Electrochemistry reactions and electricity In electrochemical - - PDF document

SLIDE 1

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Electrochemistry

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· Electrochemistry deals with relationships between reactions and electricity · In electrochemical reactions, electrons are transferred from one species to another. · Provide insight into batteries, corrosion, electroplating, spontaneity of reactions

Electrochemistry Slide 3 / 144 Electrochemical Reactions

· In electrochemical reactions, electrons are transferred between various reactant and product species in reactions. · As a result, oxidation state/number of one or more substances/species change · Oxidation number is the formal charge on the atom when it is connected to other atoms. · In order to keep track of what species loses electrons and what gains them, we assign oxidation numbers/oxidation states to individual atoms.

Slide 4 / 144 Oxidation Numbers

Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)

+1 +2

Take a look at this reaction between Zn metal and acid with assigned oxidation numbers. How do we know what number goes with each atom? Where do these numbers came from?

Slide 5 / 144 Rules for Assigning Oxidation Numbers

Elements

Elements in their elemental form have an

• xidation number of 0.

Compounds The sum of the oxidation numbers in a

neutral compound is 0.

Monoatomic ions

The oxidation number of a monatomic ion is the same as its charge.

Polyatomic ions

The sum of the oxidation numbers in a polyatomic ion is the charge on the ion.

How do we assign oxidation numbers ? Slide 6 / 144

Hydrogen -1 when bonded to a metal

+1 when bonded to a nonmetal Fluorine

Fluorine always has an oxidation number of -1.

Other halogens

Usually -1. May have positive oxidation numbers in

• xyanions.

Rules for Assigning Oxidation Numbers

For example, Cl has an oxidation number of +5 in ClO3

• .

SLIDE 2

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Nonmetals Nonmetals tend to have negative

• xidation numbers although some are

positive in certain compounds or ions.

Oxygen

Oxygen has a oxidation number of -2, except in the peroxide ion, when its

• xidation number is -1.

Rules for Assigning Oxidation Numbers Slide 8 / 144

1 What is the oxidation number of each oxygen atom in the compound MnO2 ?

A

• 2

B

• 1

C D

+1

E

+2

Slide 9 / 144

2 What is the oxidation number of the manganese atom in the compound MnO2 ?

A +3 B +2 C +1 D +4 E +7

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3 What is the oxidation number of oxygen atom in MnO4

1-, the permanganate ion?

A

• 2

B

• 1

C D +2 E +4

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4 What is the oxidation number of the manganese atom in MnO4

1-, the permanganate ion?

A +1 B +2 C +5 D +4 E +7

Slide 12 / 144

5 What is the oxidation number of sulfur in HSO4

1-,

the hydrogen sulfate ion?

A

• 2

B

+1

C

+2

D

+4

E

+6

SLIDE 3

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Oxidation-loss of electrons A species is oxidized when it loses electrons. Here, zinc loses two electrons to go from neutral Zn metal to the Zn2+ ion. Zn is also a reducing agent- provides electrons (reductant) Reducing agent loses electrons.

Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)

+1 +2

Oxidation and Reduction

LEO

The lion says

GER OIL RIG

Slide 14 / 144 Oxidation and Reduction

Reduction- gaining of electrons A species is reduced when it gains electrons. Here, each of the H+ gains an electron, and they combine to form H2. H is an oxidizing agent- accepts electrons (oxidant) An oxidizing agent gains electrons.

Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)

+1 +2

Slide 15 / 144 Oxidation and Reduction

· What is reduced is the oxidizing agent. · H+ oxidizes Zn by taking electrons from it. · What is oxidized is the reducing agent. · Zn reduces H+ by giving it electrons.

Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)

+1 +2

Slide 16 / 144 Oxidation and Reduction

Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)

+1 +2

An electrochemical reaction in which oxidation and reduction occurs is known as a REDOX reaction

Redox Reactions

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6 Which of the following is/are an oxidation- reduction (redox) reactions? (a) K2CrO4 + BaCl2 ➝ KCl + BaCrO4 (b) Pb2+ + 2 Br1- ➝ PbBr2 (c) Cu + S ➝ CuS

A a only B

b only C c only D a and c

E

b and c

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7 Which substance is oxidized in the following reaction? (First, assign oxidation numbers.) Cu + S ➝ CuS

A Cu B S C

Cu and S

D CuS E This is not a redox reaction.

SLIDE 4

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8 Which substance is the reducing agent below? Cu + S ➝ CuS

A Cu B S C

Cu and S

D CuS E This is not a redox reaction.

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9 Which substance is oxidized in the following reaction? (First, assign oxidation numbers.) Ca + Fe3+ ➝ Ca2+ + Fe

A Ca B Fe3+ C

Ca2+

D Fe E This is not a redox reaction.

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10 Which substance is the oxidizing agent below? Ca + Fe3+ ➝ Ca2+ + Fe

A Ca B Fe3+ C

Ca2+

D Fe E This is not a redox reaction.

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11 Which substance is reduced in the following reaction? (First, assign oxidation numbers.) 3 K + Al(NO3)3 ➝ Al + 3 KNO3

A K B Al C

N

D O E This is not a redox reaction.

