Chemistry 130 Electrochemistry Dr. John F. C. Turner 409 Buehler Hall [email protected]

Chemistry 130

Chemistry 130

ElectrochemistryDr. John F. C. Turner

409 Buehler Hall

[email protected]

Chemistry 130

Redox and electron transfer

A redox reaction is one where electrons are transferred between reactants and products, changing the formal oxidation states of the species involved.

In order to balance a redox reaction, we use half equations to show the change in formal oxidation state for the species involved:

Loss of Electrons is Oxidation – LEO – is a useful mnemonic for remembering the direction of redox reactions.

We term reactions such as a half-reaction or a couple.

Cuaq2+ 2e_ Cus

0 reduction

Feaq2+ − e_ Feaq

3+ oxidation

Feaq2+ − e_ Feaq

3+

Chemistry 130

Balancing redox reactions

A stoichiometric redox reaction is composed of two couples – one is an oxidation and one a reduction. Both the oxidation and reduction couple must be present in order to conserve charge.

Balancing a redox reaction is lengthy but straightforward. The key to balancing a reaction is to ensure that the electrons are balanced on both sides, using acid or base to balance the final equation with respect to acid numbers.

Chemistry 130


A stoichiometric redox reaction is composed of two couples – one is an oxidation and one a reduction. Both the oxidation and reduction couple must be present in order to conserve charge.

Balancing a redox reaction is lengthy but straightforward. The key to balancing a reaction is to ensure that the electrons are balanced on both sides, using acid or base to balance the final equation with respect to acid numbers.

Example: Write a balanced equation for the reduction of permanganate ion by thiosulphate in acid solution. The products of the reaction are sulfate ion and

The unbalanced reaction is MnO4aq

- S2O3aq2- Mnaq

2+ SO4aq2-

Mnaq2+

Chemistry 130


Example: Write a balanced equation for the reduction of permanganate ion by thiosulphate in acid solution. The products of the reaction are sulfate ion and

Step 1: Assign oxidations states to the species

Step 2: Write down the half-reactions

Mnaq2+

MnO4aq- S2O3aq

2- Mnaq2+ SO4aq

2-

MnO4aq- S2O3aq

2- Mnaq2+ SO4aq

2-

MnVII SII MnII SVI

MnO4aq- 5e- Mnaq

2+

S2O3aq2- 2SO4aq

2- 8e-

Chemistry 130



Step 2: Write down the half-reactions

Step 3: Use acid and water to balance the number of oxygen atoms present

MnO4aq- S2O3aq

2- Mnaq2+ SO4aq

2-

MnO4aq- S2O3aq

2- Mnaq2+ SO4aq

2-

MnVII SII MnII SVI

MnO4aq- 5e- Mnaq

2+

S2O3aq2- 2SO4aq

2- 8e-

MnO4aq- 5e- 8Haq

+ Mnaq2+ 4H2O

5H2O S2O3aq2- 2SO4aq

2- 8e- 10Haq+

Chemistry 130


After steps 1, 2 and 3, we have two balanced half-equations:

Step 4: We now multiply them individually to balance the number of electrons:

MnO4aq- S2O3aq

2- Mnaq2+ SO4aq

2-

MnO4aq- 5e- 8Haq

+ Mnaq2+ 4H2O

5H2O S2O3aq2- 2SO4aq

2- 8e- 10Haq+

8MnO4aq- 40e- 64Haq

+ 8Mnaq2+ 32H2O

25H2O 5S2O3aq2- 10SO4aq

2- 40e- 50Haq+

Chemistry 130


Step 5: We now add the two half- reactions together:

Step 6: Species appear on both sides of the reaction, which we cancel. I particular, 40e- occur on both sides which means that we are correct in terms of the oxidation and reductions. Also, we have

