Instrumental AnalysisInstrumental Analysis

Fundamentals of ElectrochemistryFundamentals of Electrochemistry

Instrumental AnalysisInstrumental Analysis

Fundamentals of ElectrochemistryFundamentals of Electrochemistry

Tutorial 4

Oceanographic instruments can be

powered by making a battery between the water and sediment

layers.

Bacteria oxidizes organic matter using SO4

2- ions available in the sediment layer. As a result, HS is released and acts as one of the reactant in the electrochemical cell. The other reactant is dissolved O2 molecules available at the sediment-water interface.

Electricity from the Ocean Floor

Summary of Symbols, Units, and Relationships

Relation between charge and moles

q = z . F Charge moles C/mol

(coulombs, C) of e

F is the Faraday’s constant

(96500 C/mol).

z is the total no. of moles of electrons passes in the circuit

Relation between work and voltage

Work = E . q Joules, J Volts, V coulombs

E is the electric potential difference, E is the work (J) needed to move a coulomb of positive charge from one point to the other.

Relation between free energy and potential

difference

** G = ─ W = ─ n F En is the moles of charge moved under a potential difference, E.

Ohm’s LawI = E / R

Current Volts resistance

A V Ohm,

I is the electric current, measured in ampere (A). It is the coulombs per second moving pas a point in the circuit.

Electric PowerP = work/s = E . I

Power J/s V A

(Watt, W)

P is the work per unit time done by electricity moving through a circuit.

J/mol

** the minus sign in the equation indicates that the free energy of a system decreases when the work is done on the surrounding

Relation between Current and rate

of chemical reaction

I = n . F . Rate A C/mol mol/s (C/s)

n is the number of moles of e- per 1 mole of reactant and rate is the rate of electrochemical reaction in mol/s

Nernst equation

Relation between potential of the electrode and activity of ions

E is the reduction electrode potential and E is the standard reduction electrode potential when the activities of all reactants and products are unity. n is the number of electrons in the half-cell reaction. a is the activity of reactant or product of the electrochemical reaction of the half-cell when written as reduction

Note: Nernst Equation can be applied to one half cell as well as to the reaction of the whole cell.

Relation between activity and

concentration

a is the activity of a chemical species, is the activity coefficient, c is the concentration of chemical species

aA

bB

a

alog

n

V 05916.0EE

ca

moles4.04x1.0electronsofmolesof.No

eofmoles2gainOofmolesSince

O mol 1.0mol g /32O g .23

221

2-1

2

C600,38)e C/mol005(96 mol) 4.0(Fzq -

Example (Charge)

If 3.2 g of O2 were reduced in the overall reaction with HS:

how many coulombs have been transferred from HS to O2 or how many charges pass in the circuit?

OHe2H2O 2221

e2HSHS

OHSHOHS 2221

Solution

Example (Current)

If electrons are forced into a wire which acts as anode to oxidize HS at a rate of 4.24 mmol/hr, how much current passes through?

e2HSHS

s/emoles10356.2s3600

h1

HSmol1

emol2

HSmmol10

HSmol1

h1

HSmmol24.4

s/eofmolesof.No

63

A227.0mole

couloms96500

s

emoles10356.2

mol

coulomb.

ondsec

moles

ondsec

coulombs

time

eargchCurrent

6

Solution

rate.F.nI

Generally, the current may be related to the rate of electrochemical reaction by:Rate of consumption of reactants in a unit of mol/s

Note: The electrode at which oxidation occurs is the anode while the electrode at which

reduction occurs is the cathode

n

n is the number of moles of e- per 1 mole of reactant

How much work can be done if 2.4 mmol of electrons go through a potential difference of 0.70 V in the ocean-floor battery?

Example (Work)

Solution

C 2103.2)C/mol96500()mol 3104.2(F.zq

J 101.6C)103.2)(V70.0(q.E WorkElectrical 22

• The greater the difference in potential (V), the stronger the e will be pushed around the circuit

• 12V battery “pushes” electrons through a circuit eight times harder than a 1.5V battery

Example (E from free energy change)

Calculate the voltage that will be measured by the potentiometer in the figure, Knowing that the free energy change for the net reaction is 150 kJ/mol of Cd.

)aq(Cl2)s(Ag2)aq(2Cd)s(AgCl2)s(Cd:actionReNet

e2)aq(2Cd)s(Cd:Oxidation

)aq(Cl2)s(Ag2e2)s(AgCl2:ductionRe

Solution

)C/J(V777.0)

emol

C96500()

Cdmol

emol2(

Cdmol

J310150

Fn

GE

EFnG

A spontaneous chemical reaction of negative G produces a positive voltage.

A030.0 100

V0.3

R

EI

W0.090 A)V)(0.030 0.3(I.EP

Example (Ohm’s Law)

In the following circuit, a battery generates a potential difference of 3 V and the resistor has a 100 resistance. How much current and power are delivered by the battery?

Solution

• The Nernst Equation relates the potential of the half cell to the activities of the chemical species (their concentrations)

– For the reaction (written as reduction)

– We usually calculate half-reactions at 25º C, substituting that in with the gas constant and to base 10 log gives:

]a

Aa

bBa

[lnnF

RTEE

BbenAa -

aa

balog

n

V 05916.0EE

A

B

Nernst Equation

Where a is the activities of species A and B.

n is the no. of electrons in either the electrode or cell reaction

• Write the Nernst equation for the reduction of O 2 to water:

• Note that multiplying the reaction by any factor does not affect either Eº or the calculated E:

2]H[log2

05916.023.1E

V1.23E OH2e 2H O 2-

221

2O

2

][H P

]OH[log

2

05916.023.1E

21

2

Example (Nernst Equation)

Note that:

asolid =1, agas = pressure

aH2O =1, aion = molarity

2]H[

1log

2

05916.023.1E

]H[log05916.023.1E

V1.23E OH24e 4H O 2-

2

4O

22

][H P

]OH[log

4

05916.023.1E

2

][Hlog05916.023.1E

• Find the voltage for the Ag-Cd cell and state if the reaction is spontaneous if the right cell contained 0.50 M AgNO3(aq) and the left contained 0.010 M Cd(NO3)2(aq).