Slide 23 / 144

12 Which substance is the reducing agent? 3 K + Al(NO3)3 ➝ Al + 3 KNO3

A K B Al(NO3)3 C

KNO3

D This is not a redox reaction.

Slide 24 / 144

H2S (g) + Cl2 (g) --> 2HCl (g) + S (s)

a) Assign oxidation numbers to each element above. b) Which element is oxidized? c) Which element is reduced? d) Name the reducing agent. e) Name the oxidizing agent.

Redox Practice 1

SLIDE 5

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SnCl2(aq) + 2HgCl2 (aq) --> SnCl4 (aq) + Hg2Cl2 (s)

a) Assign oxidation numbers to each element above. b) Which element is oxidized? c) Which element is reduced? d) Name the reducing agent. e) Name the oxidizing agent.

Redox Practice 2 Slide 26 / 144

13 Which element is oxidized in the reaction below?

Fe2+ + H+ + Cr2O7

2- ➝ Fe3+ + Cr3+ + H2O

Slide 27 / 144

14

H2S (g) + Cl2 (g) --> 2HCl (g) + S (s) Which is oxidized? Which is reduced?

Slide 28 / 144

15 SnCl2(aq) + 2HgCl2 (aq) --> SnCl4 (aq) + Hg2Cl2 (s)

Which is oxidized? Which is reduced?

Slide 29 / 144 Redox reactions in aqueous solutions

A large number of redox reactions occur in aqueous

solutions. Unlike acid base nutralization and precipitation reactions,most of the reaction proceed slowly. Each redox reaction is the sum of two half reactions: Consider the reaction of iodide ions and hydrogen peroxide. 2I- (aq) + H2O2(aq) + 2e- ⇒ I2 + 2OH- (aq) + 2e-

Slide 30 / 144

2I- (aq) + H2O2(aq) + 2e- ⇒ I2 + 2OH- (aq) + 2e-

• 1. Oxidation half reaction
• 2. Reduction half reaction.

2I- (aq) ⇒ I2 + 2e- oxidation H2O2(aq) + 2e- ⇒ 2OH- (aq) reduction Add the two half reactions to get the overall reaction. 2I- (aq) + H2O2(aq) + 2e- ⇒ I2 + 2OH- (aq) + 2e- How do we balance a redox reaction?

Redox reactions in aqueous solutions

This reaction involves two parts as represented below.

SLIDE 6

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Balancing Redox reactions

Half-reaction method (oxidation # method) · Assign oxidation numbers to determine what is

• xidized and what is reduced.

· Identify the oxidation and reduction process. · Write down the individual oxidation and reduction equations. · Balance these half reactions · Combine them to attain the balanced equation for the overall reaction. This method can be used in general to balance any redox reaction unless any specific condition such as acidic or basic is mentioned

Slide 32 / 144

Half-reaction method (oxidation # method) Let us consider the simple replacement reaction of Mg with AgCl +1

• 1
• 1

+2 Mg + AgCl ➝ Ag + MgCl2

Oxidation: Mg --> Mg2+ + 2 e- --------(1) Reduction: Ag+ + 1 e- --> Ag ---------(2) Since all the atoms are balanced, we need to balance

• nly electrons. Multiply equation (2) x 2

Oxidation: Mg --> Mg2+ + 2 e- --------(1) Reduction: 2Ag+ + 2 e- --> 2Ag ---------(3)

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Oxidation: Mg --> MgCl2 + 2e- Reduction: 2AgCl + 2e- --> 2Ag Adding the half-reactions (1) and (3) yields the following: Overall: Mg + 2AgCl + 2e- --> MgCl2 + 2e- + 2Ag Overall: Mg + 2AgCl + 2e- --> MgCl2 + 2e- + 2Ag and we cancel out electrons from both sides: Net equation: Mg + 2AgCl --> MgCl2 + 2Ag

Half-reaction method (oxidation # method)

Since the original equation is given with chlorine you would keep it here in the final balanced equation too.

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Redox reactions -balancing Fe3O4 +C --> Fe + CO

Fe3O4 + 4C --> 3Fe + 4CO

Practice

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Practice: SnO2 + C --> Sn + CO Redox reactions -balancing

SnO2 + 2C --> Sn + 2CO

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This diagram shows the steps involved in balancing half-reactions. · Write down the individual half reaction.

· First balance atoms other than H and O. · Balance oxygen atoms by adding H2O. · Balance hydrogen atoms by adding H+. · Balance charge by adding electrons.

· Multiply the half-reactions by integers so that the electrons gained and lost are the same.

The Half-Reaction Method

Other atoms O

H e-

In acidic medium:

SLIDE 7

Slide 37 / 144 The Half-Reaction Method

· Add the half-reactions, subtracting things that appear on both sides. · Make sure the equation is balanced according to mass. · Make sure the equation is balanced according to charge.

In acidic medium: Continued

Slide 38 / 144 The Half-Reaction Method

Consider the reaction between MnO4

− and C2O4 2− :

MnO4

− (aq) + C2O4 2− (aq) ➝ Mn2+ (aq) + CO2 (aq)

In acidic medium:

· First, we assign oxidation numbers. · We only assign oxidation numbers to elements whose

• xidation numbers CHANGES.

