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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

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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.

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Oxidation Numbers

Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)

0 +1 0+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?

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Rules for Assigning Oxidation Numbers

Elements Elements in their elemental form have an oxidation 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 oxyanions.

Rules for Assigning Oxidation Numbers

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

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NonmetalsNonmetals tend to have negative oxidation numbers although some are positive in certain compounds or ions.

OxygenOxygen has a oxidation number of -2, except in the peroxide ion, when its oxidation number is -1.

Rules for Assigning Oxidation Numbers

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

A -2

B -1

C 0

D +1

E +2

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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 0

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

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5 What is the oxidation number of sulfur in HSO41-,

the hydrogen sulfate ion?

A -2

B +1

C +2

D +4

E +6

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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)

0 +1 0+2

Oxidation and Reduction

LEO

The lion says

GER

OIL RIG

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Oxidation and Reduction

Reduction- gaining of electronsA 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)

0 +1 0+2

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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)

0 +1 0+2

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Oxidation and Reduction

Zn(s) + 2H+(aq) ➝Zn2+ + H2(g)

0 +1 0+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.

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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.

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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.

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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

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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

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13 Which element is oxidized in the reaction below?

Fe2+ + H+ + Cr2O72- ➝ Fe3+ + Cr3+ + H2O

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

Which is oxidized?

Which is reduced?

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

Which is oxidized?

Which is reduced?

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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-

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

1. Oxidation half reaction2. Reduction half reaction.

2I- (aq) ⇒ I2 + 2e- oxidationH2O2(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.

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Balancing Redox reactions

Half-reaction method (oxidation # method)

· Assign oxidation numbers to determine what is oxidized 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

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Half-reaction method (oxidation # method)

Let us consider the simple replacement reaction of Mg with AgCl

0 +1 -1-1 +20

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 only 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:

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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

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The Half-Reaction Method

Consider the reaction between MnO4− and C2O4

2− :MnO4

− (aq) + C2O42− (aq) ➝ Mn2+ (aq) + CO2 (aq)

In acidic medium:

· First, we assign oxidation numbers. · We only assign oxidation numbers to elements whose oxidation numbers CHANGES. · Here, oxygen's oxidation number remains constant at -2.

MnO4− (aq) + C2O4

2− (aq) ➝ Mn2+ (aq) + CO2 (aq)

+7 +3 +2 +4

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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:

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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:

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Oxidation Half-Reaction

C2O42− ➝ CO2

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

C2O42− ➝ 2 CO2

In acidic medium:

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Oxidation Half-Reaction

C2O42− ➝ 2 CO2

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

C2O42− ➝ 2 CO2 + 2 e−

In acidic medium:

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Reduction Half-Reaction

MnO4− ➝ Mn2+

The manganese is balanced; to balance the oxygen, we must add 4 waters to the right side.

MnO4− ➝ Mn2+ + 4 H2O

In acidic medium:

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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:

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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:

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Combining the Half-Reactions

Now we evaluate the two half-reactions together:

C2O42− ➝ 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:

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Combining the Half-Reactions

5 C2O42− ➝ 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:

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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:

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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 MethodIn acidic medium:

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Practice 2 Cu + NO3- --> NO2 + Cu2+

In acidic medium:

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Practice 3

Cr2O72- + Fe2+ + H+ --> Cr3+ + Fe 3+ + H2O

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Practice 4 :MnO4- + Br- --> Mn2+ + Br2 in acidic solution

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Practice 5 : Cr2O72- + C2H4O --> C2H4O2 + Cr3+ in acidIc

solution

Cr2O72- + 8H+ + 3C2H4O --> 3C2H4O2 + 2Cr3+ + 4H2O

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Redox reaction in basic mediumSome redox reactions requires basic medium to occur.In this case the following steps need to be performed to balance the reaction.

1- Assign the oxidation numbers2- Balance the "other atoms" involved3- Separate the half reactions4- Add water molecules to balance oxygen atom whatever side deficient in O atoms5- 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 side8- Balance the electrons lost /gained by multiplying the reactions by integers9- Add the two reactions removing any duplication if any of common species on either side.

