Electrochemistry

SCE 3109

Energetics in Chemistry

SCE 3109

Energetics in Chemistry

Course Textbooks

• Brown, T.L., Lemay, H.E. & Bursten, B.E. (2006).

Chemistry –The Central Science. 10th ed. New

Jersey: Prentice Hall.

• McMurry, J. & Fay, R.C. (2008). Chemistry. 4th ed.

New Jersey: Prentice Hall.

Additional reference:

Redox reactions

• Electrochemistry is the branch of chemistry that deal with the interconversion of electrical energy and chemical

energy.

• Electrochemical processes are redox (oxidation-reduction) reactions.

• In redox reactions, electrons are transferred from one species to another.

• Oxidation and reduction must occur together.

Oxidation is loss of electrons.

Redox reactions

Oxidation Numbers

In order to keep track of what loses electrons

and what gains them, we assign oxidation numbers.

Redox reactions

Oxidation and Reduction

• A species is oxidized when it loses electrons.

– Here, zinc loses two electrons to go from neutral zinc metal to the Zn2+ _ion.

• A species is reduced when it gains electrons.

– Here, each of the H+ _{gains an electron and they}

Redox reactions

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.}

Redox reactions

Half Reactions

• Although oxidation and reduction must take place

simultaneously, it is often convenient to consider them as separate processes.

Redox reactions

Half Reactions

• Oxidation Half-Reaction: Zn(s) → Zn2+_{(aq) + 2 e}–

Redox reactions

Half Reactions

• Reduction Half-Reaction: Cu2+_{(aq) + 2 e}–_{→ Cu(s)}

Redox reactions

Redox reactions

Balancing Redox Equations by Methods of

Half-Reaction

• The easiest way to balance the equation of an

oxidation-reduction reaction is via the half-reaction method.

• This involves treating the oxidation and reduction as two separate processes, balancing these half

reactions, and then combining them to attain the balanced equation for the overall reaction.

Redox reactions

Redox reactions

Half-Reaction Method

1. Assign oxidation numbers to determine what is oxidized and what is reduced.

2. Write the oxidation and reduction half-reactions. 3. Balance each half-reaction.

a) Balance elements other than H and O. b) Balance O by adding H₂O.

c) Balance H by adding H+.

Redox reactions

Half-Reaction Method

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

5. Add the half-reactions, subtracting things that appear on both sides.

6. Make sure the equation is balanced according to mass. 7. Make sure the equation is balanced according to

Redox reactions

Half-Reaction Method

Consider the reaction between MnO₄− _{and C}

2O42− :

MnO₄−_{(aq) + C}

Redox reactions

Half-Reaction Method

First, we assign oxidation numbers.

MnO

_{+ C}

2

O



Mn

+ CO

+7 +3 +2 +4

Since the manganese goes from +7 to +2, it is reduced. Since the carbon goes from +3 to +4, it is oxidized.

Redox reactions

Half-Reaction Method

Oxidation Half-Reaction

C₂O₄2−  CO 2

To balance the carbon, we add a coefficient of 2: C₂O₄2−  _{2 CO}

2

The oxygen is now balanced as well.

To balance the charge, we must add 2 electrons to the right side.

C₂O₄2−  2 CO

Redox reactions

Half-Reaction Method

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 H} 2O

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

8 H+ _{+ MnO}

Redox reactions

Half-Reaction Method

Reduction Half-Reaction 8 H+ _{+ MnO}

4−  Mn2+ + 4 H2O

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

5 e− _{+ 8 H}+ _{+ MnO}

Redox reactions

Half-Reaction Method

Combining the Half-Reaction C₂O₄2−  2 CO 2 + 2 e− 5 e− _{+ 8 H}+ _{+ MnO} 4−  Mn2+ + 4 H2O x 5 x 2

Redox reactions

Half-Reaction Method

Combining the Half-Reaction

5 C₂O₄2−  _{10 CO} 2 + 10 e− 10 e− _{+ 16 H}+ _{+ 2 MnO} 4−  2 Mn2+ + 8 H2O 16 H+ _{+ 2 MnO} 4− + 5 C2O42−  2 Mn2+ + 8 H2O + 10 CO2

Redox reactions

Half-Reaction Method

Balancing Redox Equations in Acidic Solution

Complete and balance the following equations using the method of half-reaction.

a) Cr₂O₇2− _{(aq) + Cl}− (aq) Cr3+ _{(aq) + Cl}

2 (g)

b) Cu (s) + NO₃− (aq)  Cu2+ _{(aq) + NO}

2 (g)

c) Mn2+ _{(aq) + NaBiO}

Redox reactions

Half-Reaction Method

• If a reaction occurs in basic solution, we can balance it as if it occurred in acid.

• Once the equation is balanced, add OH− _{to each side to}

“neutralize”the H+ _{in the equation and create water in its}

place.

• If this produces water on both sides, we have to subtract water from each side.

Balancing Redox Equations in Basic Solution

Voltaic Cells

• In spontaneous oxidation-reduction (redox) reactions, electrons are transferred and energy is released.

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

device.

• We call such a setup a voltaic

Voltaic Cells

Zn2+_{(aq) + Cu(s)}

Voltaic Cells

• A typical cell looks like this.

• The oxidation occurs at the anode.

• The reduction occurs at the cathode.

• Each of the two

compartments of a voltaic cell is called a half-cell.

Voltaic Cells

• Electrons become

available as zinc metal is oxidized at the anode.