· Here, oxygen's oxidation number remains constant at -2.

MnO4

− (aq) + C2O4 2− (aq) ➝ Mn2+ (aq) + CO2 (aq)

+7 +3 +2 +4

Slide 39 / 144 The Half-Reaction Method

· Which substance gets reduced? · Which substance gets oxidized? · Which substance is the reducing agent? · Which substance is the oxidizing agent?

MnO4

− (aq) + C2O4 2− (aq) ➝ Mn2+ (aq) + CO2 (aq)

+7 +3 +2 +4

In acidic medium:

Slide 40 / 144 The Half-Reaction Method

Since the manganese goes from +7 to +2, it is reduced. The MnO4

• ion is the oxidizing agent.

Since the carbon goes from +3 to +4, it is oxidized. The C2O4

2- ion is the reducing agent.

MnO4

− (aq) + C2O4 2− (aq) ➝ Mn2+ (aq) + CO2 (aq)

+7 +3 +2 +4

In acidic medium:

Slide 41 / 144 Oxidation Half-Reaction

C2O4

2− ➝ CO2

To balance the carbon, we add a coefficient of 2: C2O4

2− ➝ 2 CO2

In acidic medium:

Slide 42 / 144 Oxidation Half-Reaction

C2O4

2− ➝ 2 CO2

The oxygen is now balanced as well. To balance the charge, we must add 2 electrons to the right side.

C2O4

2− ➝ 2 CO2 + 2 e−

In acidic medium:

SLIDE 8

Slide 43 / 144 Reduction Half-Reaction

MnO4

− ➝ Mn2+

The manganese is balanced; to balance the

• xygen, we must add 4 waters to the right side.

MnO4

− ➝ Mn2+ + 4 H2O

In acidic medium:

Slide 44 / 144 Reduction Half-Reaction

MnO4

− ➝ Mn2+ + 4 H2O

To balance the hydrogen, we add 8 H+ to the left side.

8 H+ + MnO4

− ➝ Mn2+ + 4 H2O

In acidic medium:

Slide 45 / 144 Reduction Half-Reaction

8 H+ + MnO4

− ➝ Mn2+ + 4 H2O

To balance the charge, we add 5 e− to the left side.

5 e− + 8 H+ + MnO4

− ➝ Mn2+ + 4 H2O

In acidic medium:

Slide 46 / 144 Combining the Half-Reactions

Now we evaluate the two half-reactions together:

C2O4

2− ➝ 2 CO2 + 2 e−

5 e− + 8 H+ + MnO4

− ➝ Mn2+ + 4 H2O

To attain the same number of electrons on each side, we will multiply the first reaction by 5 and the second by 2.

In acidic medium:

Slide 47 / 144 Combining the Half-Reactions

5 C2O4

2− ➝ 10 CO2 + 10 e−

10 e− + 16 H+ + 2 MnO4

− ➝ 2 Mn2+ + 8 H2O

When we add these together, we get: 10 e− + 16 H+ + 2 MnO4

− + 5 C2O4 2− -->

2 Mn2+ + 8 H2O + 10 CO2 +10 e−

In acidic medium:

Slide 48 / 144 Combining the Half-Reactions

10 e− + 16 H+ + 2 MnO4

− + 5 C2O4 2− ➝

2 Mn2+ + 8 H2O + 10 CO2 +10 e−

The only thing that appears on both sides are the

• electrons. Subtracting them, we are left with:

16 H+ + 2 MnO4

− + 5 C2O4 2− ➝

2 Mn2+ + 8 H2O + 10 CO2

In acidic medium:

SLIDE 9

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a) Write the oxidation half reaction. b) Write the reduction half reaction. c) Write the balanced net reaction. d) Identify the oxidizing agent. e) Identify the reducing agent.

Cd(s) + NiO2 (s) --> Cd(OH)2(s) + Ni(OH)2(s)

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

Practice 1

The Half-Reaction Method

In acidic medium:

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Practice 2

Cu + NO3

• --> NO2 + Cu2+

In acidic medium:

Slide 51 / 144

Practice 3

Cr2O7

2- + Fe2+ + H+ --> Cr3+ + Fe 3+ + H2O

Slide 52 / 144

Practice 4 :MnO4

• + Br- --> Mn2+ + Br2 in acidic solution

Slide 53 / 144

Practice 5 : Cr2O7

2- + C2H4O --> C2H4O2 + Cr3+ in acidIc

solution

Cr2O7

2- + 8H+ + 3C2H4O --> 3C2H4O2 + 2Cr3+ + 4H2O

Slide 54 / 144

Redox reaction in basic medium

Some redox reactions requires basic medium to

• ccur.In this case the following steps need to be

performed to balance the reaction.

1- Assign the oxidation numbers 2- Balance the "other atoms" involved 3- Separate the half reactions 4- Add water molecules to balance oxygen atom whatever side deficient in O atoms 5- Add water molecules equal in number to the deficiency of H atoms. 6- Add same number of OH- to the other side. 7- Balance the charge by adding electrons on the appropriate side 8- Balance the electrons lost /gained by multiplying the reactions by integers 9- Add the two reactions removing any duplication if any of common species on either side. Can also be performed without splitting the two equations.

