Electrochemistry

SCE 3109Energetics in Chemistry

SCE 3109Energetics in Chemistry

Electrochemistry

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:

http://www.wwnorton.com/college/chemistry/gilbert/overview/ch17.htm

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.

Reduction is gain 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 combine to form H2.

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–

• The Zn loses two electrons to form Zn 2+ .

Redox reactions

Half Reactions

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

• The Cu2+ gains two electrons to form copper.

Redox reactions

Overall: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s)

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

Zn2+(aq) + Cu(s)Zn(s) + Cu2+(aq)

Cu(s)Cu2+(aq) + 2e-

Reduction half-reaction:

Oxidation half-reaction:

Zn2+(aq) + 2e-Zn(s)

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+.

d) Balance charge by adding electrons.

Redox reactions


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 charge.

Redox reactions


Consider the reaction between MnO4− and C2O4

2− :

MnO4−(aq) + C2O4

2−(aq) Mn2+(aq) + CO2(aq)

Redox reactions


First, we assign oxidation numbers.

MnO4− + C2O4

2- Mn2+ + CO2

+7 +3 +4+2

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

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

Redox reactions


Oxidation Half-Reaction

C2O42− CO2

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

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−

Redox reactions


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

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

8 H+ + MnO4− Mn2+ + 4 H2O

Redox reactions


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

Redox reactions


Combining the Half-Reaction

C2O42− 2 CO2 + 2 e−

5 e− + 8 H+ + MnO4− Mn2+ + 4 H2O

x 5

x 2

Redox reactions


Combining the Half-Reaction

5 C2O42− 10 CO2 + 10 e−

10 e− + 16 H+ + 2 MnO4− 2 Mn2+ + 8 H2O

16 H+ + 2 MnO4− + 5 C2O4

2− 2 Mn2+ + 8 H2O + 10 CO2

Redox reactions


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) + Cl2 (g)

b) Cu (s) + NO3− (aq) Cu2+ (aq) + NO2 (g)

c) Mn2+ (aq) + NaBiO3 (s) Bi3+ (aq) + MnO4− (aq)

Redox reactions


• 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

EXAMPLE: Brown (Pg: 854)

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 (or galvanic) cell.

Voltaic Cells


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-Zn(s)

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 attracted to the cathode.

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 reduced to Cu(s).

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 barrierseparates the two compartments, but allows migration of ions.

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 neutralisecharge 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

Voltaic Cells


Cu(s)Cu2+(aq) + 2e-

Zn2+(aq) + 2e-Zn(s)

Overall cell reaction:

Anode half-reaction:

Cathode half-reaction:

Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)

Phase boundaryPhase boundary

Electron flow

Salt bridge

Cathode half-cellAnode half-cell

Shorthand notation for galvanic 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) || Sn2+(aq) | Sn(s)

Fe2+(aq) + Sn(s)Fe(s) + Sn2+(aq)

Solution:

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.

Problem 17.1 (Page 692) (McMurry & Fay)

Example

Consider the following galvanic cell.


Example


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 nonspontaneousreaction.

Voltaic Cells

• The redox reaction between Zn and Cu2+ is spontaneousregardless 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 potential energy.


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

• It is also called the cell potential (Ecell) 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).


Cell potential is measured in volts (V).

1 V = 1 JC

voltSI unit of electric potential

jouleSI unit of energy

coulombElectric charge

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


• 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

cell.

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) Ecell = +1.10 V

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

Ecell = Ered (cathode) − Ered (anode)

Standard Reduction Potentials

Standard reduction potentials for many electrodes have been measured and tabulated.

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.


