DEPARTMENT OF CHEMISTRY PHYSICAL CHEMISTRY II

DepartmentofChemistry,PhysicalChemistry2Page1

DEPARTMENT OF CHEMISTRY PHYSICAL CHEMISTRY II

Lecture module: Electrochemistry

Lecturers: Prof K Bisetty

& Mr M Sabela

Semester 2_ 2013

Surname and Initials: _________________________________________ Student Number: ________________


Lecture 1

Electrochemistry: deals with the interconversion of electrical energy and chemical energy.

Electrical processes are redox reactions in which the energy released by a spontaneous reaction is converted to electricity or in which electricity is used to drive a non-spontaneous reaction.

Electrochemical, voltaic or galvanic cells

Voltaic Cells

• The energy released in a spontaneous redox reaction may be used to perform electrical work. Voltaic, or galvanic cells, are devices in which electron transfer occurs via an external circuit. Voltaic cells utilise spontaneous reactions.

Oxidation Numbers and Balancing Redox Equations

• Chemical reactions in which the oxidation state of more than one substance changes (in opposite directions) are called oxidation-reduction reactions (redox reactions).

Recall:

Oxidation involves loss of electrons (OIL). Reduction involves gain of electrons (RIG).

Consider the spontaneous reaction that occurs when Zn is added to HCl.

Zn(s) + 2H+(aq) → Zn2+(aq) + H2(g)

The oxidation numbers of Zn and H+ have changed. The oxidation number of Zn has increased from 0 to +2.

The oxidation number of H has decreased from +1 to 0.

Therefore, Zn is oxidised to Zn2+, while H+ is reduced to H2.

H+ causes Zn to be oxidised.

Thus, H+ is the oxidising agent, or oxidant.

Zn causes H+ to be reduced.

Thus, Zn is the reducing agent, or reductant.


Note that the reducing agent is oxidised and the oxidising agent is reduced.

• Recall the law of conservation of mass: The amount of each element present at the beginning of the reaction must be present at the end. Conservation of charge state that electrons are not lost in a chemical reaction. Some redox equations may be easily balanced by inspection.

However, for many redox reactions we need to look carefully at the transfer of electrons.

Half-Reactions

•Half-reactions are a convenient way of separating oxidation and reduction reactions.

•Consider the reaction:

Sn2+(aq) + 2Fe3+(aq) →Sn4+(aq) + 2Fe2+(aq)

• The oxidation half-reaction is: Sn2+(aq) → Sn4+(aq) +2e–

Note that electrons are a product here.

• The reduction half-reaction is:

2Fe3+(aq) + 2e– → 2Fe2+(aq)

• Note that electrons are a reactant here. • Also note that the number of electrons on each half-reaction are equal and

opposite.

Self assessment question 1: If a strip of Zn is placed in a solution of CuSO4, Cu is deposited on the Zn and the Zn

dissolves by forming Zn2+.

1.1 Write down the balanced reaction

1.2 Write down the half-cell reactions occurring at the anode and the cathode.

1.3 Identify the oxidant and reductant


Balancing Equations by the Method of Half-Reactions

Consider the titration of an acidic solution of Na2C2O4 (sodium oxalate, colourless) with KMnO4 (deep purple).

MnO4– is reduced to Mn2+ (pale pink-colourless), while the C2O4

2– is oxidised to CO2. The equivalence point is indicated by the presence of a pale pink colour. If more KMnO4 is added, the solution turns purple due to the presence of excess KMnO4. What is the balanced chemical equation for this reaction? We can determine this using the Method of Half-Reactions: Write down the two incomplete half reactions.

MnO4– (aq) →Mn2+(aq)

C2O42– (aq) → CO2(g)

Balance each half reaction. First, balance elements other than H and O.

MnO4

– (aq) → Mn2+(aq) C2O4

2– (aq) → 2CO2(g) Then balance O by adding water(l).

MnO4

– (aq) →Mn2+(aq) + 4H2O(l) C2O4

2– (aq) → 2CO2(g) Then balance H by adding H+.

8H+(aq) + MnO4– (aq) →Mn2+(aq) + 4H2O(l)

C2O42– (aq) →2CO2(g)

Finish by balancing charge by adding electrons. Note that this is an easy place to make an error! For the permanganate half-reaction, note that there is a charge of 7+ on the left and 2+ on the right. Therefore, 5 electrons need to be added to the left:

5e– + 8H+(aq) + MnO4– (aq) → Mn2+(aq) + 4H2O(l)

In the oxalate half-reaction, there is a 2– charge on the left and a 0 charge on the right, so we need to add two electrons to the products:

C2O42– (aq) → 2CO2(g) + 2e–

Multiply each half-reaction to make the number of electrons equal in each half-reaction.