SLIDE 10

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Zn + NO3

• --> Zn2+ + NH4

+ in basic medium:

Balancing in Basic Solution

Oxidation half reaction: Zn ----> Zn2+ + 2e- Reduction half reaction: NO3

• ---> NH4

+

NO3

• ---> NH4+ + 3H2O

10H2O + NO3

• ---> NH4

+ + 3H2O

10H2O + NO3

• ---> NH4

+ + 3H2O + 10OH-

8e- 10H2O + NO3

• ---> NH4

+ + 3H2O + 10OH-

4Zn ----> 4 Zn2+ + 8e-

4Zn + 1NO3

• + 7H2O--> 4Zn2+ + 1NH4+ + 10 OH-

Slide 56 / 144

Balancing in Basic Solution

Zn + NO3

• --> Zn2+ + NH4

+ in basic medium:

• 1. Assigh oxidation #s:

Zn + NO3

• --> Zn2+ + NH4

+

0 +5 2- +2 -3 +1

• 2. Balance the change in Oxidation # change on either side.

4Zn + 1NO3

• --> 4Zn2+ + 1NH4

+ Increases by 2 decreases by 8 increases by 8 decreases by 8

**

Can also be performed without splitting the two equations.

Slide 57 / 144

Balancing in Basic Solution

4Zn + 1NO3

• --> 4Zn2+ + 1NH4

+

• 3. Balance O atoms by adding H2O molecules to the side deficient in O atoms.

3 O atoms on the LHS so add 3 water on the RHS

4Zn + 1NO3

• --> 4Zn2+ + 1NH4

+ + 3H2O

• 4. The H atoms are then balanced by adding H2O to the side lacks H.

10 H on the RHS, so add 10 water on the LHS.

4Zn + 1NO3

• + 10H2O--> 4Zn2+ + 1NH4

+ + 3H2O

• 5. Add 10 OH- on the other side of the reaction to balance the extra H and O.

4Zn + 1NO3

• + 10H2O--> 4Zn2+ + 1NH4

+ + 3H2O + 10 OH-

• 6. If this produces water on both sides, you might have to subtract water from

each side.

4Zn + 1NO3

• + 7H2O--> 4Zn2+ + 1NH4+ + 10 OH-

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Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O in basic solution

+2

+3

• 1
• 2

increase by 1, 1 e- given decrease by 1 for each O atom, total 2 e- taken

2 Fe(OH)2 + H2O2 --> 2Fe(OH)3 + H2O in basic solution 2 Fe(OH)2 + 2H2O --> 2Fe(OH)3

Balance O atoms by adding 2 H2O to LHS

2 Fe(OH)2 + 2H2O --> 2Fe(OH)3 + 2H2O

Balance H atoms by adding 2 H2O to RHS Add 2 OH- on the LHS

2 Fe(OH)2 + 2H2O + 2OH- --> 2Fe(OH)3 + 2H2O

Balancing in Basic Solution Practice: 1

Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O in basic solution

Oxidation:

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Balancing in Basic Solution H2O2 --> H2O

Add 1 H2O on RHS

H2O2 --> H2O + H2O

Add 2H2O on LHS to balance H atoms 2H2O + H2O2 --> H2O + H2O

Add 2 OH- to RHS

2H2O + H2O2 --> H2O + H2O +2OH-

A

Add the two equations: 2Fe(OH)2 + H2O2 --> 2Fe(OH)3

Practice 2: Fe(OH)2 + H2O2 --> Fe(OH)3 + H2O in basic solution

Reduction

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Practice 3 : Bi(OH)3 + SnO2

• -> Bi + SnO3

2Bi(OH)3 + 3SnO2

• -> 2Bi + 3SnO3 + 3H2O

Balancing in Basic Solution

SLIDE 11

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Balancing in Basic Solution Practice 4: Cr(OH)4

• 1 + H2O2 --> (CrO4) 2- + H2O

2Cr(OH)-1 + 2OH- + 3H2O2 --> 2CrO4

2- + 8H2O

Slide 62 / 144

Voltaic Cells

The energy released in a spontaneous reaction can be used to perform electrical work. Such a set up through which we can transfer electrons is called a voltaic cell or galvanic cell or electrochemical cell

Slide 63 / 144

Voltaic Cells

In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released. In the above stup, lectron transfer takes place inside the beaker Zn + Cu2+ ➝ Zn2+ + Cu Zn metal strip placed in CuSO4

Zn metal Cu2+ 2e- Cu atom Zn2+ Cu2+ Zn metal

Note that the blue color fades as more Cu is reduced to metallic copper

single replacement

Slide 64 / 144 Voltaic Cells

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/animations/ ZnCutransfer.html

This shows what is occurring on an atomic level at the anode and the cathode.

Slide 65 / 144 Voltaic Cells

Here the Cu and Zn strips are in two different beakers Zn/ZnNO3 Zn ➝ Zn2+ + 2e- OXD- Half reaction Cu/CuNO3 Cu2+ + 2e- ➝ Cu RED- Half reaction We can use the energy to do work if we make the electrons flow through an external device.