Can also be performed without splitting the two equations.

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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 / 144Balancing 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.

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Balancing in Basic Solution4Zn + 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 + 2H2OBalance 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

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Balancing in Basic Solution

Practice 4: Cr(OH)4-1 + H2O2 --> (CrO4) 2- + H2O

2Cr(OH)-1 + 2OH- + 3H2O2 --> 2CrO42- + 8H2O

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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

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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 stripplaced 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

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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.

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Voltaic CellsHere the Cu and Zn strips are in two different beakers

Zn/ZnNO3 Zn ➝ Zn2+ + 2e- OXD- Half reaction

Cu/CuNO3Cu2+ + 2e- ➝ CuRED- Half reaction

We can use the energy to do work if we make the electrons flow through an external device.

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Voltaic CellsHere 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- ➝ CuRED- Half reaction

salt bridge

e-

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

Voltaic Cells

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Voltaic Cells

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

Zn2+

NO3-

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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 Zn2+ are produced

more NO3- are

created in solution

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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.

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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+

Cu2+

Zn metal

e-

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Voltaic Cells

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

Zn metal Cu2+

2e- Cu atom

Zn2+

Cu2+

Zn metal

e-

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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+

Cu2+

Zn metal

e-

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Voltaic Cells

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

This shows how a typical voltaic cell works

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16 The electrode at which oxidation occurs is called the _______________.

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

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18 Which element is oxidized in the reaction below?

Fe2+ + H+ + Cr2O72- ➝ Fe3+ + Cr3+ + H2O

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19 Fe2+ + H+ + Cr2O72- ➝ 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.

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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.

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

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23 The cation concentration increases in the solution where oxidation occurs.

Yes

No

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

True

False

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25 The salt bridge ions may react with the Ions in the cell compartments to form a precipitate.

Yes

No

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

A KNO3

B Na2SO4

C LiC2H3O2

D PbCl2

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

A AgBr B KClC BaF2 D CuS

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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/CuNO3Cu2+ + 2e- ➝ CuRED- Half reaction

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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

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· 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

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· 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

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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

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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

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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) -3.05Na+(aq) + e- ➝ Na(s) -2.71Al3+(aq) + 3e- ➝ Al(s) -1.662H+(aq) + 2e- ➝ H2(g 0Cu2+(aq) + 2e- ➝ Cu(s) + 0.34F2(g) + 2e- ➝ 2F-(aq) + 2.87

The more negative value indicate that, reduction is unlikely at that electrodeThe 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)

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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

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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

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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)

FeZn

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

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

A CuB NaC Cu and Na cannot make a voltaic cell.

F2(g) + 2e- ➝ 2F-(aq) + 2.87Cu2+(aq) + 2e- ➝ Cu(s) + 0.342H+(aq) + 2e- ➝ H2(g 0Al3+(aq) + 3e- ➝ Al(s) -1.66Na+(aq) + e- ➝ Na(s) -2.71Li+ (aq) + e- ➝ Li(s) -3.05

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

A LiB AlC Li and Al cannot make a voltaic cell.

F2(g) + 2e- ➝ 2F-(aq) + 2.87Cu2+(aq) + 2e- ➝ Cu(s) + 0.342H+(aq) + 2e- ➝ H2(g 0Al3+(aq) + 3e- ➝ Al(s) -1.66Na+(aq) + e- ➝ Na(s) -2.71Li+ (aq) + e- ➝ Li(s) -3.05

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

Ecell = Eored.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

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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

CuZn

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

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Cell Potentials

E°cell = E°

red(cathode) - E°red(anode)

= +0.34V - (-0.76V)

E°red = +1.10V

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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

E0cell = 0.34 - (-0.76)

= + 1.10vE°

cell = +0.34V - (-0.76V)

= +1.10V

Cell Potentials

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

cell)?