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

E°cell = 0.34 V


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:


2H1+(aq) + Cu(s)H2(g) + Cu2+(aq)

2H1+(aq) + 2e-H2(g)



Cathode half-reaction: Cu(s)Cu2+(aq) + 2e-



Zn(s)Zn2+(aq) + 2e-

0.76 V = 0 V ― Eored (anode)

Eored = -0.76 V

As a standard reduction potential:

Eored (anode) = - 0.76 V


H2(g) + Zn2+(aq)2H1+(aq) + Zn(s)

Zn2+(aq) + 2e-Zn(s)

H2(g)2H1+(aq) + 2e-



Cathode half-reaction:

Using Standard Cell Potentials

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



• For the oxidation in this cell,

• For the reduction,

Ered = −0.76 V

Ered = +0.34 V


= +0.34 V − (−0.76 V) = +1.10 V


• 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 VZn2+(aq) + 2 e–→ Zn(s) E°= –0.76 V

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


= +0.80 V − (−0.76 V) = +1.56 V


Example

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


= +0.536 V − (−1.66 V) = +2.196 V

2Al3+(aq) + 6I― (aq)2Al(s) + 3I2(s)


Example

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


= +1.33 V − (0.536 V) = +0.794 V

2Cr3+(aq) + 3I2 (s) + 7H2OCr2O72-(aq) + 14H+ (aq) + 6I―(aq)


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+.


In3+(aq) + 2e-In+(aq)

2Br―(aq)Br2(l) + 2e-

= +1.065 V − Eored (anode)1.46 V

Eored (anode) = - 0.395 V


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-


Solution:

a)

b) Ecell = Ered (cathode) − Ered (anode)

Cd2+(aq) + 2e-Anode: Cd(s)

= (- 0.136 V) − (- 0.403 V) = 0.267 V

Sn(s)Cathode: Sn2+(aq) + 2e-


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-

(b) + 0.499 V


2Ag2+(g) + Cu(s)2Ag(s) + Cu2+(aq)

Ag(s)]2 x [Ag1+(aq) + e-

Cu2+(aq) + 2e-Cu(s)

E° = 0.46 V

E° = 0.80 V

E° = -0.34 V

Zn2+(g) + Cu(s)Zn(s) + Cu2+(aq)

Cu(s)Cu2+(aq) + 2e-

Zn2+(aq) + 2e-Zn(s)

E° = 1.10 V

E° = -(-0.76 V)

E° = 0.34 V

Half-cell potentials are intensive properties.


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?

Answer: (a) +0.337 V (b) +3.169 V


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) + H2(g)(a) Cu(s) + 2H+(aq)

Cu(s) + 2H+(aq)(b) Cu2+(aq) + H2(g)

2Cl-(aq) + I2(s)(c) Cl2(g) + 2I-(aq)


Solution: Cu2+(aq) + H2(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: Cu2+(aq) + 2e-Cu(s)

Eored = 0 V

Eored = +0.34 V

E = Ered (cathode) − Ered (anode)

= 0 V − (0.34 V) = -0.34 V

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


Solution:

In this reaction, H2 is oxidized to H+ and Cu2+ is reduced to Cu.

Cu(s)Cu2+(aq) + 2e-Reduction:

Oxidation: 2H+(aq) + 2e-H2(g)

Eored = +0.34 V

Eored = 0 V


= 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) + H2(g)


Solution:

I2(s) + 2e-2I-(aq)

Reduction:

Oxidation:

2Cl-(aq)Cl2(g) + 2e- Eored = +1.36 V

Eored = +0.54 V


= 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) + I2(s)(c) Cl2(g) + 2I-(aq)


Predicting whether a redox reaction is spontaneous

Practice Exercise: Brown (Pg 869)

McMurry & Fay (Pg 701)(Problem 17.8)

Oxidizing and Reducing Agents

• The strongest oxidizers have the most positive reduction potentials.

• The strongest reducers have the most negative reduction potentials.



Example:

a) Rank the following ions in order of increasing strength as oxidizing agents: NO3

- (aq), Ag+ (aq), Cr2O72- (aq).

Answer: Ag+ < NO3- < Cr2O7

2-

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

Answer: Al (s) > Fe (s) > I- (aq)


Example:

c) Arrange the following oxidizing agents in order of increasing strength under standard-state conditions: Br2(aq), Fe3+ (aq), Cr2O7

2- (aq).