Slide 66 / 144 Voltaic Cells

Here the Cu and Zn strips are in two different beakers · The salt bridge allows the migration of the ions to keep electrical neutrality · Electrons are generated at the anode and flows through the external line to the cathode. Zn/ZnNO3 Zn ➝ Zn2+ + 2e- OXD- Half reaction Cu/CuNO3 Cu2+ + 2e- ➝ Cu RED- Half reaction

salt bridge

e-

SLIDE 12

Slide 67 / 144

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/ animations/CuZncell.html

Voltaic Cells Slide 68 / 144

Voltaic Cells

· A typical cell looks like this. · The oxidation occurs at the anode. · The reduction occurs at the cathode.

Zn2+ NO3

• Slide 69 / 144

Voltaic Cells

Once even one electron flows from the anode to the cathode, the charges in each beaker would not be balanced and the flow of electrons would stop.

more Zn

2+ are

produced more NO3

• are

created in solution

Slide 70 / 144

Voltaic Cells

· Therefore, we use a salt bridge, usually a U- shaped tube that contains a gel of a salt solution, to keep the charges balanced. · Cations move toward the cathode. · Anions move toward the anode.

more Zn2+ are produced more NO3

• are in

solution

The increase in Zn2+ and NO3

• ions in the two compartment create

electrical imbalance. The salt bridge ions will neutralize these ions and create neutrality.

Slide 71 / 144

Voltaic Cells

· In the cell, then, electrons leave the anode and flow through the wire to the cathode. · As the electrons leave the anode, the cations formed dissolve into the solution in the anode

Zn metal Cu2+ 2e- Cu atom Zn2+ Cu

2+

Zn metal

e-

Slide 72 / 144

Voltaic Cells

· As the electrons reach the cathode, cations in the cathode are attracted to the now negative cathode.

Zn metal Cu

2+

2e- Cu atom Zn2+ Cu

2+

Zn metal

e-

SLIDE 13

Slide 73 / 144

Voltaic Cells

· The electrons are taken by the cation, and the neutral metal atoms are deposited onto the cathode.

Zn metal Cu2+ 2e- Cu atom Zn2+ Cu

2+

Zn metal

e-

Slide 74 / 144 Voltaic Cells

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/ flashfiles/electroChem/volticCell.html

This shows how a typical voltaic cell works

Slide 75 / 144

16 The electrode at which

• xidation occurs is called

the _______________.

Slide 76 / 144

17 In a voltaic cell, electrons flow from the ______ to the ________.

Slide 77 / 144

18 Which element is oxidized in the reaction below?

Fe2+ + H+ + Cr2O7

2- ➝ Fe3+ + Cr3+ + H2O

Slide 78 / 144

19 Fe2+ + H+ + Cr2O7 2- ➝ Fe3+ + Cr3+ + H2O

If a voltaic cell is made with Fe and Cr electrode in contact with their own solution, the electrons will flow from ------ to --------- electrode.

SLIDE 14

Slide 79 / 144

A) maintain electrical neutrality in the half-cells via migration of ions. B) provide a source of ions to react at the anode and cathode. C) provide oxygen to facilitate oxidation at the anode. D) provide a means for electrons to travel from the anode to the cathode. E) provide a means for electrons to travel from the cathode to the anode.

20 The purpose of the salt bridge in an electrochemical cell is to ________________.

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21 A cell was made with Mg and Cu as two electrodes. The

electrons will flow from ------- to --------- electrode.

Slide 81 / 144

22 The electrode where reduction is taking place is the ----------

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23 The cation concentration increases in the solution where

• xidation occurs.

Yes No

Slide 83 / 144

24 the cations move towards the anode and anions move towards the cathode in a voltaic cell. True False

Slide 84 / 144

25 The salt bridge ions may react with the Ions in the cell compartments to form a precipitate. Yes No

SLIDE 15

Slide 85 / 144

26 Which of the following substances would NOT provide a suitable salt bridge?

A KNO3 B

Na2SO4

C

LiC2H3O2

D

PbCl2

Slide 86 / 144

27 Which of the following substances would provide a suitable salt bridge?

A AgBr B

KCl

C

BaF2

D

CuS

Slide 87 / 144

28 In a Cu-Zn voltaic cell, which of the following is true?

A Both strips of metal will increase in mass. B

Both strips of metal will decrease in mass.

C

Cu will increase in mass; Zn will decrease.

D

Cu will decrease in mass; Zn will increase.

E

Neither metal will change its mass, since electrons have negligible mass.

Zn/ZnNO3 Zn ➝ Zn2+ + 2e- OXD- Half reaction Cu/CuNO3 Cu2+ + 2e- ➝ Cu RED- Half reaction

Slide 88 / 144

29 In any voltaic cell, which of the following is true?

A The cathode will always increase in mass. B

The anode strip will always decrease in mass.

C

The anode strip will always increase in mass.

