CHEMISTRY 59-320CHEMISTRY 59-320ANALYTICAL CHEMISTRYANALYTICAL CHEMISTRY

Fall - 2010Fall - 2010

Lecture

Chapter 14: Fundamentals of Electrochemistry

14-1 Basic concepts

• Oxidation:• Reduction:• Reducing agent:• Oxidizing agent:• Electric charge, q, is measured in coulombs (C), q = n*F (F is

Faraday constant)• Electric current: the quantity of charge flowing each second through

a circuit, I = electric charge/time• Electric potential, E (volts)• Work = E*q• The free Gibbs energy equals the maximum possible electrical work

that can be done by the reaction on its surroundings, work = -∆G; ∆G = - E*q = -nFE

• Ohm’s law I = E/R• Power, work done per unit time (SI unit: J/s, known as watt)

• Exercise A: A mercury cell formerly used to power heart pacemakers has the following reaction:

If the power required to operate the pacemaker is 0.010 0 W, how many kilograms of HgO (FM 216.59) will be consumed in 365 days?

The cell voltage will be 1.35 V (because ????) Use the equation:

I = P/E = 0.010 0 W/1.35 V = 7.41 × 10−3 C/s I = (7.41 × 10−3 C/s)/(9.649 × 104 C/mol)

= 7.68 × 10−8 mol e−/s for 365 days: I = 2.42 mol e−, which is 1.21 mol HgO The formula mass HgO = 216.6, thus we need 1.21 x 216.6 g = 0.262 kg HgO

P power E Volts I Ampere

14.2 Galvanic cells

• Galvanic cell: (also called a voltaic cell) uses a spontaneous chemical reaction to generate electricity.

• Batteries and fuel cells are galvanic cells that consume their reactants to generate electricity. A battery has a static compartment filled with reactants. In a fuel cell, fresh reactants flow past the electrodes and products are continuously flushed from the cell.

• Half reaction: describes a process occurring at each electrode, involving only oxidation or only reduction.

• Salt bridge: a conducting ionic medium in contact with two electrolyte solutions. It maintains electroneutrality throughout the cell.

• Anode: Electrode at which oxidation occurs. In electrophoresis, it is the positively charged electrode.

• Cathode: Electrode at which reduction occurs. In electrophoresis, it is the negatively charged electrode.

Line notation of an electrochemical cell

Line diagram for the above cell:

Cd(s)|CdCl2(aq)|AgCl(s)|Ag(s)

Line diagram for the above cell:

Cd(s)|CdCl2(aq)||AgNO3(aq)|Ag(s)

14-3 Standard potentials• Standard reduction potential: E° The voltage that would be

measured when a hypothetical cell containing the desired half-reaction (with all species present at unit activity) is connected to a standard hydrogen electrode anode.

• E° defines the tendency for a reduction process to occur at an electrode.

We will write all half-reactions as reductions. In this way, the voltageof an electrochemical cell is the cathode potential – anode potential.

• All ionic species present at a=1 (approximately 1 M).• All gases are at 1 bar (approximately 1 atm).• No metallic substance is indicated, the potential is established on an inert metallic electrode (ex. Pt).

Application of standard potentials

Solution: Because standard potential describes the tendency of a reduction to occur. As a result of complexation, the standard potential of ferric ion decreases from 0.771 to 0.356 V. It means that the ion becomes more difficult to be reduced. Therefore, Fe(III) is stabilized more.

If the potential values provided above are not standard, can one still reachthe same conclusion?

Problem: 14-13. (a) Cyanide ion causes E° for Fe(III) to decrease: Fe3+ + e- ↔ Fe2+ E°= 0.771 V Fe(CN)6

3+ + e- ↔ Fe(CN) 63+ E° = 0.356 V

Which ion, Fe3+, or Fe2+, is stabilized more by complexing with CN?

14-4 Nernst Equation

Problem 14-16: Write the Nernst equation for the following half-reaction and findE when pH = 3.00 and PasH3 = 1.0 mbar. As(s) + 3H+ + 3e- ↔ AsH3(g) Eo = -0.238 V

Answer: E = Eo – [RT/(nF)]*ln{(PasH3 /(aH+)3} Since no temperature is provide, we can assume T = 25 oC, if so we willUse equation 14-15 to carry out the calculation E = E0 – (0.05916/3)*log {0.001/ /(aH+)3} = - 0.238 – (0.05916/3)*6 = -0.356 V.