To balance the 5 electrons for permanganate and 2 electrons for oxalate, we need 10 electrons for both. Multiplying MnO4

-/Mn2+ by 2 and C2O42–/CO2 by 5 gives:

10e– + 16H+(aq) + 2MnO4– (aq) →2Mn2+(aq) + 8H2O(l)

5C2O42– (aq) → 10CO2(g) + 10e–

Now add the reactions and simplify. 16H+(aq) + 2MnO4

– (aq) + 5C2O42– (aq) → 2Mn2+(aq) + 8H2O(l) + 10CO2(g)

The equation is now balanced!

Note that all of the electrons have cancelled out!


Balancing Equations for Reactions Occurring in Basic Solution The same method as above is used, but OH– is added to “neutralise” the H+ used. The equation must again be simplified by cancelling like terms on both sides of the equation. 1. Electrode Process A Molecular View of the Electrode Process

“Rules” of voltaic cells: At the anode electrons are products. Oxidation occurs at the anode. At the cathode electrons are reagents. Reduction occurs at the cathode. The flow of electrons from anode to cathode requires an external wire. The transfer of ions through a salt bridge maintains overall charge balance for

the two compartments.

Passage of electricity through a chemical solution promotes reaction in the vicinity of the electrodes and is referred to as electrolysis.

If on the other hand reaction are chosen so that it involves a loss of electrons (oxidation) and a gain of electrons (reduction) and the arrangement is such that electrons are transferred from one reaction vessel to the other via a metallic conductor then such an arrangement is found to serve as a source of electrical energy. This is known as electrochemical and galvanic cell.


Worked Example I

The following sketch of a galvanic cell is based on the Self assessment question1

(i) The Zn rod/plate dipping into a solution of Zn2+ is called a Zn half cell or Zn electrode. Similarly the Cu/Cu2+ is referred to as the Cu half cell.

(ii) The solutions are connected indirectly through a salt bridge e.g. KCl or NH4NO3 in a U-tube. If the solutions were allowed direct contact, energy of the reaction would be liberated as heat.

→

no measurement of electrical energy.

(iii) The voltmeter (V) indicates the cell voltage or the electrometotive force (e.m.f.) produced by the two half cells combined.

(iv) Reaction at the electrodes: At the LHE

→ 2 At the RHE

2 →


Self assessment question 2 Sketch a galvanic cell involving the reactions ZnSO4 and standard hydrogen electrode (where H2(g) at 1 atm is passed over platinum electrode in contact with 1M HCl solution. Explain how it works. Why is a salt bridge needed?


Lecture 2

Half Cell voltage (Standard Electrode Potentials)

Since the cell voltage arises through the combined action of the two half-cells, the contribution from each is referred to as the half-cell voltage or electrode potential. An element dipping into a 1M solution of its ions is called a standard half-cell or electrode and the voltage associated with the latter is known as the standard half cell voltage or electrode potential.

Electrode potentials cannot be measured directly. However, they may be determined with reference to a standard hydrogen gas electrode (i.e. pure hydrogen gas at 1 atm pressure in contact with 1 M hydrogen ions and at a temperature of 298.15 K) which is arbitrarily fixed at zero potential. Consequently, in any cell which contains the hydrogen electrode, the entire measured voltage is attributed to the half reaction at the other electrode.

Standard State Conditions.

Solution is 1 M or more precisely the activities (ideal concentrations) of solutes and gases equal 1. Therefore 1 M solution refers to an ideal solution. The gas pressure is equal to 1 atm and the temperature is usually 298.15 K or 25 0C.

Cell emf under Standard Conditions •The flow of electrons from anode to cathode is spontaneous. What is the “driving force”? • Electrons flow from anode to cathode because the cathode has a lower electrical

potential energy than the anode.

Potential difference is the difference in electrical potential. The potential difference is measured in volts. One volt (V) is the potential difference required to impart one joule (J) of

energy to a charge of one coulomb (C):

1V = 1 JC-1

Electromotive force (emf) is the force required to push electrons through the external circuit.

Cell potential: Ecell is the emf of a cell. This is known as the cell voltage. Ecell is > 0 for a spontaneous reaction – For 1 M solutions or 1 bar pressure for gases at 25C (standard conditions),

the standard emf (standard cell potential) is called Ecell. • For example, for the reaction:

Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) • Ecell = +1.10 V


NOTE:

(i) When the Zn/Zn2+ is coupled to the standard hydrogen electrode (SHE) the electrons flow from the former to the latter: i.e.