Answer: Fe3+ < Br2 < Cr2O72-

d) Arrange the following reducing agents in order of increasing strength under standard-state conditions : Al(s), Na(s), Zn(s).

Answer: Zn < Al < Na


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


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 delivers the highest voltage, and calculate Eo for the reaction.


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 B2+.

(c) A3+ + 2D A+ + 2 D+

2.85 V

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.


• 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.

1 F = 96,485 C/mol = 96,485 J/V-mol


• 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 G both indicate that a reaction is spontaneous.


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.


• Under standard conditions,

G = −nFE


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


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°


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:

Answer: -270 kJ/mol


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.

Solution: G° = -nFE° Eo = ?

Cell EMF Under NonstandardConditions

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

• 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 Grather than G to predict the direction of the reaction.


• 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


is -33.2 kJ/mol at 25°C.

In certain experiment, the initial pressures are PH2= 0.250

atm, PN2= 0.870 atm, and PNH3

= 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)N2(g) + 3H2(g)


Solution:

G = G°+ RT ln Q

= G°+ RT ln(PH2

)3 (PN2)

(PNH3)2

= -33.2 x 103 J/mol + (8.314 J/Kmol)(298 K) ln(0.250)3 (0.870)

(12.9)2

= -9.9 x 103 J/mol

= -9.9 kJ/mol

Because ΔG is negative, the net reaction proceed from left to right.


• 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 −RTnF ln Q

or, using base-10 logarithms,

E = E −2.303 RT

nF log Q

Using: G = -nFE and G° = -nFE°

Nernst Equation

At room temperature (298 K),

Thus the equation becomes

E = E −0.0592

n log Q

2.303 RTF = 0.0592 V

Nernst Equation

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

Cu2+(aq) + 2Fe2+(aq)Cu(s) + 2Fe3+(aq)

Example

Consider a galvanic cell that uses the reaction:

[Fe2+] = 0.20 M[Fe3+] = 1.0 x 10-4 M [Cu2+] = 0.25 M

Nernst Equation

Calculate E°:

Fe2+(aq)Fe3+(aq) + e-

Cu2+(aq) + 2e-Cu(s) Eored = 0.34 V

Eored = 0.77 V

E°cell = 0.77 V –(0.34 V) = 0.43 V

Solutionlog Q

n

0.0592 VE = E° -

Nernst Equation

Calculate E:

log Qn

0.0592 VE = E° -

E = 0.25 V

log(1.0 x 10-4)2

(0.25)(0.20)2= 0.43 V -

2

0.0592 V

log[Fe3+]2

[Cu2+][Fe2+]2

n

0.0592 VE = Eo -

Solution

Nernst Equation

when [Cr2O72- ] = 2.0 M, [H+] = 1.0 M, [I-] = 1.0 M, [Cr3+] =

1.0 x 10-5 M.

2Cr3+(aq) + 3I2(s) + 7H2O(l)Cr2O72- (aq) + 14H+(aq) + 6I-(aq)

Example

Calculate the emf at 298 K generated by the cell:

Answer: E = 0.89 V

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


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 Qn

0.0592 VE = E° -

log= 1.10 V -2

0.0592 V

[Cu2+]

[Zn2+]

Nernst Equation


logE = 1.10 V -2

0.0592 V

[Cu2+]

[Zn2+]

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 [Zn2+] is 2.0 M,

E = 1.05 V

Nernst Equation


logE = 1.10 V -2

0.0592 V

[Cu2+]

[Zn2+]

As reactants are converted to products, the value of Edecreases, 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:


Ni2+(0.001 M)Ni2+(1.00 M)

Ni2+(0.001 M) + 2e-Ni(s)

Overall:

Anode:

Cathode: Ni(s)Ni2+(1.00 M) + 2e-

Ni(s) | Ni2+(0.001 M) || Ni2+(1.00 M) | Ni(s)

= -0.28 V –(-0.28 V) = 0 V

Concentration Cells

Ni2+(0.001 M)Ni2+(1.00 M)Overall:

log1.00

0.001

n

0.0592 VE = 0 -

= 0 –(- 0.0888) = +0.0888 V

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, would be 0, but Q would not.Ecell