A procedure for writing a net cell reaction and finding its voltage

• Step 1: Write reduction half-reactions for both half-cells and find E° for each in Appendix H. Multiply the half-reactions as necessary so that they both contain the same number of electrons.

(When you multiply a reaction, you do not multiply E°, why? ).

• Step 2: Write a Nernst equation for the right half-cell, which is attached to the positive terminal of the potentiometer. This is E+.

• Step 3 Write a Nernst equation for the left half-cell, which is attached to the negative terminal of the potentiometer. This is E_.

• Step 4 Find the net cell voltage by subtraction: E = E+ - E_.

• Step 5 To write a balanced net cell reaction, subtract the left half-reaction from the right half-reaction. (Subtraction is equivalent to reversing the left-half reaction and adding.)

• Solution: The only substance we need to consider the effect of

activity is OH-. However, when you write the Nernst equation for the cell reaction, OH- disappears from the equation, therefore there is no change in potential during the discharge process.

Problem 14-18. A nickel-metal hydride rechargeable battery for laptop computers is based on the following chemistry:

The anode material, MH, is a transition metal hydride or rare earth alloy hydride. Explain why the voltage remains nearly constant during the entire discharge cycle.

14-5 Eo and the equilibrium constant

• At the equilibrium point, the cell potential is 0! the equilibrium constant can be calculated from rearranging Nernst equation.

• Example: using standard potential to find the equilibrium constant for the reaction Cu(s) + 2Fe3+ ↔ 2Fe2+ + Cu2+

• Solution: step 1: identify cathode and anode

step 2: find the standard potential for half reactions

step 3: calculation Eo for the cell reaction

Finding K for net reactions that are not redox reaction

• The general practice is to arbitrarily design one hald reaction (for cathode) and then use this one to subtract the cell reaction to obtain the other half reaction (for the anode).

• Problem 14-28: Calculate Eo for the half reaction

Pd(OH)2(s) + 2e- ↔ Pd(s) + 2OH-

given that Ksp for Pd(OH)2 is 3x10-28 and Eo = 0.915 V for the reaction Pd2+ + 2e- ↔ Pd(s)

• Solution: Pd(OH)2(s) ↔ Pd2+ + 2OH- Ksp

Ksp = [Pd2+]([OH-])2

from Ksp one can get Eo for the above reaction (eq 14-23)

Eo = -0.814 V

Since the above reaction can be derived by using Pd(OH)2(s) + 2e- ↔ Pd(s) + 2OH- to subtract Pd2+ + 2e- ↔ Pd(s)

we have -0.814 V = Eo - 0.915 V

Eo = 0.101 V

14-6 Cells as chemical probes• It is important to distinguish two types of equilibrium in a galvanic

cell: equilibrium between the two half-cells and equilibrium within each half cell.

A galvanic cell that can beused to measure the formationconstant of Hg(EDTA)2-

Hg2+ + 2e- ↔ Hg(l) Eo = 0.852 V

14-7 Biochemists use Eo’

• Formal potential: Potential of a half-reaction (relative to a standard hydrogen electrode) when the formal concentrations of reactants and products are unity. Any other conditions (such as pH, ionic strength, and concentrations of ligands) also must be specified.

Finding the formal potential

• Reduction potential of ascorbic acid, showing its dependence on pH.

(a) Graph of the function labeled formal potential.

(b) Experimental polarographic half-wave reduction potential of ascorbic acid in a medium of ionic strength = 0.2 M.

At high pH (>12), the half-wave potential does not level off to a slope of 0, as Equation 14-34 predicts. Instead, a hydrolysis reaction of ascorbic acid occurs and the chemistry is more complex.

• Example. Cadmium electrode immersed in a solution that is 0.015M in Cd2+ .

Solution: Cd2+ + 2e Cd(s) 0 = -0.403

• Example: Calculate the potential for a platinum electrode immersed in a solution prepared by saturating a 0.015 M solution of KBr with Br2

Br2(l) + 2e 2Br- E0 = 1.065 V

Solution:

o

2+

0.0591 1ε = ε - log

2 Cd

0.0591 1ε = -0.403 - log 0.457

2 0.0150V

2

0.0591E= 1.065 - log

2 1.0

Br

20.0591E = 1.065 - log 0.0150 1.173

2V

1. When you multiply a reaction by a coefficient, the potential ofsuch a reaction does not change!2. When you subtract or add reactions, the Gibbs energy of those reactionscan always be directly added or subtracted, not their potentials!

Important notes