→ 2 2 ↔ 2 2 → 2 2 ↔

But with a Cu/Cu2+ system the electrons flow in the opposite direction: i.e.

→ 2 2 2 2 ↔

2 → 2 ↔ thustheelectrodepotentialmusthaveoppositesigns.

(ii) A standard half-cell will have negative voltage if the electrons flow from

the metal (containing the electrode or half-cell) through the wire to the H2 electrode (anode is where oxidation takes place, therefore electrons flow from anode to cathode).

(iii) A negative voltage indicates a tendency of an electrode to give up electrons. Thus when two half-cells are coupled to form an electrochemical cell, the half-cell with the more negative exhibits a stronger tendency to give up electrons whilst the electrode with the less negative voltage will take up electrons.

(iv) When the half-cell reaction is written as a reduction process the voltage

associated with it is called the reduction potential. (v) The complete cell voltage is given by the expression:

Where Eocell values represent reduction potential (in the arrangement of

an electrochemical cell, the half-cell with the most negative value is put on the LHS) In the case of a voltaic cell consisting of Zn/Zn2+ (1 M) and Cu/Cu2+ arranged as shown in the diagram and for which the E0 values are

2 ↔ , 0.76 2 ↔ , 0.34

The complete cell voltage is given by:

0.34 0.76 1.10

(vi) A table of E0cell values (i.e. standard reduction potentials) gives the

half-cell voltages in order of decreasing negative values. That is


positive denotes a decreasing tendency to give up electrons (refer to tables in the text book and tutorials). This is also referred to as the electromotive series from which one can predict whether an element will oxidize or reduce. Standard electrode potentials are useful in determining the strength of oxidizing and reducing agents under standard-state conditions.

2. Significance of Eocell values

The sign of Eocell may be used to predict whether a given chemical reaction

resulting from the combination of any two half-cells will occur spoteneously or not. It can be shown from thermodynamics that if Eo

cell is positive the reaction will occur spontaneously.

Worked Example 2 Given 2 standard electrodes: viz:

2 ↔ ; 0.76 ↔ ; 0.80

(i) Arrange the electrodes to form a cell according to conversion. (ii) Write the electrode reactions. (iii) Write the overall cell reaction (iv) Calculate the cell voltage (v) State whether the rxn will occur spontaneously as written. (vi) Denote the polarity of the electrodes in the diagram in (i).

Solution:

(i)

(ii) At LHE : 2 ↔ (oxidation)

RHE : ↔ (reduction)

(iii) → 2 2 2 → 2 Balance to “cancel” electrons. 2 → 2

(iv) E0cell = EoRHE - Eo

LHE = 0.8 – (-0.76) = 1.56V

(v) E0cell is +ve. Therefore reaction will occur as denoted in (iii).

(vi) Zn electrode is –ve (anode where oxidation occurs)

Cu electrode is +ve.


NOTE: The electrode potential is an intensive property. This means that its value is independent of the amount of species in the reaction. Thus, the electrode potential for the half reaction

2 4 → 2

Is the same as for

2 →

The emf corresponds to the free energy change per faraday, so that all emf’s are on the same basis. Therefore the emf’s are subtracted/added directly. Only the coefficients of equations may change to “cancel out electrons.

Strengths of Oxidising and Reducing Agents

Consider a table of standard reduction potentials.

We can use this table to determine the relative strengths of reducing (and oxidising) agents.

The more positive the E red, the stronger the oxidising agent (written in the table as a reactant).

The more negative the E red, the stronger the reducing agent (written as a product in the table).

We can use this to predict if one reactant can spontaneously oxidise another. Any substance on the right will reduce any substance higher than it on the left

- E cell > 0

Self assessment question 3

State whether the following are TRUE or FALSE

We can use this table to predict if one reactant can spontaneously reduce another.

F2 can oxidise H2 or Li and Ni2+ can oxidise Al(s).

Li can reduce F2.


Thermodynamics of cells

Electrical work is performed when a reaction takes place in a galvanic cell. Since this is reversible work at constant temperature and pressure, it is therefore the free energy change. i.e.

∆

∆

∆

.

96500 /

E = cell emf (volts)

E(volts) x F(coulumbs) = EF volt- coulomb

= EF joules

Units for G is joules.

NOTE: a +ve cell emf therefore corresponds to a spontaneous cell reaction and hence to a –ve free energy change.

e.g. calculate G0 for the reaction:

2 → 2

Note the details in the earlier example

∆

2 96500 1.56

3.01x105J

∆ .