• Then, electrons leave the anode and flow

through the wire to the cathode.

Zn2+_{(aq) + 2e}

Voltaic Cells

• As the electrons leave the anode, the cations formed dissolve into the solution in the anode

compartment.

• The zinc electrode loses mass. • The concentration of the Zn2+ _solution increases. Zn2+_{(aq) + 2e} -Zn(s)

Voltaic Cells

Cu(s) Cu2+_{(aq) + 2e}

-• As the electrons reach the cathode, cations in the solution are

Voltaic Cells

Cu(s) Cu2+_{(aq) + 2e}

-• The electrons are taken by the cation, and the neutral metal, Cu, is deposited on the

cathode.

• The Cu electrode gains mass.

• The Cu2+ _solution

becomes less

concentrated as Cu2+ _is

Voltaic Cells

• For a voltaic cell to work, the solutions in the two half-cells must remain electrically neutral.

Voltaic Cells

• As Zn is oxidized in the anode compartment, Zn2+ _{ions enter the}

solution.

• Thus, there must be some means for

positive ions to migrate out of the anode

compartment or for

negative ions to migrate in to keep the solution electrically neutral.

Voltaic Cells

• Similarly, the reduction of Cu2+ _{at the cathode}

removes positive charges from the solution, leaving an excess of negative

charges in that half-cell. • Thus, positive ions must

migrate into the compartment or

negative ions must migrate out.

Voltaic Cells

• Therefore, we use a porous glass barrier or a salt bridge to maintain the electrical neutrality of the solutions.

• Porous glass barrier separates the two compartments, but

Voltaic Cells

• A salt bridge, usually a U-shaped tube that

contains a salt solution, such as NaNO₃(aq),

whose ions will not

react with other ions in the cell or with the

electrode materials. • The salt solution is

often incorporated into a pasta or gel so that the salt solution does not pour out.

Voltaic Cells

• As oxidation and

reduction proceed at the electrodes, ions from the salt bridges migrate to neutralise charge in the cell compartment.

– Cations move toward the cathode.

– Anions move toward the anode.

Voltaic Cells

• In any voltaic cell the electrons flow from the anode through the

external circuit to the cathode.

• The anode in a voltaic cell is labeled with a negative sign and the cathode with a positive sign.

Voltaic Cells

Overall cell reaction:

Anode half-reaction:

Cathode half-reaction:

Zn(s) | Zn2+_{(aq) || Cu}2+_{(aq) | Cu(s)}

Phase boundary Phase boundary Electron flow Salt bridge Cathode half-cell Anode half-cell

Example

Write a balanced equation for the overall cell reaction, and give a brief description of a galvanic cell represented by the following shorthand notation:

Fe(s) | Fe2+_{(aq) || Sn}2+_{(aq) | Sn(s)}

Fe2+_{(aq) + Sn(s)}

Fe(s) + Sn2+_(aq)

Example

Design a galvanic cell that uses the redox reaction 2Ag(s) + Ni2+_(aq)

2Ag+_{(aq) + Ni(s)}

Identify the anode and cathode half-reactions, and sketch the experimental setup. Label the anode and cathode,

indicate the direction of electron and ion flow, and identify the sign of each electrode.

Example

Consider the following galvanic cell.

Example

Problem 17.4 (Page 693) (McMurry & Fay)

a) Complete the drawing by adding any components essential for a functioning cell.

b) Label the anode and cathode, and indicate the direction of ion flow.

c) Write a balanced equation for the cell reaction. d) Write the shorthand notation for the cell.

Electrochemical Cells

Electrochemical cells are of two basic types:

Galvanic (Voltaic) Cell:

A spontaneous chemical reaction which generates an electric current.

Electrolytic Cell:

An electric current which drives a nonspontaneous

Voltaic Cells

• The redox reaction between Zn and Cu2+ _{is spontaneous}

regardless of whether they

react directly or in the separate compartments of a voltaic cell.

Voltaic Cells

• Electrons move through the external circuit from zinc anode to the

copper cathode.

• Why do electrons

transfer spontaneously from Zn anode to Cu cathode?

Electromotive Force (emf)

• Water only

spontaneously flows one way in a waterfall. • Likewise, electrons

only spontaneously flow one way in a

redox reaction — from higher to lower

Electromotive Force (emf)

• The potential difference between the anode and cathode in a cell is called the electromotive force (emf).

• It is also called the cell potential (E_cell) or the cell voltage.

• The potential difference between the two electrodes of a voltaic cell provides the driving force that pushes

electrons through the external circuit.

Electromotive Force (emf): The force or electrical potential that pushes the negatively charged electrons away from the anode (- electrode) and pulls them toward the cathode (+ electrode).

Electromotive Force (emf)

Cell potential is measured in volts (V).

1 V = 1

_C

J

volt

SI unit of electric potential

joule

SI unit of energy

coulomb

Electric charge

1 coulomb is the amount of charge transferred when a current of 1 ampere flows for 1 second.

Electromotive Force (emf)

• The emf of a paricular voltaic cell depends on the

 specific reaction that occur at the cathode and anode  the concentrations of reactants and products

 temperature

• Under standard conditions (1 M concentrations for reactants and products in solution; 1 atm pressure for those that are gases; solids and liquids in pure form), the emf is called the

standard emf, or the standard cell potential, and is denoted Eo

Standard Cell Potentials

• For the following Zn-Cu voltaic cell, the standard cell potential at 25°C is +1.10 V.