• 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

G = G + RT ln Q

G = 0 Q = K

0 = G0 + RT lnK

G0 = RT lnK

At Equilibrium

E = E −RTnF ln Q

At Equilibrium

E = 0 Q = K

0 = E −RTnF ln K

ln KnF

RTE°=


G0 = RT lnK


Three methods to determine equilibrium constants:

3. K from electrochemical data:

K =[A]a[B]b

[C]c[D]d1. K from concentration data:

RT

-G°ln K =2. K from thermochemical data:

RT

nFE°ln K =

ln KnF

RTE°=

or



Example:

The equilibrium constant for the reaction

Sr(s) + Mg2+(aq) Sr2+ (aq) + Mg (s)

is 2.69 x 1012 at 25°C. Calculate E° for a cell made up of Sr/Sr2+ and Mg/Mg2+ half-cells.

Solution:ln K

nF

RTE°=

Answer: 0.368 V


Example:

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

Sn(s) + 2Cu2+(aq) Sn2+ (aq) + 2Cu+ (aq)

Solution: ln K

nF

RTE°=

Answer: 6.5 x 109

Eo ?


Example:

Calculate ΔGo for the following process at 25°C:

AgCl (s) Ag+ (aq) + Cl- (aq)

The Ksp of silver chloride is 1.6 x 10-10.


Solution:

Ksp = [Ag+][Cl-] = 1.6 x 10-10

Go = RT lnK

Go = (8.314 J/Kmol)(298 K) ln 1.6 x 10-10

= 5.6 x 10-4 J/mol

= 56 kJ/mol


Example:

Calculate ΔG o and Kc for the following reaction at 25°C:

Mg(s) + Pb2+(aq) Mg2+(aq) + Pb(s)

Answer: -432 kJ/mol, 5 x 1075

Applications of Oxidation-Reduction Reactions

Batteries

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

Batteries

Lead Storage Battery

Batteries


2PbSO4(s) + 2H2O(l)Pb(s) + PbO2(s) + 2H1+(aq) + 2HSO41-(aq)

PbSO4(s) + 2H2O(l)PbO2(s) + 3H1+(aq) + HSO41-(aq) + 2e-

PbSO4(s) + H1+(aq) + 2e-Pb(s) + HSO41-(aq)

Overall:

Anode:

Cathode:

Batteries


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 engine.

Batteries


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

Dry-Cell Batteries

Leclanchécell

Mn2O3(s) + 2NH3(aq)+ H2O(l)2MnO2(s) + 2NH41+(aq) + 2e-

Zn2+(aq) + 2e-Zn(s)Anode:

Cathode:

Batteries

Dry-Cell Batteries

Alkaline cell

Mn2O3(s) + 2OH1-(aq)2MnO2(s) + H2O(l) + 2e-

ZnO(s) + H2O(l) + 2e-Zn(s) + 2OH1-(aq)Anode:

Cathode:

The alkaline dry cell is a modified version of the Leclanchécell in which the acidic NH4Cl electrolyte of the Leclanchécell is replaced by a basic electrolyte, either NaOH or KOH.

Batteries

Alkaline 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 H2(g).

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

Batteries

Mercury 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) MHab(s) + NiO(OH)(s)

Ni(OH)2(s) + OH1-(aq)NiO(OH)(s) + H2O(l) + e-

M(s) + H2O(l) + e-MHab(s) + OH1-(aq)

Overall:

Anode:

Cathode:

Batteries

Lithium and Lithium Ion Batteries

LiCoO2(s)Li1-xCoO2(s) + xLi1+(soln) + xe-

xLi1+(soln) + 6C(s) + xe-LixC6(s)Anode:

Cathode:

LixMnO2(s)MnO2(s) + xLi1+(soln) + xe-

xLi1+(soln) + xe-xLi(s)Anode:

Cathode:

Lithium Ion

Lithium

Batteries

Lithium Ion 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

Hydrogen-Oxygen Fuel Cell

Fuel Cells

2H2O(l) 2H2(g) + O2(s)

4OH1-(aq)O2(g) + 2H2O(l) + 4e-

4H2O(l) + 4e-2H2(g) + 4OH1-(aq)

Overall:

Anode:

Cathode:

Hydrogen-Oxygen Fuel Cell

Corrosion

Corrosion

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


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 reduce Fe2+ to Fe.