MnO4aq- S2O3aq

2- Mnaq2+ SO4aq

2-

8MnO4aq- 40e- 64Haq

+ 25H2O 5S2O3aq2-

8Mnaq

2+ 32H2O 10SO4aq2- 40e- 50Haq

+

64−50Haq+ and 32−25H2O

Chemistry 130


Step 6: The final reaction is

MnO4aq- S2O3aq

2- Mnaq2+ SO4aq

2-

8MnO4aq- 14Haq

+ 5S2O3aq2- 8Mnaq

2+ 10SO4aq2- 7H2O

8MnO4aq- 40e- 64Haq

+ 25H2O 5S2O3aq2-

8Mnaq

2+ 32H2O 10SO4aq2- 40e- 50Haq

+

Chemistry 130


Example: In basic solution, bromine disproportionates to give bromide ion and bromate ion. Write a balanced equation for this reaction.

The unbalanced reaction is


Step 2: Write down the half-reactions:

Br2 l Braq- BrO3aq

-

Br2 l Braq- BrO3aq

-

Br0 Br−I Br−V

12

Br2 l e- Braq-

12

Br2 l BrO3aq- 5e-

or Br2l 2e- 2Braq

-

Br2 l 2BrO3aq- 10e-

Chemistry 130


Step 3: Use acid and water to balance the number of oxygen atoms present

We now have two individually balanced half reactions.

Step 4: We now multiply them individually to balance the number of electrons:

Br2 l Braq- BrO3aq

-

Br2l 2e- 2Braq-

6H2O Br2l 2BrO3aq- 10e- 12Haq

+

5Br2l 10e- 10Braq-


+

Chemistry 130


Step 5: We now add the two half- reactions together:

This is the balanced reaction in acid solution. We need the reaction in basic solution so we add sufficient hydroxide ion to remove the hydronium ion.

Br2 l Braq- BrO3aq

-

5Br2l 10e- 10Braq-


+

6Br2l 10e- 6H2O 10Braq- 2BrO3aq

- 10e- 12Haq+

or

3Br2l 3H2O 5Braq- BrO3aq

- 6Haq+

Chemistry 130


This is the balanced reaction in acid solution. We need the reaction in basic solution so we add sufficient hydroxide ion to remove the hydronium ion.

Which is the final, balanced equation in basic solution.

Br2 l Braq- BrO3aq

-

Acid

3Br2l 3H2O 5Braq- BrO3aq

- 6Haq+

Base

3Br2l 6OHaq- 3H2O 5Braq

- BrO3aq- 6Haq

+ 6OHaq-

3Br2l 6OHaq- 3H2O 5Braq

- BrO3aq- 6H2O

3Br2l 6OHaq- 5Braq

- BrO3aq- 3H2O

Chemistry 130

Redox reactions and electrochemical cells

Redox reactions involve the transfer of electrons from one species to another. For the two previous examples in acidic and basic solution, there reactions take place implicitly in the same system.

If we separate the two reaction physically, then no reaction will obviously take place unless there is a way that the electrons can pass between the oxidative and reductive parts of the reaction – the half-reactions.

If we connect the two reactions via a conducting wire, the external electrical contact, then electrons can move through the wire.

This is known as a voltaic cell.

A salt bridge connects the rest of the circuit – the internal electrical contact.

Chemistry 130

Redox reactions and electrochemical cells

We can harness the current that moves through the cell to do work and when we do so, we are converting chemical energy into work of some description – a light bulb or a motor are two obvious applications.

In a simple cell based on copper and silver, the half reactions are:

The reaction between copper metal and silver ion occurs because the electrons on copper are less tightly bound – copper is more electropositive than silver.

Spontaneously, the reaction will proceed at the Cu surface to form silver metal and copper ion.

Agaq+ e_ Ags

0 reduction

Cus0 − 2e_ Cuaq

2+ oxidation

Chemistry 130

Electrochemical cells

An electrochemical cell is system that contains an oxidation couple and a reduction couple, together with two electrodes that connect the two reactions and allow electrons to pass and a salt bridge that allows the necessary charge neutrality of the system to be maintained.