For any cell reaction that proceeds spontaneously, such as that in a voltaic cell, the cell potential will be positive.

Zn2+_{(aq, 1 M) + Cu(s)}

Zn(s) + Cu2+_{(aq, 1 M) E}

Standard Cell Potentials

• The emf or cell potential of a voltaic cell depends on the particular cathode and anode half-cells involved.

• The standard cell potential

Standard Reduction Potentials

Standard Hydrogen Electrode

• The standard reduction potentials (often called half-cell potentials) are determined from the difference between two electrodes.

• The reference point is called the standard hydrogen

electrode (S.H.E.) and consists of a platinum electrode in contact with H₂ gas (1 atm) and aqueous H+ _{ions (1 M).}

• The standard hydrogen electrode is assigned an arbitrary value of exactly 0.00 V.

Standard Reduction Potentials

The standard hydrogen electrode (S.H.E.) has been chosen to be the reference electrode.

Standard Reduction Potentials

Cu(s) Cu2+_{(aq) + 2e}

-0.34 V = E°_red (cathode) ― 0 V

E°_red = 0.34 V A standard reduction potential can be defined:

E_cell = E_red (cathode) − E_red (anode)

2H1+_{(aq) + Cu(s)}

H₂(g) + Cu2+_(aq)

2H1+_{(aq) + 2e}

-H₂(g)

Overall cell reaction:

Anode half-reaction:

Standard Reduction Potentials

Eo

red (anode) = - 0.76 V

E_cell = E_red (cathode) − E_red (anode)

H₂(g) + Zn2+_(aq) 2H1+_{(aq) + Zn(s)} Zn2+_{(aq) + 2e} -Zn(s) H₂(g) 2H1+_{(aq) + 2e}

-Overall cell reaction:

Anode half-reaction:

Using Standard Cell Potentials

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

Using Standard Cell Potentials

• For the oxidation in this cell,

• For the reduction,

E_red = −0.76 V

E_red = +0.34 V

E_cell = E_red(cathode) − E_red (anode) = +0.34 V − (−0.76 V) = +1.10 V

Using Standard Cell Potentials

• When selecting two half-cell reactions the more negative value will form the oxidation half-cell (anode).

• Consider the reaction between zinc and silver: Ag+_{(aq) + e}– _{→ Ag(s) E°= 0.80 V}

Zn2+_{(aq) + 2 e}–_{→ Zn(s) E°= –0.76 V}

• Therefore, zinc forms the oxidation half-cell (anode).

E_cell = E_red(cathode) − E_red (anode) = +0.80 V − (−0.76 V) = +1.56 V

Using Standard Cell Potentials

Example

• Using data in Appendix E, calculate the standard emf for a cell that employs the following overall cell reaction:

E_cell _{= Ered}_{(cathode) − Ered} _(anode)

= +0.536 V − (−1.66 V) = +2.196 V

2Al3+_{(aq) + 6I}― _(aq)

Using Standard Cell Potentials

Example

• Using data in Appendix E, calculate the standard emf for a cell that employs the following overall cell reaction:

E_cell _{= Ered}_{(cathode) − Ered} _(anode)

= +1.33 V − (0.536 V) = +0.794 V

2Cr3+_{(aq) + 3I}

2 (s) + 7H2O

Using Standard Cell Potentials

Example

• A voltaic cell is based on the half reactions:

The standard emf for this cell is 1.46 V. Using data in Appendix E, calculate Eo

red for the reduction of In3+ to In+.

E_cell = E_red(cathode) − E_red (anode)

In3+_{(aq) + 2e} -In+_(aq) 2Br―_(aq) Br₂(l) + 2e -= +1.065 V − Eo red (anode) 1.46 V Eo red (anode) = - 0.395 V

Using Standard Cell Potentials

Example

• A voltaic cell is based on the following two standard half reactions:

By using data in Appendix E, determine

a) the half-reactions that occur at the cathode and the anode

b) the standard cell potential. Cd(s)

Cd2+_{(aq) + 2e}

-Sn(s) Sn2+_{(aq) + 2e}

-Using Standard Cell Potentials

Solution:

a)

b) _E

cell = Ered(cathode) − Ered (anode)

Cd2+_{(aq) + 2e}

-Anode: Cd(s)

= (- 0.136 V) − (- 0.403 V) = 0.267 V Sn(s)

-Using Standard Cell Potentials

Example

• A voltaic cell is based on Co2+_{/Co half-cell and an}

AgCl/Ag half-cell.

a) Write the half-reaction occurs at the anode? b) What is the standard cell potential?

Answers: (a) Co  Co2+ _{+ 2e}

Using Standard Cell Potentials

Using Standard Cell Potentials

Example

a) What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO₃)₂ solution and a Cr electrode in a 1.0 M Cr(NO₃)₃ solution?

b) What is the standard emf of an electrochemical cell made of a Mg electrode in a 1.0 M Mg(NO₃)₂ solution and a Ag electrode in a 1.0 M AgNO₃ solution?

Using Standard Cell Potentials

Predicting whether a redox reaction is spontaneous

Using the standard reduction potentials listed in Appendix E, determine whether the following reactions are

spontaneous under standard conditions. Cu2+_{(aq) + H}

2(g)

(a) Cu(s) + 2H+_(aq)

Cu(s) + 2H+_(aq) (b) Cu2+_{(aq) + H} 2(g) 2Cl-_{(aq) + I} 2(s) (c) Cl₂(g) + 2I-_(aq)

Using Standard Cell Potentials

Solution: Cu2+_{(aq) + H}

2(g)

(a) Cu(s) + 2H+_(aq)

In this reaction, Cu is oxidized to Cu2+ _{and H}+ _{is reduced to}

H₂.