D

Both A and B

Slide 89 / 144

· Water only spontaneously flows one way in a waterfall. · Likewise, electrons only spontaneously flow one way in a redox reaction—from higher to lower potential energy. · The accumulation of large number of electrons at the anode create higher potential at the anode. · Natural flow will occur to cathode where there is less potential · Higher - to - lower

Electro motive force Slide 90 / 144

· The potential difference between the anode and cathode in a cell is called the electromotive force (emf). · It is also called the cell potential and is designated Ecell.

Electro motive force

SLIDE 16

Slide 91 / 144

The difference in potential energy /electon charge is measured in volts. 1 volt is the potential required to impart 1joule energy to a charge of 1coulomb 1v = 1J / 1C The potential difference between the electrodes is the driving force that pushes the electrons - so called EMF In a voltaic cell, EMF = Ecell

Electro motive force Slide 92 / 144 Electromotive Force (emf)

In a spontaneous reaction, Ecell is positive EMF depends on the cell reaction involved Standard condition: 1M, 1atm and 25°C E°cell = standard cell potential Cell potential is measured in volts (V).

1V = 1J/C Slide 93 / 144

Standard Reduction Potentials

Reduction potentials for many electrodes have been measured and tabulated. By convention, the process is viewed as a reduction and the values are reported as reduction potential

Li+ (aq) + e- ➝ Li(s) -3.05 Na+(aq) + e- ➝ Na(s) -2.71 Al3+(aq) + 3e- ➝ Al(s) -1.66 2H+(aq) + 2e- ➝ H2(g 0 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 F2(g) + 2e- ➝ 2F-(aq) + 2.87

The more negative value indicate that, reduction is unlikely at that electrode The more positive the value is, reduction is highly likely at that electrode. This parallels their activity in single replacement reaction.

Electrode potential: The tendency of an electrode to lose or gain electrons is called electrode potential ( oxidation or reduction potential)

Slide 94 / 144

Standard Hydrogen Electrode ( SHE)

· By definition, the reduction potential for hydrogen is 0 V: 2 H+ (aq, 1M) + 2 e− ➝ H2 (g, 1 atm)

Pt H2, 1 atm HCl, 1M

Standard Reduction Potentials

Slide 95 / 144

How did we measure the reduction potential of all lements?

Pt H2, 1 atm HCl, 1M Zn Zn(NO3)2

Their values are referenced to a Standard Hydrogen Electrode (SHE). The metal electrode will be connected to the SHE By definition, the reduction potential for hydrogen is 0 V: The reduction potential measured will be that of the metal

Standard Reduction Potentials

Slide 96 / 144

30 If a volatic cell is made with iron and zinc, which metal will be reduced?

Use the reduction potential table and compare the values. The more positive the value is, that is where reduction takes place, is the cathode. Oxidation - at Anode (vowels) Reduction - at Cathode (consonants)

Fe Zn 0.1M Zn(NO3)2 0.1M Fe(NO3)2

SLIDE 17

Slide 97 / 144

31 If a volatic cell is made with Cu and Na, which metal will be the cathode?

A Cu B

Na

C

Cu and Na cannot make a voltaic cell. F2(g) + 2e- ➝ 2F-(aq) + 2.87 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 2H+(aq) + 2e- ➝ H2(g 0 Al3+(aq) + 3e- ➝ Al(s) -1.66 Na+(aq) + e- ➝ Na(s) -2.71 Li+ (aq) + e- ➝ Li(s) -3.05

Slide 98 / 144

32 If a volatic cell is made with Li and Al, which metal will be the anode?

A Li B

Al

C

Li and Al cannot make a voltaic cell. F2(g) + 2e- ➝ 2F-(aq) + 2.87 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 2H+(aq) + 2e- ➝ H2(g 0 Al3+(aq) + 3e- ➝ Al(s) -1.66 Na+(aq) + e- ➝ Na(s) -2.71 Li+ (aq) + e- ➝ Li(s) -3.05

Slide 99 / 144

The cell potential at standard conditions can be found through this equation:

Ecell = Eo

red.pot (cathode) − Eo

red.pot (anode)

Because cell potential is based on the potential energy per unit of charge, it is an intensive property. This means that it does not depend on the amount of substance (e.g. mass or moles).

Cell Potentials Slide 100 / 144

Cell Potentials

· For the oxidation in this cell, · E°

red = - 0.76v

· For the reduction, · E°

red = + 0.34v

A cell with Cu and Zn electrodes

1M Zn(NO3)2 1M Cu(NO3)2 Cu Zn Zn(s) ➝ Zn2+ + 2e- Cu2+ (aq) + 2e-➝ Cu(s)

Slide 101 / 144

Cell Potentials

E°

cell = E° red(cathode) - E° red(anode)

= +0.34V - (-0.76V)

E°

red = +1.10V

Slide 102 / 144

The greater the difference between the two electrode potential, the greater the voltage of the cell.

Cu2+ + 2e- --> Cu Zn --> Zn2+ + 2e- More positive

• 0.76

+ 0.34

E0

cell = 0.34 - (-0.76)

= + 1.10v E°

cell

= +0.34V - (-0.76V)

= +1.10V

Cell Potentials

SLIDE 18

Slide 103 / 144

33 Which of the following volatic cells would yield the greatest voltage (Eo

cell)?