A Cu-AlB Cu-NaC Al-Li

F2(g) + 2e- ➝ 2F-(aq) + 2.87Cu2+(aq) + 2e- ➝ Cu(s) + 0.342H+(aq) + 2e- ➝ H2(g 0Al3+(aq) + 3e- ➝ Al(s) -1.66Na+(aq) + e- ➝ Na(s) -2.71Li+ (aq) + e- ➝ Li(s) -3.05

D F2 - Cu

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

cell)?

A Cu-AlB Al-NaC Na-Li

F2(g) + 2e- ➝ 2F-(aq) + 2.87Cu2+(aq) + 2e- ➝ Cu(s) + 0.342H+(aq) + 2e- ➝ H2(g 0Al3+(aq) + 3e- ➝ Al(s) -1.66Na+(aq) + e- ➝ Na(s) -2.71Li+ (aq) + e- ➝ Li(s) -3.05

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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 valuesIncr

easi

ng s

treng

th o

f oxi

dizi

ng a

gent

Incr

easi

ng s

treng

th o

f red

ucin

g ag

ent

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

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Class Practice:

Ni Sn

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

Identify:CathodeAnodeOxidation half reactionReduction half reactionCombined cell reaction (balanced) E°

cell =

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Class Practice 2

Fe Sn

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

Identify:CathodeAnodeOxidation half reactionReduction half reactionCombined cell reaction (balanced) E°

cell

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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

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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

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Free Energy

Under standard conditions,ΔG° = -nFE°

Standard condition: 250C, 1 atm and 1M

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Nernst Equation

· Remember that ΔG = ΔG° + RT ln Q

· This means−nFE = −nFE° + RT ln Q

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

or, using base-10 logarithms,

E = E° - [8.31 x 298] x 2.303 log Q

[n x 96500 ]

E = E° - lnQRTnF

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Nernst Equation

At room temperature (298 K),

Thus, the equation becomes

E = E° - logQ0.0592n

= 0.0592 V2.303 RT

F

E = E° - [8.31 x 298] x 2.303 log Q

[n x 96500 ]

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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° - logQ0.0592n

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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 __________________.

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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 /e lectroChem/voltaicCellEMF.html

Ni Ni

1M [Ni2+]0.001M [Ni2+]

Ni Ni

0.5M [Ni2+]0.5M [Ni2+]

E = E° - logQ0.0592n

**

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

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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° - logQ0.0592n

= -0.0529/2 log0.01= - 0.0591

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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

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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

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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, _____________________.

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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 / 144Electrolysis

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

Current (A)and time

Quantity of charge(Coulombs)

Moles of electrons(Faradays)

Moles of substance oxidizedor reduced

Grams ofsubstance

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.

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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

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A) joule B) coulomb C) calorieD) NewtonE) Mole

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

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41 How many coulombs result from a current of 50 amps (A) applied for 20 seconds?

A 2.5

B 10

C 70

D 700

E 1000 A Cs=

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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

Cs=

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

A 1 atmB 25oCC 0.0821 L-atm/mol-KD 96,500 C

E 760 mm Hg

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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

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

A 0.25 B 0.5C 1D 48,250

E 96,500

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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- = 96500CHow 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 aspectElectrolysis

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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

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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

ElectrolysisSlide 134 / 144

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

A 0.33 molB 0.5 mol

C 1.0 molD 3.0 mol

Cr3+(aq) + 3 e- --> Cr(s)

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47 One mole of electrons would allow electroplating of __________ mol of zinc.

A Zn cannot be electroplated.

B 0.5 mol

C 1.0 mol

D 2.0 molZn2+

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

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48 How many minutes will it take to plate out 16.22 g of Al metal from a solution of Al3+ using a current of12.9 amps in an electrolytic cell?

A 60.1 minB 74.9 minC 173 minD 225 minE 13,480 min

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Electrochemistry -Applied

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

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Batteries

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Alkaline Batteries

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Hydrogen Fuel Cells

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Corrosion and…

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…Corrosion Prevention

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In Biology

electron transport in Mitochondria ......

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