H₂(g) 2H+_{(aq) + 2e}

-Reduction:

Oxidation: _Cu(s) Cu2+_{(aq) + 2e}

-Eo

red = 0 V

Eo

red = +0.34 V

E  = E_red(cathode) − E_red (anode) = 0 V − (0.34 V) = -0.34 V

Because EO _{is negative, the reaction is not spontaneous in the} direction written.

Using Standard Cell Potentials

Solution:

In this reaction, H₂ is oxidized to H+ _{and Cu}2+ _{is reduced to}

Cu. Cu(s) Cu2+_{(aq) + 2e} -Reduction: Oxidation: _H 2H+_{(aq) + 2e} -2(g) Eo red = +0.34 V Eo red = 0 V

E  = E_red(cathode) − E_red (anode) = 0.34 V − (0 V) = 0.34 V

Because EO _{is positive, the reaction is spontaneous and could be} used to build a voltaic cell.

Cu(s) + 2H+_(aq)

(b) Cu2+_{(aq) + H}

Using Standard Cell Potentials

E  = E_red(cathode) − E_red (anode) = 1.36 V − (0.54 V) = +0.82 V

Because EO _{is positive, the reaction is spontaneous and could be} used to build a voltaic cell.

2Cl-_{(aq) + I}

2(s)

Using Standard Cell Potentials

Predicting whether a redox reaction is spontaneous

Practice Exercise: Brown (Pg 869)

Oxidizing and Reducing Agents

• The strongest oxidizers have the most positive reduction potentials.

• The strongest reducers have the most negative reduction potentials.

Oxidizing and Reducing Agents

Example:

a) Rank the following ions in order of increasing strength as oxidizing agents: NO₃- _{(aq), Ag}+ _{(aq), Cr}

2O72- (aq).

Answer: Ag+ _{< NO}

3- < Cr2O7

2-b) Rank the following species from the strongest to the weakest reducing agent: I- _{(aq), Fe (s), Al (s).}

Oxidizing and Reducing Agents

Example:

c) Arrange the following oxidizing agents in order of

increasing strength under standard-state conditions: Br₂(aq), Fe3+ _{(aq), Cr}

2O72- (aq).

Answer: Fe3+ _{< Br}

2 < Cr2O7

2-d) Arrange the following reducing agents in order of

increasing strength under standard-state conditions : Al(s), Na(s), Zn(s).

Oxidizing and Reducing Agents

The greater the difference between the two, the

greater the voltage of the cell.

Oxidizing and Reducing Agents

Consider the following table of standard reduction potentials:

a) Which substance is the strongest reducing agent? Which is the strongest oxidizing agent?

b) Which substances can be oxidized by B2+_{? Which can be}

reduced by C?

c) Write a balanced equation for the overall cell reaction that

Oxidizing and Reducing Agents

Answers:

(a) D is the strongest reducing agent. A3+ _{is the strongest oxidizing agent.}

(b) B2+ _{can oxidize C and D.}

C can reduce A3+ _{and B}2+_.

(c) A3+ _{+ 2D  A}+ _{+ 2 D}+

Free Energy & Redox Reactions

• Two quantitative measures of the tendency for a chemical reaction to occur:

(a) the cell potential, E (elecrochemical quantity)

(b) the free-energy change, G (themochemical quantity)

• The change in Gibbs free energy, G, is a measure of the spontaneity of a process that occurs at constant temperature and pressure.

Free Energy & Redox Reactions

• The relationship between emf, E and the free-energy change, G is

G = −nFE

where n is the number of moles of electrons transferred, and F is a constant, the Faraday.

Note:

Faraday’s constant is the quantity of electrical charge on one mole of electrons.

Free Energy & Redox Reactions

• G for a redox reaction can be found by using the equation

G = −nFE

If the free-energy change, G, for a reaction is negative, the reaction is spontaneous.

The greater the negative value of G, the greater the tendency for the reaction to occur.

The unit of G is J/mol.

Note: A positive value of E and a negative value of



Free Energy & Redox Reactions

For a constant-temperature process:

The change of Gibbs free energy (ΔG) G = −nFE

G < 0 The reaction is spontaneous in the forward direction. G > 0 The reaction is nonspontaneous as written. The

reaction is spontaneous in the reverse direction. G = 0 The reaction is at equilibrium.

Free Energy & Redox Reactions

• Under standard conditions,

Free Energy & Redox Reactions

Calculate the standard free-energy change for this reaction at 25 °C.

Zn2+_{(aq) + Cu(s)}

Zn(s) + Cu2+_(aq)

The standard cell potential at 25 °C is 1.10 V for the reaction: G° = -212267 J/mol = -212.267 kJ/mol (1.10 V) = -(2) V mol 96,485 J G° = -nFE°

Free Energy & Redox Reactions

What is the standard free-energy change for this reaction at 25 °C?

Al3+_{(aq) + Cr(s)}

Al(s) + Cr3+_(aq)

Example:

The standard cell potential at 25 °C is 0.92 V for the reaction:

Free Energy & Redox Reactions

3Ca2+_{(1 M) + 2Al(s)}

3Ca(s) + 2Al3+_{(1 M)}

Example:

Calculate the standard free-energy change for the following reaction at 25°C:

Answer: 7 x 102 _kJ/mol

Comment: The positive value of ΔG° tell us that the reaction is not spontaneous under standard-state conditions at 25°C.