V 799.0E )s(AgeAg

V 402.0E )s(Cde22Cd

V 781.00.50

1log

1

05916.0799.0E

V461.00.010

1log

2

05916.0402.0E

V 242.1)461.0(781.0EEEcell

Example (Ecell)

For Ag/Ag+

electrode

For Cd/Cd2+

electrode

For the Cell

)s(Ag2e2Ag2

e2CdCd 2

)s(Ag2CdAg2)s(Cd 2

Note that you will get the same value of cell potential if you apply the Nernst equation directly to the overall cell reaction.

V242.1

]5.0[

]01.0[log

2

05916.0201.1E

]5.0[

]01.0[log

2

05916.0))402.0(799.0(E

]Ag[

]Cd[log

2

05916.0)EE(E

]Ag][Cd[

]Ag][Cd[log

2

05916.0EE

2cell

2cell

2

2

Cd/CdAg/Agcell

2

22

cellcell

2

Since Ecell is found to be +ve quantity, the reaction as written is spontaneous

Example (Ecell)

A cell was prepared by dipping a Cu wire and a saturated calomel electrode

(ESCE = 0.241 V) into 0.10 M CuSO4 solution (ECu/Cu2+ = 0.339 V). The Cu wire

was attached to the positive terminal and the calomel to the negative terminal

of the potentiometer.

1- Write the half-cell reaction of Cu electrode.2- Write the Nernst equation for the Cu electrode.3- Calculate the cell voltage.4- What would happen if the [Cu2+] were 4.864x10-4 M?

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

2-

3- Ecell = ECu/Cu2+ ESCE = 0.309 V 0.241 V = 0.068 V

V309.0)10.0(log2

05916.0339.0E

solution

]Cu[log2

05916.0EE 20

Electrode connected to the positive terminal of the potentiometer is the cathode and the other is the anode

Example (Ecell)A 100.0 ml solution containing 0.100 M NaCl was titrated with 0.100 M AgNO3 and the voltage of the cell shown in the figure was monitored.

a) Calculate the voltage after the addition of 65.0 ml of AgNO3. (EAg/Ag+ = 0.799 V, ESCE = 0.241 V)

b) Explain why a silver electrode can be used as anindicator electrode for both Ag+ and for halides.

Solution:

a) The titration reaction is: Ag+(aq) + Cl(aq) AgCl(s). Equivalence point is reached upon addition of 100.0 ml titrant (in this case).

After addition of 65.0 ml of AgNO3, we can say:

(M V)reacted Cl- = (M V)Ag+ added (note that Ag+ and Cl react in the ratio of 1:1)

(M V)Cl- remains unreacted = (M V)total Cl- (M V)Ag+ = (0.1 x 100) (0.1 x 65)

(M V)Cl- remains unreacted = 3.5 mmole MCl- = 3.5 mmol/165 = 0.0212 mol/L

NaCl

• To find the cell voltage we need to know Ag+:

where Ksp is the solubility product of insoluble AgCl

so Ag+ = Ksp / [Cl] [Ag+] depends on the concentration of Cl ion

= 1.8 x 10-10 M / 0.0212 M

= 8.5 x 10-9 M

• The potential of Ag electrode is given by:Ag electrode reaction: Ag+ + e- Ag+

EAg/Ag+ = 0.799 + 0.05916 log (8.5 10-9) = 0.322 V

• The cell potential: Ecell = ESCE EAg/Ag+ = 0.241 0.322 = 0.081 V

b) We can see from the example that the silver electrode can act indirectly as halide indicator electrode if solid insoluble silver halide is present. The solubility of silver halide will be affected by whatever halide ion is present and in turn the concentration of Ag+ ion as well as the potential of the Ag electrode.

spK]Cl[]Ag[

]Ag[log05916.0E]Ag[

1log05916.0E

]Ag[

]Ag[log

1

05916.0EE

]Cl[

spKlog05916.0

Ag/AgE]Ag[log05916.0

Ag/AgEAg/AgE

Exercise 1

A mercury cell used to power heart pacemaker runs on the following reaction:

Zn(s) + HgO(s) ZnO(s) + Hg( ) E = 1.35 V

If the power required to operate the pace maker is 0.010 W, how many kilogram of HgO (molar mass = 216.59) will be consumed in 365 days?

Exercise 2

Calculate the voltage of each of the following cells:

(a) Fe(s) / FeBr2(0.010 M) // NaBr(0.050 M) / Br2( ) / Pt(s)

(b) Hg( ) / Hg2Cl2(s) / KCl(0.060 M) // KCl(0.040 M) / Cl2(g, 0.50 bar) / Pt(s)

(EFe/Fe2+ = 0.44 V, EBr2/Br = 1.078 V, EHg/Hg2Cl2 = 0.268 V, ECl2/Cl = 1.360 V)

If you are unable to solve these problems or need to revise your answer, please refer to the “Solution Manual for Quantitative Chemical Analysis”, by D.C. Harris (GUC library)

Try to solve problems 14-3, 14-4, 14-5, 14-7, 14-12, 14-14, 14-17, 14-18, and 14-19Try to solve problems 14-3, 14-4, 14-5, 14-7, 14-12, 14-14, 14-17, 14-18, and 14-19

Harris text book, p308-310Harris text book, p308-310