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


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: 2H2O(l)O2(g) + 4H1+(aq) + 4e-

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 nonspontaneousreaction.

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 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.



• 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.





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)4H2O(l) + 4e-


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.



Cathode:

The positive electrode attracts Cl- anions, which replenish the electrons removed by the battery and are thereby oxidized to chlorine gas.



Cathode:

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



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.


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.


Answer:

AnodeCathode


Answer:

4K(l) + O2(g) + 2H2O(l)4K+(l) + 4OH-(l)

4K(l)4K+(l) + 4e-

O2(g) + 2H2O(l) + 4e-4OH-(l)

Overall:

Anode:

Cathode:


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.



• 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 overvoltage.



• 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 instead of O2.



• 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.



Cl2(g) + H2(g) + 2OH-(aq)2Cl-(l) + 2H2O(l)

H2(g) + 2OH-(aq)2H2O(l) + 2e-


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.



• The useful by-product NaOH can be obtained by evaporating the aqueous solution at the end of electrolysis.


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?

(a) LiCl (b) CuSO4 (c) K2SO4


Answer (a) LiCl

The anode half-reaction might be:


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-


E°red = -3.05 V


Answer (a) LiCl

Cl2(g) + H2(g) + 2OH-(aq)2Cl-(l) + 2H2O(l)

H2(g) + 2OH-(aq)2H2O(l) + 2e-


Overall:

Anode:

Cathode:


Answer (b) CuSO4

The anode half-reaction:

O2(g) + 4H+(aq) + 4e-2H2O(l) E°red = 1.23 V


Cu(s)Cu2+(aq) + 2e-


E°red = +0.34 V

S2O82-(aq) + 2e-2SO4

2-(aq) E°red = 2.01 V


Answer (b) CuSO4

2Cu(s) + O2(g) + 4H+(aq)2Cu2+(aq) + 2H2O(l)Overall:

Anode:

Cathode:

O2(g) + 4H+(aq) + 4e-2H2O(l)

Cu(s)Cu2+(aq) + 2e-


Answer (c) K2SO4


O2(g) + 4H+(aq) + 4e-2H2O(l) E°red = 1.23 V


K(s)K+(aq) + e-


E°red = -2.93 V

S2O82-(aq) + 2e-2SO4

2-(aq) E°red = 2.01 V


Answer

2H2(g) + O2(g)2H2O(l)Overall:

Anode:

Cathode:

O2(g) + 4H+(aq) + 4e-2H2O(l)

(c) K2SO4

H2(g) + 2OH-(aq)2H2O(l) + 2e- x 2


Example

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

Answer:

Anode, O2; Cathode, H2


Solution


O2(g) + 4H+(aq) + 4e-2H2O(l) E°red = 1.23 V


Mg(s)Mg2+(aq) + 2e-


E°red = -2.37 V

• nitrate ions migrate to the anode but because they are too stable to be oxidised.


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-, CO3

2-, NO3-, PO4

3- 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 oxidisedto O2 and H+.

Commercial Applications ofElectrolysis


• Extraction of metal (Na, Al)

• Manufacture of chemicals (Cl2, NaOH)

• Purification metal

• Electroplating

• Electroplating plastic

• Anodizing

• Effluent treatment


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).



• 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.



• 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.


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).




A Membrane Cell for Electrolytic Production of Cl2 and NaOH


Production of Cl2 and NaOH

• Chlorine is used in water and sewage treatment, as a bleaching agent in manufacturing paper, and in the manufacture of plastics such as polyvinyl chloride (PVC).

• Sodium hydroxide is employed in making paper, textiles, soaps and detergents.