Cell EMF Under Nonstandard

Conditions

• Suppose we start a reaction in solution with all the reactants in their standard states (that is, all at 1 M concentration.

• As soon as the reaction starts, the standard-state condition no longer exists for the reactants or the

products because their concentrations are different from 1 M.

• Under nonstantard-state conditions, we must use G rather than G to predict the direction of the reaction.

Cell EMF Under Nonstandard

Conditions

• The relationship between G and G is G = G°+ RT ln Q

R is the gas constant (8.314 J/K •mol)

T is the absolute temperature (K) Q is the reaction quotient

Cell EMF Under Nonstandard

Conditions

is -33.2 kJ/mol at 25°C.

In certain experiment, the initial pressures are P_H2 = 0.250 atm, P_N2 = 0.870 atm, and P_NH3 = 12.9 atm.

Calculate ΔG for the reaction at these pressure, and predict the direction of reaction?

Example:

The standard free-energy change for the reaction 2NH₃(g)

Cell EMF Under Nonstandard

Conditions

Cell EMF Under Nonstandard

Conditions

• The emf generated under nonstandard conditions can be calculated by using an equation first derived by Walter Nernst (1864-1941), a German chemist who established many of the theoretical foundations of electrochemistry.

Nernst Equation

G = G + RT ln Q

This mean −nFE = −nFE + RT ln Q

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

E = E − RT_nF ln Q

or, using base-10 logarithms,

E = E − 2.303 RT_nF log Q

Nernst Equation

At room temperature (298 K),

Thus the equation becomes

E = E − 0.0592_n log Q

2.303 RT

Nernst Equation

What is the potential of a cell at 25 °C that has the following ion concentrations?

Cu2+_{(aq) + 2Fe}2+_(aq)

Cu(s) + 2Fe3+_(aq)

Example

Consider a galvanic cell that uses the reaction:

[Fe2+_{] = 0.20 M}

Nernst Equation

-Nernst Equation

Nernst Equation

when [Cr₂O₇2- _{] = 2.0 M, [H}+_{] = 1.0 M, [I}-_{] = 1.0 M, [Cr}3+_{] =}

1.0 x 10-5 _M.

2Cr3+_{(aq) + 3I}

2(s) + 7H2O(l)

Cr₂O₇2- _{(aq) + 14H}+_{(aq) + 6I}-_(aq)

Example

Calculate the emf at 298 K generated by the cell:

Nernst Equation

Why the emf of a voltaic cell drops as the cell discharges? Consider the following reaction:

Zn2+_{(1 M) + Cu(s)}

Zn(s) + Cu2+_{(1 M)}

E°_cell = 1.10 V

The standard cell potential at 25 °C is 1.10 V.

Nernst Equation

Zn2+_{(aq) + Cu(s)}

Zn(s) + Cu2+_(aq)

As the voltaic cell is discharged, the reactants are consumed and the products are generated, so the concentrations of these substances change.

The emf produced under non-standard conditions at 298 K: log Q n 0.0592 V E = E° -log = 1.10 V -2 0.0592 V [Cu2+_] [Zn2+_]

Nernst Equation

As the cell operates, the concentration of reactant (Cu2+₎

decreases and the concentration of product (Zn2+₎

increases relative to standard conditions. Thus, the emf decreases.

For example, when [Cu2+_{] is 0.05 M and [Zn}2+_{] is 2.0 M,}

Nernst Equation

As reactants are converted to products, the value of E decreases, eventually reaching E = 0 (the cell is “dead”). Because G = −nFE, it follows that G = 0 when E = 0. Recall thet a system is at equilibrium when G = 0.

Thus, when E = 0, the cell reaction has reached equilibrium and no net reaction is occurring.

Concentration Cells

EXAMPLE

• Calculate the emf of the following concentration cell:

E_cell = E_red (cathode) − E_red (anode)

Ni2+_{(0.001 M)} Ni2+_{(1.00 M)} Ni2+_{(0.001 M) + 2e} -Ni(s) Overall: Anode: Cathode: Ni2+_{(1.00 M) + 2e}- Ni(s) Ni(s) | Ni2+_{(0.001 M) || Ni}2+_{(1.00 M) | Ni(s)} = -0.28 V –(-0.28 V) = 0 V

Concentration Cells

The concentration cell generates an emf of 0.0888 V even though E°= 0.

The difference in concentration provides the driving force for the cell.

Concentration Cells

• The Nernst equation implies that a cell could be created that has the same substance at both electrodes.

• For such a cell, E_cell would be 0, but Q would not. • Therefore, as long as the concentrations are

different, E will not be 0.

• When the concentrations in the two compartments become the same, the value of Q = 1 and E = 0.

Equilibrium Constants

Equilibrium Constants

Equilibrium Constants

Three methods to determine equilibrium constants:

3. K from electrochemical data:

K =

[A]a_[B]b

[C]c_[D]d

1. K from concentration data:

RT -G° ln K =

2. K from thermochemical data:

RT nFE° ln K = ln K nF RT E°= or

Equilibrium Constants

Example:

The equilibrium constant for the reaction

Sr(s) + Mg2+_{(aq) Sr}2+ _{(aq) + Mg (s)}

is 2.69 x 1012 _{at 25°C. Calculate E° for a cell made up of}

Sr/Sr2+ _{and Mg/Mg}2+ _half-cells. Solution: ln K nF RT E°= Answer: 0.368 V

Equilibrium Constants

Example:

Calculate equilibrium constant for the following reaction at 25°C:

Sn(s) + 2Cu2+_{(aq) Sn}2+ _{(aq) + 2Cu}+ _(aq)

Solution: ln K nF RT E°= Answer: 6.5 x 109 Eo _?

