ + 2e-

reduction 2H+ + 2e- --> H2

cell reaction Zn + 2H+ --> Zn2+ + H2

therefore,E0 = E0

red - E0Oxid

E0 = E0H+ - E0

Zn2+

+0.76V = 0.00V - E0Zn2+

E0Zn2+ = -0.76V

The value is negative because the tendency for zinc ions to be reduced to zinc metal is less than the tendency of hydrogen ions to be reduced to hydrogen gas. Therefore it is negative because

Zinc is oxidized.

therefore,E0 = E0

red - E0Oxid

E0 = E0H+ - E0

Zn2+

+0.76V = 0.00V - E0Zn2+

E0Zn2+ = -0.76V

The value is negative because the tendency for zinc ions to be reduced to zinc metal is less than the tendency of hydrogen ions to be reduced to hydrogen gas. Therefore it is negative because

Zinc is oxidized.

Second example this time for copper.Reduction Cu2+ + 2e- --> CuOxidation H2 --> 2H+ + 2e-

Cell reaction Cu2+ + H2 --> Cu + 2H+

Cell potential measured at +0.76V.So, E0 = E0

red - E0Oxid

+0.34V = E0Cu2+ - 0.00H+

E0Cu2+ = +0.34V

The potential for Cu ions to be reduced is higher than the potential for H ions to be reduced. Therefore it is positive because

Copper is reduced. Table 23.2 in your book on page 688 has a

list of all the 1/2 cell potentials.

Second example this time for copper.Reduction Cu2+ + 2e- --> CuOxidation H2 --> 2H+ + 2e-

Cell reaction Cu2+ + H2 --> Cu + 2H+

Cell potential measured at +0.76V.So, E0 = E0

red - E0Oxid

+0.34V = E0Cu2+ - 0.00H+

E0Cu2+ = +0.34V

The potential for Cu ions to be reduced is higher than the potential for H ions to be reduced. Therefore it is positive because

Copper is reduced. Table 23.2 in your book on page 688 has a

list of all the 1/2 cell potentials.

Calculating Standard Cell Potentials


To function a cell must be constructed of 2 half-cells.

If the cell potential for a given redox reaction is positive then the reaction is spontaneous.

To function a cell must be constructed of 2 half-cells.

If the cell potential for a given redox reaction is positive then the reaction is spontaneous.



ExampleDetermine the cell reaction, the

standard cell potential, and the half-cell that acts as the cathode for a voltaic cell composed of the following half-cells.

Fe3+ + e- --> Fe2+ E0Fe3+ = +0.77V

Ni2+ + 2e- --> Ni E0Ni2+ = -0.25V

Answer:Reduction takes place in the Fe3+ half

cell. So, this cell is the Cathode

ExampleDetermine the cell reaction, the

standard cell potential, and the half-cell that acts as the cathode for a voltaic cell composed of the following half-cells.

Fe3+ + e- --> Fe2+ E0Fe3+ = +0.77V

Ni2+ + 2e- --> Ni E0Ni2+ = -0.25V

Answer:Reduction takes place in the Fe3+ half

cell. So, this cell is the Cathode

The 1/2 reactions are:Ni --> Ni2+ + 2e-

2[Fe3+ + e- --> Fe2+]The cell reaction is:Ni + 2Fe3+ --> Ni2+ + 2Fe2+

Cell potential is:E0 = E0

red - E0Oxid

+0.77V - (-0.25V) = +1.02VThe reaction is spontaneous.

The 1/2 reactions are:Ni --> Ni2+ + 2e-

2[Fe3+ + e- --> Fe2+]The cell reaction is:Ni + 2Fe3+ --> Ni2+ + 2Fe2+

Cell potential is:E0 = E0

red - E0Oxid

+0.77V - (-0.25V) = +1.02VThe reaction is spontaneous.

Your Turn: Calculate E0

cell to determine whether the following redox reaction is spontaneous as written:Ni + Fe2+ --> Ni2+ + FeFrom table 23.2 in your textE0

Ni2+ = -0.25V E0Fe2+ = -0.44V

answerNickel is oxidized Iron is reducedE0= -0.44V - (-0.25V) = -0.19Reaction will proceed in the reverse

Your Turn: Calculate E0

cell to determine whether the following redox reaction is spontaneous as written:Ni + Fe2+ --> Ni2+ + FeFrom table 23.2 in your textE0

Ni2+ = -0.25V E0Fe2+ = -0.44V

answerNickel is oxidized Iron is reducedE0= -0.44V - (-0.25V) = -0.19Reaction will proceed in the reverse

Electrolytic Cells23.3

Electrolytic Cells23.3

Electrolysis - making a nonspontaneous reaction go.

ex - silverplated dishes and utensils, gold-plated jewelry and chrome-plated auto parts

The electrolytic cell is an electrochemical cell used to cause a chemical change through the application of electrical energy (DC Current).

Similarities between voltaic and electric cells:electrons flow from the anode to the cathodereduction occurs at cathodeoxidation anode

Electrolysis - making a nonspontaneous reaction go.

ex - silverplated dishes and utensils, gold-plated jewelry and chrome-plated auto parts

The electrolytic cell is an electrochemical cell used to cause a chemical change through the application of electrical energy (DC Current).

Similarities between voltaic and electric cells:electrons flow from the anode to the cathodereduction occurs at cathodeoxidation anode

DifferencesVoltaic is spontaneousElectrolytic is result of outside

push of electronsCathode is negative electrode

(connected to negative electrode of battery) and vice versa for anode.

DifferencesVoltaic is spontaneousElectrolytic is result of outside

push of electronsCathode is negative electrode

(connected to negative electrode of battery) and vice versa for anode.

Electrolysis of water produces hydrogen and oxygen gas.

An electrolyte is often used to help conduct an electric current. The electrolyte is usually redoxed.

The electrolysis of Brine (salt water) produces Cl2 + H2 + NaOH

Electrolysis of molten NaCl produces:liquid Na used in sodium vapor lamps and as

the coolant in some nuclear reactors.chlorine gas used to sterilize drinking water

and the manufacturing of PVC and pesticides.

Which one comes off of which electrode?sodium gains and electron - cathode

Electrolysis of water produces hydrogen and oxygen gas.

An electrolyte is often used to help conduct an electric current. The electrolyte is usually redoxed.

The electrolysis of Brine (salt water) produces Cl2 + H2 + NaOH

Electrolysis of molten NaCl produces:liquid Na used in sodium vapor lamps and as

the coolant in some nuclear reactors.chlorine gas used to sterilize drinking water

and the manufacturing of PVC and pesticides.

Which one comes off of which electrode?sodium gains and electron - cathode

ElectroplatingElectroplatingThe depositing of a thin layer of a

metal on an objectCommon metal used to plate:

Ag, Au,Cu, Ni, CrAn object to be silver plated is made

the cathode. Why?Anode is the silver to be depositedElectrolyte is a silver salt.Also used to purify metals and

various other processes you will read about.

The depositing of a thin layer of a metal on an object

Common metal used to plate:Ag, Au,Cu, Ni, Cr

An object to be silver plated is made the cathode. Why?

Anode is the silver to be depositedElectrolyte is a silver salt.Also used to purify metals and

various other processes you will read about.

Redox Reactions oxidation reduction reactions. Ch 22 sec 1 From the combustion of gasoline to the...

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