The two couples are termed half cells and in general, an oxidative and a reductive cell can be assembled to form the full electrochemical cell.

There is a special nomenclature that is used to describe electrochemical cells.

Oxidation takes place at the anode

Reduction takes place at the cathode

Chemistry 130

Cell diagrams

Cell diagrams are used to represent the cell and is a method that represents the full cell reaction. We denote phase boundaries in the cell with a vertical line: |

and the boundary between the two cells, which contains the external electrical connection (the wire) and the internal electrical connection (the salt bridge) by a double line:

For the silver and copper half-reactions, we write the cell as

∥

Agaq+ e_ Ags

0

Cus0 − 2e_ Cuaq

2+

Cus0 ∣ Cuaq

2+

Anode∥ Agaq

+ ∣ Ags0

CathodeOxidation Reduction

Chemistry 130

Cell diagrams

A standard cell is written with the anode on the left and the cathode on the right.

The standard cell potential is then written as

Agaq+ e_ Ags

0

Cus0 − 2e_ Cuaq

2+

Cus0 ∣ Cuaq

2+

Anode∥ Agaq

+ ∣ Ags0

CathodeOxidation Reduction

E ° = E1/2° right − E1/2

° left

E ° = E1/2° cathode − E1/2

° anode

Chemistry 130

Potentials and the Gibbs function

For each half reaction, we can write a half-cell potential, such that we can calculate the potential of the full cell by adding these half-cell potentials together.

The half-cell potential is related to the standard Gibbs function for the half reaction via

F is Faraday's constant and which is the charge on 1 mole of electrons. n is the number of electrons passed in the half-cell reaction.

Manipulating half-cell or full cell potentials is equivalent to manipulating the standard Gibbs function for the half-cell or full cell reaction.

E1/2°

G ° = −nF E1/2°

F = 96,485 C mol−1

Chemistry 130

Standard potentials

The zero point for electrode potentials is the Standard Hydrogen Electrode (SHE)

Platinum acts as a catalyst for the hydrogen-hydronium oxidation, which is why there are two phase boundaries in the cell.

All other electrode potentials are measured against this reaction, with the pressure of hydrogen being 1 atm, all concentrations 1 M and the temperature being 298 K.

Standard half-cell reactions are written as reductions by convention.

Pts ∣ H2g ∣ H3Oaq+ ∥ E1/2

° = 0.00 V

Chemistry 130

Standard potentials

Example: Calculate the cell potential for the zinc-copper cell, given the following cell potentials and construct a full cell diagram for the spontaneous reaction.

Step 1: Calculate the change in standard Gibbs function for each reaction

We do not need to include the value of F as we will convert back at the end of the calculation.

Cuaq2+ 2e_ Cus

0 E1/2° = 0.340 V

Znaq2+ 2e_ Zns

0 E1/2° = −0.763 V

Cuaq2+ 2e_ Cus

0 E1/2° = 0.340 V G° = −nF E1/2

° = −0.68F

Znaq2+ 2e_ Zns

0 E1/2° = −0.763 V G° = −nF E1/2

° = 1.526F

Chemistry 130

Standard potentials


Step 2: Arrange the equations to give a balanced stoichiometric reaction, changing as necessary

is positive, so as written, the reaction is non-spontaneous, so we reverse the reaction.

Cuaq2+ 2e_ Cus

0 E1/2° = 0.340 V G° = −nF E1/2

° = −0.68F

Znaq2+ 2e_ Zns

0 E1/2° = −0.763 V G° = −nF E1/2

° = 1.526F

G °

Cus0 Cuaq

2+ 2e_ G° = 0.68F

Znaq2+ 2e_ Zns

0 G° = 1.526F

Cus0 Znaq

2+ Cuaq2+ Zns

0 G° = 2.206F

G °

Chemistry 130

Standard potentials


We now have a spontaneous reaction as is negative.