Equilibrium Constants

Example:

Calculate ΔGo _{for the following process at 25°C:}

AgCl (s) Ag+ _{(aq) + Cl}- _(aq)

Equilibrium Constants

Equilibrium Constants

Example:

Calculate ΔG o _{and K}

c for the following reaction at 25°C:

Mg(s) + Pb2+_{(aq) Mg}2+_{(aq) + Pb(s)}

Applications of

Oxidation-Reduction Reactions

Batteries

Batteries, which consist of one or more voltaic cells, are used widely as self-contained power sources.

Batteries

Batteries

Lead Storage Battery

2PbSO₄(s) + 2H₂O(l) Pb(s) + PbO₂(s) + 2H1+_{(aq) + 2HSO}

41-(aq)

PbSO₄(s) + 2H₂O(l) PbO₂(s) + 3H1+_{(aq) + HSO}

41-(aq) + 2e -PbSO₄(s) + H1+_{(aq) + 2e} -Pb(s) + HSO₄1-_(aq) Overall: Anode: Cathode:

Batteries

Lead Storage Battery

Because the reaction product, solid PbSO₄, adheres to the surface of the electrodes, a “run-down”lead storage battery can be recharged by using an external sources of direct current to drive the cell reaction in the reverse,

nonspontsneous direction.

In an automobile, the battery is continuously recharged by a device called an alternator, which is driven by the

Batteries

Lead Storage Battery

A lead storage battery typically provides good service for several years, but eventually the spongy PbSO₄ deposits on the electrodes turn into hard, crystalline form that can’t be converted back to Pb and PbO₂.

Then it’s n longer possible to recharge the battery, and it must be replaced.

Batteries

Dry-Cell Batteries Leclanchécell

Batteries

Batteries

The alkaline dry cell is a modified version of the Leclanchécell in which the acidic NH₄Cl electrolyte of the Leclanchécell is

Batteries

Batteries

Dry-Cell Batteries

Corrosion of the zinc anode is a significant side reaction under acidic conditions because zinc reacts with H+_{(aq) to}

give Zn2+_{(aq) and H}

2(g).

Under basic conditions, however, the cell has a longer life because zinc corrodes more slowly.

Batteries

Batteries

Nickel-Cadmium (“ni-cad”) Batteries

Ni(OH)₂(s) + OH1-_(aq) NiO(OH)(s) + H₂O(l) + e -Cd(OH)₂(s) + 2e -Cd(s) + 2OH1-_(aq) Anode: Cathode:

Batteries

Nickel-Metal Hydride (“NiMH”) Batteries

M(s) + Ni(OH)₂(s) MH_ab(s) + NiO(OH)(s) Ni(OH)₂(s) + OH1-_(aq) NiO(OH)(s) + H₂O(l) + e -M(s) + H₂O(l) + e -MH_ab(s) + OH1-_(aq) Overall: Anode: Cathode:

Batteries

Lithium and Lithium Ion Batteries

LiCoO₂(s) Li_1-xCoO₂(s) + xLi1+_{(soln) + xe}

-xLi1+_{(soln) + 6C(s) + xe}

-Li_xC₆(s)

Anode:

Cathode:

Li_xMnO₂(s) MnO₂(s) + xLi1+_{(soln) + xe}

-xLi1+_{(soln) + xe} -xLi(s) Anode: Cathode: Lithium Ion Lithium

Batteries

Fuel Cells

A fuel cell is a galvanic cell in which one of the reactants is a fuel such as hydrogen or methanol.

A fuel cell differs from an ordinary battery in that the

reactants are not contained within the cell but instead are continuously supplied from an external reservoir.

Fuel Cells

Fuel Cells

Corrosion

Prevention of Corrosion

• Iron is often covered with a coat of paint or another metal such as tin, chromium or zinc to protect its surface

against corrosion.

• Covering the surface with paint or tin simply a means of preventing oxygen and water from reaching the iron

surface.

• If the coating is broken and the iron is exposed to oxygen and water, corrosion will begin.

• Galvanized iron, which is iron coated with a thin layer of zinc, uses the principles of electrochemistry to protect

the iron from corrosion even after the surface coat is broken.

Corrosion

Prevention of Corrosion

1. Galvanization: The coating of iron with zinc.

As the potentials indicate, zinc is oxidized more easily than iron: Fe(s) Fe2+_{(aq) + 2e} -Zn(s) Zn2+_{(aq) + 2e}- _E° red = -0.76 V E°_red = -0.45 V

When some of the iron is oxidized (rust), the process would be reversed immediately because zinc can

Corrosion

Prevention of Corrosion

1. Galvanization: The coating of iron with zinc.

As long as the zinc and iron are in contact, the zinc protects the iron from oxidation even if the zinc layer becomes scratched.

Corrosion

Prevention of Corrosion

Attaching a magnesium stake to iron will corrode the magnesium instead of the iron. Magnesium acts as a

sacrificial anode.