A saturated aqueous solution of sodium chloride flows into the anode compartment, where Cl- is oxidized to Cl2 gas.



Water enters the cathode compartment, where it is converted to H2 gas OH- ions.



Between the anode and cathode compartments is a special plastic membrane that is permeable to cations but not to anions or water.



The membrane keeps the Cl2 and OH- ions apart but allows a current of Na+ ions to flow into cathode compartment, thus maintaining electrical neutrality in both compartments.



The Na+ and OH- ions flow out of the cathode compartment as an aqueous solution of NaOH.


Electrorefining of copper metal

• The purification of a metal by means of electrolysis is called electrorefining.

• For example, impure copper obtained from ores is converted to pure copper in an electrolytic cell that has impure copper as the anode and pure copper as the cathode.





• At the impure Cu anode:

Copper is oxidized along with more easily oxidized metallic impurities such as Zn and Fe.

Less easily oxidized impurities such as Ag, Au and Pt fall to the bottom of the cell as anode mud.

• At the pure Cu cathode:

Cu2+ ions are reduced to pure copper metal.

Less easily reduced metal ions (Zn2+, Fe2+ and so forth) remain in the solution.



Cu(s)Cu2+(aq) + 2e-M2+(aq) + 2e-M(s)

(M = Cu, Zn, Fe)


Electroreplating

• Closely related to electrorefining is electroplating, the coating of one metal on the surface of another using electrolysis.

• For example, steel automobile bumpers are sometimes plated with chromium to protect them from corrosion, and silver-plating is commonly used to make items of fine table service.

• The object to be plated is carefully cleaned and then set up as the cathode of an electrolytic cell that contains a solution of ions of the metal to be deposited.


Electroreplating

Problem 17.21 (McMurry & Fay, page 722)

Sketch an electrolytic cell suitable for electroplating a silver spoon. Describe the electrodes and the electrolyte, label the anode and cathode, and indicate the direction of electron and ion flow. Write balanced equations for anode and cathode half-reactions. What is the overall cell reaction?


Electroplating plastic

• Plastics can be metallized though, through numerous plating processes.

• The parts must be perfectly clean from any oil, grease or any plastic injection mold compounds. If the parts are not cleaned properly, the metal will peel off over time from the plated plastic part.



• Then process the plastic part in a very aggressive chromic/sulfuric acid bath to etch the plastic surface (make small pits on the surface).

• Then place the plastic part in a palladium chloride bath to place metal particles in the previous pits made on the plastic surface.

• After this palladium metal deposition, one can electroplate the part with copper metal and then chrome plate or many other various metals such as nickel or gold.



• By definition, electroplating requires that an electric current flow between the part being plated and the chemical solution.

• If the part is non-conductive then an electric current cannot flow. How can one plate nonconductive materials like plastic?



• The answer is "electroless" plating. It is a way to coat one metal with another without passing current.

• Electroless plating is applied first to get a conductive surface, then the electroless plated parts are electroplated.

In the elecroless plating process, chemical baths are used to deposit metal coatings but do not need electricity.

After an electroless coating is produced, a normal electroplating bath can be used to make the coating even thicker.



• Many automotive parts, including grilles and all manner of decorative trim have been plated plastic.


Anodizing

• Anodizing is an electrolytic process used to increase the thickness of the natural oxide layer on the surface of metal parts.

• The process is called "anodizing" because the part to be treated forms the anode electrode of an electrical

circuit.


Anodizing

• Anodizing is an operation performed mainly on aluminium. The effect is to develop an oxide coating on the aluminium. Again, this is a conversion coating: the surface is converted from aluminium to aluminiumoxide.

• Unlike plating, the part to be coated is connected to the anode - not the cathode.

• The part is immersed in a dilute solution of sulfuric acid, current is passed, and the part is anodized an oxide coating forms.


Anodizing

• The oxide coating is harder and more corrosion resistant than bare aluminium.

• Some aluminum pots and pans are anodized, and door and window frames are often anodized.