Step 3: Convert the calculated back to the electrode potential:

Cus0 Cuaq

2+ 2e_ G° = 0.68F

Znaq2+ 2e_ Zns

0 G° = 1.526F

Cus0 Znaq

2+ Cuaq2+ Zns

0 G° = 2.206F

Cuaq2+ Zns

0 Cus0 Znaq

2+ G° = −2.206F

G °

Ecell° = −

G°

nF= −

−2.206F2F

=2.206F

2F= 1.103 V

G °

Chemistry 130

Standard potentials


Step 4: Construct the cell diagram from the balanced, spontaneous reaction:

Cuaq2+ Zns

0 Cu s0 Znaq

2+ G° = −2.206F Ecell° = 1.103 V

Cuaq2+ Zns

0 Cus0 Znaq

2+ Ecell° = 1.103 V

Zns0 ∣ Znaq

2+

Anode∥ Cuaq

+ ∣ Cus0

CathodeEcell

° = 1.103 V

Oxidation Reduction

Chemistry 130

Standard potentials

This method always gives the correct reaction and the correct direction for the spontaneous cell reaction, even when the cell reaction is not obvious. It is equivalent to the Right – Left rule but is simpler as you don't need to identify the cathodic and anodic reaction initially.

Chemistry 130

Thermodynamics and electrochemistry

The relationships between the equilibrium constant, the cell potential and the change in the standard Gibbs functions allows us to write a relationship between the cell potential and the equilibrium constant G° = −nF Ecell

°

G° = −RT lnKeq

−nF Ecell° = −RT ln Keq

Ecell° = RT

nFlnKeq

Chemistry 130

Non-standard cell potentials

The dependency of the Gibbs function with concentration is

where Q is the reaction quotient. Including this in the equation for the cell potential gives the Nernst equation:

G = G ° RT lnQ

G = G° RT lnQ

−nF Ecell = −nF Ecell° RT lnQ

nF Ecell = nF Ecell° − RT lnQ

Ecell = Ecell° − RT

nFlnQ

Nernstequation

Chemistry 130


The Nernst equation allows us to relate changes in concentration to the cell potential. For the reaction

is written for concentrations of 1 mol l-1. In general, the potential of the cell will vary with concentration as

Cuaq2+ Zns

0 Cus0 Znaq

2+ Q =[Znaq

2+ ]

[Cuaq2+ ]

Ecell = Ecell° − RT

nFlnQ

Ecell = Ecell° −

RTnF

ln{[Znaq2+ ]

[Cuaq2+ ] }

Cuaq2+ Zns

0 Cu s0 Znaq

2+ G° = −2.206F Ecell° = 1.103 V

Ecell°

Chemistry 130


Because of the logarithmic relationship between concentration and the cell potential, we can measure very precisely very large or very small quantities.

Values of Keq that span 50 or 100 orders of magnitude result in

changes in the cell potential of a few volts, allowing us to measure very subtle changes in concentration very accurately.

In order to use the Nernst equation, we first calculate the standard cell potential for the balanced stoichiometric reaction.

Chemistry 130


Example: Calculate the cell potential based on the couples when

Step 1: Write down the half reactions for the two couples:

Step 2: From the balanced half cell reactions, calculate the change in the standard Gibbs function in terms of F

Cu2+ / Cu0 and Zn2+ / Zn0

[Znaq2+ ] = 2 mol l−1 and [Cuaq

2+ ] = 0.05 mol l−1

Cuaq2+ 2e_ Cus

0 E1/2° = 0.340 V

Znaq2+ 2e_ Zns

0 E1/2° = −0.763 V

Cuaq2+ 2e_ Cus

0 E1/2° = 0.340 V G° = −nF E1/2

° = −0.68F

Znaq2+ 2e_ Zns

0 E1/2° = −0.763 V G° = −nF E1/2

° = 1.526F

Chemistry 130



Step 3: Arrange the equations to give a balanced stoichiometric reaction, changing as necessary