Mg2+_{(aq) + 2e}

-Mg(s)

Anode:

Cathode: O₂(g) + 4H1+_{(aq) + 4e}- 2H₂O(l)

2. Cathodic Protection: Instead of coating the entire surface of the first metal with a second metal, the second metal is placed in electrical contact with the first metal:

Electrochemical Cells

Electrochemical cells are of two basic types:

Galvanic (Voltaic) Cell:

A spontaneous chemical reaction which generates an electric current.

Electrolytic Cell:

An electric current is used to drives a nonspontaneous

reaction.

The processes occurring in voltaic (galvanic) and electrolytic cells are the reverse of each other.

Electrolysis and Electrolytic Cells

Electrolysis: The process of using an electric current to bring about chemical change.

Electrolysis and Electrolytic Cells

Electrolysis of Water

• Water in a beaker under atmospheric conditions (1 atm and 25°C) will not

spontaneously decompose to form H₂ and O₂.

• However, this reaction can be induced in a electrolytic cell.

• The electrolytic cell consists of a pair of electrodes made of a non-reactive metal, such as platinum, immersed in water.

Electrolysis and Electrolytic Cells

Electrolysis of Water

• When the electrodes are connected to the battery,

nothing happens because there are not enough ions in pure water to carry much of an electric current.

(Remember that at 25°C, pure water has only 1 x 10-7

M OH- _ions.)

• The reaction is occurs readily in a 0.1 M H₂SO₄ solution because there are a sufficient number of ions to conduct electricity.

Electrolysis and Electrolytic Cells

Electrolysis and Electrolytic Cells

Electrolysis of Water

2H₂(g) + O₂(g) + 4H+_{(aq) + 4OH}-_(aq)

6H₂O(l) O₂(g) + 4H+_{(aq) + 4e} -2H₂O(l) Overall: Anode: Cathode:

If the anode and cathode solutions are mixed, the H+ _and

OH- _{ions react to form water:}

4H₂O(l) 4H+ _{(aq) + 4OH}-_(aq)

The net electrolysis reaction is therefore the decomposition of water:

2H₂(g) + O₂(g) 2H₂O(l)

2H₂(g) + 4OH-_(aq)

-Electrolysis and Electrolytic Cells

Electrolysis of Molten Sodium Chloride

Cathode:

The negative electrode attracts Na+ _{cations, which combine}

with the electrons supplied by the battery and are thereby reduced to liquid sodium metal.

Electrolysis and Electrolytic Cells

Electrolysis of Molten Sodium Chloride

Cathode:

The positive electrode attracts Cl- _{anions, which replenish}

the electrons removed by the battery and are thereby oxidized to chlorine gas.

Electrolysis and Electrolytic Cells

Electrolysis of Molten Sodium Chloride

Cathode:

The signs of the electrodes are opposite for galvanic and electrolytic cell.

Electrolysis and Electrolytic Cells

Electrolysis of Molten Sodium Chloride

2Na(l) + Cl₂(g) 2Na+_{(l) + 2Cl}-_(l) 2Na(l) 2Na+_{(l) + 2e} -Cl₂(g) + 2e -2Cl-_(l) Overall: Anode: Cathode:

As in a galvanic cell, the anode is the electrode where oxidation take place, and the cathode is the electrode where reduction take place.

Electrolysis and Electrolytic Cells

Example

Metallic potassium was first prepared by Humphrey Davy in 1807 by electrolysis of molten potassium hydroxide:

a) Label the anode and cathode, and show the direction of ion flow.

b) Write balanced

equations for the anode, cathode and overall cell reactions.

Electrolysis and Electrolytic Cells

Answer:

Anode

Electrolysis and Electrolytic Cells

Electrolysis and Electrolytic Cells

Electrolysis of Aqueous Sodium Chloride • When an aqueous salt solution is electrolyzed, the

electrode reactions may differ from those for electrolysis of the molten salt because water may be involved.

Electrolysis and Electrolytic Cells

Electrolysis of Aqueous Sodium Chloride • The anode half-reaction might be:

Cl₂(g) + 2e

-2Cl-_(aq)

O₂(g) + 4H+_{(aq) + 4e}

-2H₂O(l) E°_red = 1.23 V

E°_red = 1.36 V

• Both E°_red are not very different, but the values do

suggest that H₂O should be preferentially oxidized at the anode.

• However, by experiment the observed product at the anode is Cl₂ because of a phenomenon called

Electrolysis and Electrolytic Cells

Electrolysis of Aqueous Sodium Chloride

• In studying electrolytic processes, we sometimes find that the voltage required for a reaction is always higher than the electrode potential indicates.

• The additional voltage required is the overvoltage.

• The additional voltage for O₂ formation is quite high (can be as large as 1 V). Therefore, under normal operating conditions Cl₂ gas is actually formed at the anode

Electrolysis and Electrolytic Cells

Electrolysis of Aqueous Sodium Chloride • The cathode half-reaction might be:

Na(s) Na+_{(aq) + e} -H₂(g) + 2OH-_(aq) 2H₂O(l) + 2e- _E° red = -0.83 V E°_red = -2.71 V

Water is reduced preferentially because the standard reduction potential is much less negative.

Electrolysis and Electrolytic Cells

Electrolysis of Aqueous Sodium Chloride

Cl₂(g) + H₂(g) + 2OH-_(aq) 2Cl-_{(l) + 2H} 2O(l) H₂(g) + 2OH-_(aq) 2H₂O(l) + 2e -Cl₂(g) + 2e -2Cl-_(aq) Overall: Anode: Cathode:

• The minimum potential required to force this

nonspontaneous reaction to occur under standard-state conditions is 2.19 V plus the overvoltage.