• The most common anodizing processes, for example sulfuric acid on aluminium, produce a porous surface which can accept dyes easily. The number of dye colors is almost endless; however, the colors produced tend to vary according to the base alloy.


Anodizing

These inexpensive decorative carabiners have an anodized aluminium surface that has been dyed and are made in many colors.


Effluent treatment

• Electrolytic effluent treatment is based on the anodic dissolution of metals, which form their hydroxides, and the pollutants are removed by sorption (absorption and adsorption) , coagulation, and other processes occurring in the space between the electrodes.

Reference: Malkin, V.P. (2003). Industrial Ecology: Electrolytic Effluent Treatment. Chemical and Petroleum Engineering. 39(1–2): 46-50.


Effluent treatment

• Basic processes in removing heavy-metal ions by electrocoagulation from industrial effluents:

– the hydroxides of these metals are formed by the alkali produced at the cathode, where hydrogen is formed and excess of hydroxyl ions is formed in the cathode layer;

– the heavy-metal hydroxides coprecipitate with the coagulant (e.g. iron hydroxides) formed by electrolytic oxidation of an appropriate anode material (e.g. iron);

– the heavy-metal ions are sorbed on the coagulant.

Quantitative Aspects ofElectrolysis


• The stoichiometry of a half-reaction shows how many electrons are needed to achieve an electrolytic process.

Na(s)Na+(aq) + e-

(One mole of electrons are required to produce 1 mol of Na from Na+ .)

Al(s)Al3+(aq) + 3e-

(Three mole of electrons are required to produce 1 mol of Al from Al3+ .)


• To find out how many moles of electrons pass through a cell in a particular experiment, we need to measure the electric current and the time that the current flows.

• The number of coulombs of charges passed through the cell:

Charge(C) = Current(A) x Time(s)

• Because the charge on 1 mol of electrons is 96,485 C, the number of moles of electrons passed through the cell is

Moles of e- = Charge(C) x96,485 C

1 mol e-


Moles of e- = Charge(C) x


96,485 C

1 mol e-

Faraday constant


Example

Calculate the mass of aluminium produced in 1.00 h by the electrolysis of molten AlCl3 if the electrical current is 10.0 A.


Moles of e- = Charge(C) x96,485 C

1 mol e-

= (10.0 A) x (3600 s) = 3.60 x 104 C

= (3.60 x 104 C)96,485 C

1 mol e-

= 0.373 mol e-


Moles of e- = 0.373 mol e-

Al3+(aq) + 3e- Al(s)

From the half-reaction equation, three mole of electrons are required to produce 1 mol of Al from Al3+ .

Therefore, 0.373 mole of electrons will produce mol of Al.

Moles of Al =

Mass of Al = 3.36 g


Example

A constant current of 30.0 A is passed through an aqueous solution of NaCl for a time of 1.00 h. How many grams of NaOH and how many liters of Cl2 gas are produced at STP?

Answer: 12.5 L


• Faraday's 1st Law of Electrolysis - The mass of a substance altered at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the circuit.

• Faraday's 2nd Law of Electrolysis - For a given quantity of electricity (electric charge), the mass of an elemental material altered at an electrode is directly proportional to the element's equivalent mass. The equivalent mass of a substance is its molar mass divided by the number of electrons transferred per ion.


• Faraday's laws can be summarized by

where

m is the mass of the substance liberated at an electrode

Q is the total electric charge passed through the substance

F = 96,485 C mol-1 is the Faraday constant

M is the molar mass of the substance

n is the number of electrons transferred per ion

n


• Note that M / n is the same as the equivalent mass of the substance altered.

• For Faraday's first law, M, F, and n are constants, so that the larger the value of Q the larger m will be.

• For Faraday's second law, Q, F, and n are constants, so that the larger the value of M / n (equivalent weight) the larger m will be.

n


• In the simple case of constant-current electrolysis, Q = Itleading to

n

Electrical Work

• We have seen that a positive value of E is associated with a negative value for the free-energy change and, thus, with a spontaneous process.