A Cu-Al B

Cu-Na

C

Al-Li F2(g) + 2e- ➝ 2F-(aq) + 2.87 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 2H+(aq) + 2e- ➝ H2(g 0 Al3+(aq) + 3e- ➝ Al(s) -1.66 Na+(aq) + e- ➝ Na(s) -2.71 Li+ (aq) + e- ➝ Li(s) -3.05

D

F2 - Cu

Slide 104 / 144

34 Which of the following volatic cells would yield the lowest voltage (Eo

cell)?

A Cu-Al B

Al-Na

C

Na-Li F2(g) + 2e- ➝ 2F-(aq) + 2.87 Cu2+(aq) + 2e- ➝ Cu(s) + 0.34 2H+(aq) + 2e- ➝ H2(g 0 Al3+(aq) + 3e- ➝ Al(s) -1.66 Na+(aq) + e- ➝ Na(s) -2.71 Li+ (aq) + e- ➝ Li(s) -3.05

Slide 105 / 144

Oxidizing and Reducing Agents

· The strongest oxidizers have the most positive reduction potentials. · The strongest reducers have the most negative reduction potentials. F is a strong oxidizing agent than Cl

F2(g) + 2e- --> 2F- 2.87v Cl2(g) + 2e- --> 2Cl- 1.36v I2(s) + 2e- --> 2I- 0.53v Rb+ + e- --> Rb(s) -2.92v

Most positive values Most negative values

Increasing strength of oxidizing agent Increasing strength of reducing agent

Slide 106 / 144

35 The more _______ the value of Ered, the greater the driving force for reduction.

Slide 107 / 144

Class Practice:

Ni Sn 1M Ni (NO3)2 1M Sn(NO3)2

Identify: Cathode Anode Oxidation half reaction Reduction half reaction Combined cell reaction (balanced)

E°

cell =

Slide 108 / 144

Class Practice 2

Fe Sn 1M Fe(NO3)3 1M Sn(NO3)2

Identify: Cathode Anode Oxidation half reaction Reduction half reaction Combined cell reaction (balanced)

E°

cell

SLIDE 19

Slide 109 / 144

36 Calculate E° for the following reaction:

Sn4+(aq) + 2K(s) --> Sn2+(aq) + 2K+(aq) A) +6.00 V B) -3.08 V C) +3.08 V D) +2.78 V E) -2.78 V

Slide 110 / 144

Free Energy

ΔG for a redox reaction can be found by using the equation ΔG = −nFE where n is the number of moles of electrons transferred, and F is a constant, the Faraday. 1 F = 96,500 C/mol = 96,485 J/V-mol

Slide 111 / 144

Free Energy

Under standard conditions, ΔG° = -nFE° Standard condition: 250C, 1 atm and 1M

Slide 112 / 144 Nernst Equation

· Remember that ΔG = ΔG° + RT ln Q · This means −nFE = −nFE° + RT ln Q

Slide 113 / 144 Nernst Equation

Dividing both sides by −nF, we get the Nernst equation:

• r, using base-10 logarithms,

E = E° - [8.31 x 298] x 2.303 log Q [n x 96500 ]

E = E° - lnQ RT nF Slide 114 / 144 Nernst Equation

At room temperature (298 K), Thus, the equation becomes

E = E° - logQ 0.0592 n = 0.0592 V 2.303 RT F

E = E° - [8.31 x 298] x 2.303 log Q [n x 96500 ]

SLIDE 20

Slide 115 / 144 Equilibrium Constant

· When E=0, -nFE = 0 · 0= E° – logK (0.0592/n) · · E° = logK (0.0592/n) · · logK = nE°/0.0592 · · Equilibrium constant for a redox reaction can be calculated using the above . ΔG = ΔG° + RT ln Q E = E° - logQ 0.0592 n

Slide 116 / 144

A) ΔG = -nF/E B) ΔG = -E/nF C) ΔG = -nFE D) ΔG = -nRTF E) ΔG = -nF/ERT

37 The relationship between the change in Gibbs free energy and the emf of an electrochemical cell is given by __________________.

Slide 117 / 144

Concentration Cells

· Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes. For such a cell, E° would be 0, but Q would not. · Therefore, as long as the concentrations are different, E will not be 0.

http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/electroChem/ voltaicCellEMF.html

Ni Ni 1M [Ni2+] 0.001M [Ni2+] Ni Ni 0.5M [Ni2+] 0.5M [Ni2+]

E = E° - logQ 0.0592 n **

[dilute] log Q = log ------------- [concent]

Slide 118 / 144

38

A cadmium rod is placed in a 0.010M solution of cadmium sulfate at 298K. Calculate the potential of the electrode at the is temperature.