Step 4: Ensure that is negative and therefore the reaction is spontaneous



2+ ] = 0.05 mol l−1

G °

Cus0 Cuaq

2+ 2e_ G° = 0.68F

Znaq2+ 2e_ Zns

0 G° = 1.526F

Cus0 Znaq

2+ Cuaq2+ Zns

0 G° = 2.206F

Cus0 Znaq

2+ Cuaq2+ Zns

0 G° = 2.206F

Cuaq2+ Zns

0 Cus0 Znaq

2+ G° = −2.206F

G °

Chemistry 130



Step 5: Calculate the standard cell potential from the change in standard Gibbs function.

Step 6: Calculate the value of the reaction quotient for the reaction



2+ ] = 0.05 mol l−1

Ecell° = −

G°

nF= −

−2.206F2F

=2.206F

2F= 1.103 V

Cuaq2+ Zns

0 Cus0 Znaq

2+ Q =[Znaq

2+ ]

[Cuaq2+ ]

= 20.05

= 40

Chemistry 130



Step 7: Use the Nernst equation to calculate the cell potential



2+ ] = 0.05 mol l−1

Ecell = Ecell° −

RTnF

ln{ [Znaq2+ ]

[Cuaq2+ ] }

Ecell° = 1.103 V

Ecell = 1.103 − 8.314×2982×96,485

ln{ 20.05 } = 1.103 − 0.013ln {40}

Ecell = 1.103 − 0.013ln {40} = 1.055 V

Chemistry 130

Chemistry 130

Nuclear ChemistryDr. John F. C. Turner

409 Buehler Hall

[email protected]

Chemistry 130

Radioactivity

The majority of the chemical elements are stable in that their nuclei do not show any form of decay.

In general, from hydrogen (Z = 1) to bismuth (Z = 83) all elements have at least one stable nucleus with the exceptions of technetium (Z = 43) and promethium (Z = 61).

After Bi, all the elements do not have a stable nucleus.

Even for the elements that have at least one stable isotope, many radioactive isotopes are also known and are produced artificially.

Examples include

H13

tritiumC6

14 P1532

Chemistry 130

The nucleus

The nucleus is a very complicated object. Part of the difficulty of describing the nucleus is the strength of the forces involved and the level at which they couple to each other.

The two forces in the nucleus that are important in terms of stability are the strong nuclear force and the electromagnetic force.

Electrons are bound in the atom by the electromagnetic force and the equilibrium size of the atom represents the magnitude of the force involved – atoms have a radius of ~ 1-2 Å or ~ 10-10 m.

The nuclear radius is ~ 1-2 fm or ~ 10-15 m or 5 orders of magnitude smaller than the atom.

Chemistry 130

The nucleus

The constituents of the nucleus are the neutron and the proton.

The neutron has no electric charge, a spin of ½ and a mass very similar to that of the proton.

The proton has a single positive charge and a spin of ½ as well. Both the neutron and the proton are therefore fermions and obey the Pauli principle in a manner analogous to electrons in the atom:

The wavefunction of a fermion must change sign on interchange of particles

or

No more than two fermions can occupy the same state

Chemistry 130

The nucleus

Though a description of the nucleus is extremely hard, several features are immediately obvious.

There is a force that is attractive and extremely powerful – the strong nuclear force.

The strong force is powerful enough to overcome the electrostatic repulsion of the protons in the nucleus and so the nucleons in the nucleus are bound – elements after hydrogen are usually stable up to Z = 83.

The size of the nucleus is dictated by the balance between these two forces.

The definition of the nuclear surface and the nuclear radius and size is ambiguous and depends on the particle.

Chemistry 130

This week

Revise kinetics and acids, in addition to the electrochemistry

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Chemistry 130 Electrochemistry Dr. John F. C. Turner 409 Buehler Hall [email protected]