• For any spontaneous process, ΔG is a measure of the maximum useful work, wmax, that can be extracted from the process:

ΔG = wmax

• Because ΔG = -nFE, The maximum amount of useful electrical work produced by a voltaic cell:

wmax = -nFE

Electrical Work

wmax = -nFE

• The cell emf, E is a positive number for a voltaic cell, so wmax will be a negative number for a voltaic cell.

• The negative value for wmax means that a voltaic cell does work on its surroudings. (Remember that work done by a system on its surroundings is indicated by a negative sign for w.)

Electrical Work

• In an electrolytic cell we use an external source of energy to bring about a nonspontaneous electrochemical process.

• In this case, ΔG is positive and Ecell is negative.

• To force the process to occur, we need to apply an external potential, Eext, which must be larger in magnitude than Ecell:

Eext > -Ecell

For example, if a nonspontaneous process has Ecell = -0.9 V, then the external potential Eext must be greater than 0.9 V in order for the process to occur.

Electrical Work

• When an external potential Eext is applied to a cell, the surroundings are doing work on the system.

• The work performed in an electrolytic cell:

w = nFEext

Note:

a) The work will be a positive number because the suroundings are doing work on the system.

b) The quantity n is the number of mole of electrons forced into the system by the external potential.

c) nF is the total electrical charge supplied to the system by the external source of electricity.

Electrical Work

• Electrical work can be expressed in energy units of watts times time.

The watt (W) is a unit of electrical power (that is, the rate of energy expenditure).

1 W = 1 J/s

Thus, a watt-second is a joule.

• The unit employed by electrical utilities is the kilo-watt-hour (kWh), which equals 3.6 x 106 J.

1 kWh = (1000 W)(1 hr)1 hr

3600 s

1 W

1 J/s = 3.6 x 106 J

Electrical Work

SUMMARY

• The maximum amount of electrical work produced by a voltaic cell is given by the product of the total charge delivered, nF, and the emf, E:

wmax = -nFE

• The work performed in an electrolysis:

w = nFEext

where Eext is the applied external potential.

Electrical Work

EXAMPLE

A voltaic cell is based on the reaction

Under standard conditions, what is the maximum electricalwork, in joule, that the cell can accomplish if 75.0 g of Sn isconsumed?

Sn2+(aq) + 2I-(aq)Sn(s) + I2(s)

Answer: -8.19 x 104 J

wmax = -nFE

Electrical Work

SOLUTIONwmax = -nFE

E =

n =75.0 g

118.71 g mol= 1.26 mol

Note: The (-) sign indicates that work is done by the cell.

x 2

= -8.19 x 104 J

wmax = -nFE = - (1.26 mol) (0.672 V)V mol

96,485 J

0.672 V

Electrical Work

EXAMPLE

A voltaic cell is based on the reaction

Under standard conditions, what is the maximum electricalwork, in joule, that the cell can accomplish if 50.0 g of copperis plated out?


Answer: -1.67 x 105 J

Electrical Work

EXAMPLE

• Calculate the number of kilowatt-hours of electricity required to produce 1.0 x 103 kg of aluminium by electrolysis of Al3+ if the applied voltage is 4.50 V.

w = nFEext

n = ? w = ?

3.6 x 106 J = 1 kWh

_______ J = ? kWh

Electrical Work

SOLUTIONw = nFEext

n =(1.0 x 103)(1000) g

27 g/mol= 1.1 x 105 mol

w = nFEext = (1.1 x 105 mol) (4.50 V)V mol

96,485 J

= 4.82 x 1010 J

Kilowatt-hours =

x 3

4.82 x 1010 J

3.6 x 106 Jx 1 kWh = 1.34 x 104 kWh

Electrical Work

EXAMPLE

• Calculate the number of kilowatt-hours of electricity required to produce 1.00 kg of Mg from electrolysis of molten MgCl2 if the applied emf is 5.00 V.

Answer: 11.0 kWh

Key Concept Summary

Date post:	24-Nov-2014
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Electrochemistry

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