SELF ASSESSMENT QUESTION

Consider the reaction of nickel with silver ion:

Ni(s) + 2Ag+(aq) → Ni2+(aq) + 2Ag(s)

Predict whether the reaction is spontaneous or not?

Calculate the work done by the Ni-Ag cell


Lecture 3

NERNST EQUATION

Dependence of e.m.f. on concentration

The e.m.f. of a cell depends on the concentration of ion and on gas pressures. We can relate cell e.m.f’s for various concentrations of ions and various gas pressures to standard electrode potentials by means of the Nernst equation. The free-energy change, G, is related to the standard free-energy change, G0, by the equation:

G = G0 + RT ln Q..............................(1)

Here Q is the thermodynamic reaction quotient. We can apply this equation to a galvanic cell: Since ∆ and ∆ , we can write the above equation (1) in the form:

ln ……………… 1.1

/ ln ……………… 1.2

This is the Nernst equation which relates the cell emf (Ecell) to its standard emf (E0cell)

and the reaction quotient, Q.

NOTE: R=8.314J/mol.K; F=96 500C/mol; n = number of electrons ; T = absolute temp (K); ln x = 2.303 log x

therefore at 250C or 298K

.

. .

=

. log

(volt-coulomb = J, therefore V=J/C i.e. value in volts)

(i.e. at 98K, (RTln10)/F = 0.0591V)

Therefore : E = E0cell – 0.0591/n log Q at 250C ----------------------------(1.3)

The Nernst eqn. can be used to calculate the cell emf for non-standard conditions.


Worked Example 3:

What is the emf of the following cell at 298 K?

Given : 2 ↔ , 0.76

2 ↔ , 0.34

(RTln 10)/F = 0.0591V

LHE : ↔ 2 , 0.76

RHE : 2 ↔ , 0.34

Overall: ↔ 1.10

2

1.0 10

0.11.00 10

now . log

1.10 0.05912

log 1.0 10

= 1.22 V

Equilibrium Constant (K)

G = G0 + RT ln Q refer to equation (1) At equilibrium G = 0 When the galvanic cell reaction has reached equilibrium, Ecell = 0

Therefore 0 = G0 + RTlnQ

At equilibrium conditions, the quotient Q is denoted by K: Q ⇛K, equilibrium constant.

Therefore, we can write:

G0 = -RT ln K = -nFE0cell

at equilibrium

or ln K = nFE0cell/ RT

or Log K = nFE0cell/2.303RT

We can now also calculate the equilibrium constant for a galvanic cell reaction, given the appropriate information.


Concentration Cells • A concentration cell is one whose emf is generated solely because of a concentration difference.

Worked Example 4:

Consider a cell with two compartments, each with a Ni(s) electrode but with different

concentrations of Ni2+(aq).

• One cell has [Ni2+] = 1.0 M and the other has [Ni2+] = 0.001 M.

• The standard cell potential is zero – since both half-reaction are the same.

• But this cell is operating under non-standard conditions!

• The driving force is the difference in Ni2+ concentrations.

• Anode (dilute Ni2+): Ni(s) → Ni2+(aq) + 2e-

• Cathode (concentrated Ni2+): Ni2+(aq) + 2e- → Ni(s)

• Using the Nernst equation we can calculate a cell potential of +0.0888 V for this concentration cell.

SELF ASSESSMENT QUESTIONS

1.Calculate the standard Gibbs free energy change, Go, in J/mol at 250C for the following reaction from standard electrode potentials.

4 2 → 3 2

2.Use the standard cell potential to calculate the value of the equilibrium constant, K, at 25 0C for the following reaction.

2 → 2 2


3.A cell is constructed at 250C as follows. One half-cell consists of a chlorine/chloride, Cl2/Cl-, electrode with the partial pressure of Cl2 = 0.100 atm and [Cl-] = 0.100M. The other half-cell involves the MnO4

-/ Mn2+ couple in acidic solution with [MnO4

-] = 0.100M, [Mn2+] = 0.10M, and [H+] = 0.100M. Apply the Nernst equation to the overall cell reaction to determine the cell potential for this cell.


Lecture 4

Quantitative Aspects of Electrolysis

• We want to know how much material we obtain with electrolysis.

• Consider the reduction of Cu2+ to Cu.

Cu2+(aq) + 2e- → Cu(s).

• 2 mol of electrons will plate 1 mol of Cu.

• The charge of one mol of electrons is 96,500 C (1 F).

• A Coulomb is the amount of charge passing a point in one second when the current is one Ampere.