E = E° - logQ 0.0592 n = -0.0529/2 log0.01 = - 0.0591

Slide 119 / 144 Elcectrolytic cell/ Electrolysis

· Voltaic cells work as a result of a spontaneous reaction · We can use electricity from outside source to make a nonspontaneous reaction to become spontaneous. · A chemical reaction by using outside electricity is known as electrolysis. Such a cell is known as an electrolytic cell

Slide 120 / 144

Electrolytic Cell Voltaic (Electrochemical) Cell

Energy is absorbed to drive a nonspontaneous redox reaction Energy is released from a spontaneous redox reaction Surroundings (battery or power supply) do work on the system (cell) System (cell) does work on the surroundings (e.g. light bulb)

Elcectrochemical/voltaic cell

SLIDE 21

Slide 121 / 144

A) an electric current is produced by a chemical reaction B) electrons flow toward the anode C) a nonspontaneous reaction is forced to occur D) gas is produced at the cathode E) oxidation occurs at the cathode

39 One of the differences between a voltaic cell and an electrolytic cell is that in an electrolytic cell, _____________________.

Slide 122 / 144

Electroplating

· Uses an active electrode to deposit a thin layer of one metal to another metal object · Item to be coated is cathode (metal ions get reduced at the (-) electrode)

e-

Ag

+

Ag

+

Ag

Slide 123 / 144

Electrolysis

This flowchart shows the steps relating the quantity of electrical charge used in electrolysis to the amounts of substances

• xidized or reduced.

Current (A) and time Quantity of charge (Coulombs) Moles of electrons (Faradays) Moles of substance

• xidized
• r reduced

Grams of substance

A typical problem will give the current (amperes) that is applied for a specific amount of time (seconds). You would be asked to solve for the mass of metal that can be produced through electroplating. Alternatively, you might be asked for either the time or amount of current that is needed to produce a specific amount (given mass) of metal.

Slide 124 / 144

The quantity of charge passing through is measured in coulombs 1 mole of electrons passage = 96500C = 1Faraday 1coulomb = 1 ampere passing in 1 second Coulombs (C ) = ampere x seconds

Electrolysis Slide 125 / 144

A) joule B) coulomb C) calorie D) Newton E) Mole

40 The quantity of charge passing a point in a circuit in one second when the current is one ampere is called a ___________________.

Slide 126 / 144

41 How many coulombs result from a current

• f 50 amps (A) applied for 20 seconds?

A 2.5 B

10

C

70

D

700

E

1000

A C s =

SLIDE 22

Slide 127 / 144

42 How many seconds must a current of 25 A be applied in order to produce a charge of 100 C?

A 0.25 B

0.4

C

4

D

75

E

125

A C s =

Slide 128 / 144

43 What amount of charge is required to release one mole of electrons?

A 1 atm B

25oC

C

0.0821 L-atm/mol-K

D

96,500 C

E

760 mm Hg

Slide 129 / 144

44 How many moles of electrons would be released by a charge of 158,000 C?

A 96,500 / 158,000 B

158,000 / 96,500

C

158,000*96,500

D

158,000 - 96,500

Slide 130 / 144

45 How many moles of electrons would be released by a charge of 48,250 C?

A 0.25 B

0.5

C

1

D

48,250

E

96,500

Slide 131 / 144

If 10.0 A passes through molten AlCl3 for 60 minutes, how much of Al will be deposited? total charge C = 10.0 A x 60min x 60sec. = 3.6 x104 C Remember!! 1mole e- = 96500C How many moles of e- are we talking about in here????? Moles of e- = 3.6 x 104 /96500 = 0.373 moles of e- Al3+ + 3e- ➝ Al 1 mol Al = 3 mols e- Moles of Al = 0.373 x 1 mole Al/3 mole e- = 0.124 mol Al How many grams of Al ? = 27g x 0.124 mol = 3.36g Al

Quantitative aspect Electrolysis

Slide 132 / 144

Calculate the number of grams of aluminum produced in 30.0 minutes by electrolysis of at a current of 12.0 A.

practice:1

Electrolysis

SLIDE 23

Slide 133 / 144

How many minutes will it take to plate out 6.36 g of Cu metal

from a solution of Cu2+ using a current of 12 amps in an electrolytic cell?

practice:2

Electrolysis Slide 134 / 144

46 Plating out 1 mol of chromium requires _______ of electrons.

A 0.33 mol B

0.5 mol

C

1.0 mol

D

3.0 mol

Cr3+

(aq) + 3 e- --> Cr(s)

Slide 135 / 144

47 One mole of electrons would allow electroplating

• f __________ mol of zinc.

A Zn cannot be electroplated. B

0.5 mol

C

1.0 mol

D

2.0 mol

Zn2+

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

Slide 136 / 144

48

How many minutes will it take to plate out 16.22 g of Al metal from a solution of Al3+ using a current of 12.9 amps in an electrolytic cell? A 60.1 min

B

74.9 min

C

173 min

D

225 min

E

13,480 min

Slide 137 / 144 Electrochemistry -Applied

· Batteries · Hydrogen fuel cells · Corrosion · Corrosion prevention · Biology

Slide 138 / 144 Batteries

SLIDE 24

Slide 139 / 144 Alkaline Batteries Slide 140 / 144 Hydrogen Fuel Cells Slide 141 / 144 Corrosion and… Slide 142 / 144 …Corrosion Prevention Slide 143 / 144 In Biology

electron transport in Mitochondria ......

Slide 144 / 144