E°_red = 1.36 V E°_red = -0.83 V

• Na+ _{acts as a spectator ion and is not involved in the}

electrode reactions. Thus NaCl solution is converted to NaOH solution as the electrolysis proceeds.

Electrolysis and Electrolytic Cells

Electrolysis of Aqueous Sodium Chloride • The useful by-product NaOH can be obtained by

Electrolysis and Electrolytic Cells

Example

Predict the half-cell reactions that occur when aqueous solutions of the following salts are electrolyzed in a cell with inert electrodes. What is the overall cell reaction in each case?

Electrolysis and Electrolytic Cells

Answer (a) LiCl

The anode half-reaction might be: Cl₂(g) + 2e

-2Cl-_(aq)

O₂(g) + 4H+_{(aq) + 4e}

-2H₂O(l) E°_red = 1.23 V

E°_red = 1.36 V

The cathode half-reaction might be: Li(s) Li+_{(aq) + e} -H₂(g) + 2OH-_(aq) 2H₂O(l) + 2e- _E° red = -0.83 V E°_red = -3.05 V

Electrolysis and Electrolytic Cells

Answer (a) LiCl

Cl₂(g) + H₂(g) + 2OH-_(aq) 2Cl-_{(l) + 2H} 2O(l) H₂(g) + 2OH-_(aq) 2H₂O(l) + 2e -Cl₂(g) + 2e -2Cl-_(aq) Overall: Anode: Cathode:

Electrolysis and Electrolytic Cells

Answer (b) CuSO₄ The anode half-reaction:

O₂(g) + 4H+_{(aq) + 4e}

-2H₂O(l) E°_red = 1.23 V

The cathode half-reaction might be: Cu(s) Cu2+_{(aq) + 2e} -H₂(g) + 2OH-_(aq) 2H₂O(l) + 2e- _E° red = -0.83 V E°_red = +0.34 V S₂O₈2-_{(aq) + 2e}

-2SO₄2-_{(aq) E°}

Electrolysis and Electrolytic Cells

-Electrolysis and Electrolytic Cells

Answer (c) K₂SO₄ The anode half-reaction:

O₂(g) + 4H+_{(aq) + 4e}

-2H₂O(l) E°_red = 1.23 V

The cathode half-reaction might be: K(s) K+_{(aq) + e} -H₂(g) + 2OH-_(aq) 2H₂O(l) + 2e- _E° red = -0.83 V E°_red = -2.93 V S₂O₈2-_{(aq) + 2e}

-2SO₄2-_{(aq) E°}

Electrolysis and Electrolytic Cells

Electrolysis and Electrolytic Cells

Example

An aqueous solution of Mg(NO₃)₂ is electrolyzed. What are the products at the anode and cathode?

Answer:

Electrolysis and Electrolytic Cells

Solution

The anode half-reaction:

O₂(g) + 4H+_{(aq) + 4e}

-2H₂O(l) E°_red = 1.23 V

The cathode half-reaction might be: Mg(s) Mg2+_{(aq) + 2e} -H₂(g) + 2OH-_(aq) 2H₂O(l) + 2e- _E° red = -0.83 V E°_red = -2.37 V • nitrate ions migrate to the anode but because they are

Electrolysis and Electrolytic Cells

Factors that affects the electrolysis products of an aqueous ionic solutions:

1. Due to the over-voltage from O₂ formation, ions like I-_,

Br-_{, and Cl}- _{are oxidised when compare to water}

molecules.

2. Oxoanions like SO₄2-_{, CO}

32-, NO3-, PO43- are not

oxidised by water molecule because the central non-metal of these oxoanions (eg. S, C, N and P) is

already in its highest oxidation state. Water is oxidised to O₂ and H+_.

Commercial Applications of

Electrolysis

Commercial Applications of

Electrolysis

• Extraction of metal (Na, Al)

• Manufacture of chemicals (Cl₂, NaOH) • Purification metal

• Electroplating

• Electroplating plastic • Anodizing

Commercial Applications of

Electrolysis

Commercial Applications of

Electrolysis

Down’s Cell for the Production of Sodium Metal

• Sodium metal is produced commercially in a Downs cell by electrolysis of a molten mixture of sodium chloride and calcium chloride.

• The presence of CaCl₂ allows the cell to be operated at a lower temperature because the melting point of the NaCl-CaCl₂ mixture (about 580°C) is depressed well below that of pure NaCl (801°C).

Commercial Applications of

Electrolysis

Down’s Cell for the Production of Sodium Metal

• The liquid sodium produced at the cylindrical steel cathode is less dense than the molten salt and thus

floats to the top part of the cell. Where it is drown off into a suitable container.

Commercial Applications of

Electrolysis

Down’s Cell for the Production of Sodium Metal

• Chlorine gas forms at the graphite anode, which is

separated from the cathode by an iron screen to keep the highly reactive sodium and chlorine away from each other.

Commercial Applications of

Electrolysis

Commercial Applications of

Electrolysis

Hall-Heroult Process for the Production of Aluminum • The Hall-Heroult process involves electrolysis of a

molten mixture of aluminium oxide (Al₂O₃) and cryolite (Na₃AlF₆) at about 1000°C in a cell with graphite

electrodes.

• Electrolysis of pure Al₂O₃ is impractical because it melts at a very high temperature (2045°C).

Commercial Applications of

Electrolysis

Commercial Applications of

Electrolysis