• The amount of Cu can be calculated from the current (I) and time (t) required to plate.

Q = I t

Electrolytic Cells An electrolytic cell is an electrochemical cell in which an electric current drives a nonspontaneuos reaction. The process of producing a chemical change in an electrolytic cell is called electrolysis.

Electrolysis

When electrodes connected to the poles of a battery that are placed in a solution of an acid, base or a salt, the positive (+) and negative (-) ions each migrate towards the electrode which is of opposite sign to their own, and become discharged when they come into contact with the electrodes. This process is called electrolysis and the acids, bases and salts are called electrolytes. The positive ions travel to the cathode and are called cations. The negative ions travel to the anode and are called anions.

When the ions are discharged at the electrodes they always go into other reactions, either with each other, with the electrode, with the water or with substances in solution. These reactions are called secondary reactions. A great variety of products can thus be obtained by electrolytic process.

Electrolysis of Molten Salts

Figure: shows a simple electrolytic cell. Wires from a battery from a battery are connected to electrodes that dip into molten sodium chloride (NaCl melts at 801oC). At the electrode connected to the negative pole of the battery, globules of sodium metal form; chloride gas evolves from the other electrode. The half-reactions are

Na+(l) + e- Na (l)

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Stoichiometry of electrolysis

Faraday showed that the amounts of substances released at the electrodes during electrolysis are related to the total charge that has flowed in the electric circuit.

Faraday’s Laws

(1) The quantity of a substance liberated at an electrode is directly proportional to the quantity of electric charge that has flowed in the circuit.

(2) For a given quantity of electric charge, the amount of any metal that is deposited is proportional to its atomic mass divided by the charge on the ion. Quantity of electricity is measured in coulombs(C). A current which delivers 1C/s has a magnitude of ampere (1A). Hence quantity of electricity q = i x t where i is the current and t is time in seconds e.g. quantity of electricity involved when a current of 2A flows for 3mins is q = i x t = 2 x 3 x 60 C = 360C. Standard amount of substance is the mole and the corresponding quantity of electricity is the Faraday, F = 96 500 coulombs. This is the electrical charge on one mole of electrons. Na+ + e- → Na 1 mole of e- produces 1 mole of Na atoms. 1 mole of e- has a charge of 1F = 96 500 C/mol Cl- → 1/2Cl2 + e-

So the quantity of charge is a measure of the quantity of substance produced.

Therefore:

Where q is the number of coulombs of electricity flowing, n is the number of moles of chemical substance either oxidized or reduced, and z is the charge per ion.


e.g. (i) What mass of Ag will be deposited at the cathode when a current of 0.5A flows 10.0 mins through a solution of AgNO3? (Ag = 107.9 g.mol).

→ 0.5 10 60

(ii) What current would have flowed for 1 hour through a cell containing silver nitrate if 2.158g of silver had been deposited? What mass of zinc would be deposited if the same current were passed through zinc chloride solution? (1F = 96500 C/mol, Ag =107.9g/mol, Zn = 65.4g/mol) To deposite 1 mole Ag+ i.e. 107.9 requires 96500C

.

.? ?

.e. .

.96500 1932

193260 60

0.5364

→ 2

1 2 Thus 0.5 mole .. .. 96 500C ? mole .. .. 1 932C

Or

= 0.01001 mole Zn = 0.01001 x 65.4g Zn = 0.6548g

Or .

.

. / ; 0.6548

This last method can be used for cells connected in series (any number of cells).


SELF ASSESSMENT QUESTIONS

1. In the industrial production of aluminium, aluminium oxide, Al2O3, is electrolysed at 100 oC. Assume that the cathode reaction is

3 →

How many coulombs of electricity are required to produce 5.12kg of aluminium?

2. A steady current of 1.5A passed through an electrolytic cell containing AgNO3 until 1.45g of silver were deposited at the cathode.

(i) How long did the current flow? (ii) What mass of zinc would be deposited if the same quantity of

electricity passed through cell containing ZnSO4?

Further Reading:

1. Marten J. ten Hoor, “Redox Balancing without Puzzling,” J. Chem. Educ., Vol. 74, 1997, 1376-1368.

2. P. J. Moran and E. Gileadi, “Alleviating the Common Confusion Caused by Polarity in Electrochemistry,” J. Chem. Educ., Vol. 66, 1989, 912.

3. Ron DeLorenzo, “Electrochemical Errors,” J. Chem. Educ., Vol. 62, 1985, 424-425.

Departm

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DEPARTMENT OF CHEMISTRY PHYSICAL CHEMISTRY II