Analytical electrochemistry

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Transcript Analytical electrochemistry

CHEMISTRY 59-320 ANALYTICAL CHEMISTRY Fall - 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 I

=

P

= /

E

= 0.010 0 W/1.35 V = 7.41 × 10 −3 (7.41 × 10 −3 C/s)/(9.649 × 10 4 C/mol) C/s = 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

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)|CdCl 2 (aq)|AgCl(s)|Ag(s) Line diagram for the above cell: Cd(s)|CdCl 2 (aq)||AgNO 3 (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

.

• 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).

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

Application of standard potentials

Problem: 14-13. (a) Cyanide ion causes E ° for Fe(III) to decrease: Fe 3+ + e ↔ Fe 2+ E °= 0.771 V Fe(CN) 6 3+ + e ↔ Fe(CN) 6 3+ E ° = 0.356 V Which ion, Fe 3+ , or Fe 2+ , is stabilized more by complexing with CN?

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 reach the same conclusion?

14-4 Nernst Equation

Problem 14-16

: Write the Nernst equation for the following half-reaction and find E when pH = 3.00 and P asH3 As(s) + 3H + + 3e = 1.0 mbar.

↔ AsH 3 (g) E o = -0.238 V Answer: E = E o – [RT/(nF)]*ln{(P asH3 /(a H+ ) 3 } Since no temperature is provide, we can assume T = 25 o C, if so we will Use equation 14-15 to carry out the calculation E = E 0 – (0.05916/3)*log {0.001/ /(a H+ ) 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.)

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.

• 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.

14-5 E

o

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) + 2Fe 3+ ↔ 2Fe 2+ + Cu 2+ • Solution: step 1: identify cathode and anode step 2: find the standard potential for half reactions step 3: calculation E o 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 E o for the half reaction Pd(OH) 2 (s) + 2e ↔ Pd(s) + 2OH given that K sp reaction Pd 2+ for Pd(OH) 2 + 2e is 3x10 ↔ Pd(s) -28 and E o = 0.915 V for the • Solution: Pd(OH) 2 (s) ↔ Pd 2+ K sp = [Pd 2+ ]([OH ]) 2 + 2OH Ksp from K sp one can get E o for the above reaction (eq 14-23) E o = -0.814 V + 2e Since the above reaction can be derived by using Pd(OH) 2 (s) ↔ Pd(s) + 2OH to subtract Pd 2+ + 2e ↔ Pd(s) we have -0.814 V = E o - 0.915 V E o = 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 be used to measure the formation constant of Hg(EDTA) 2 Hg 2+ + 2e ↔ Hg(

l

) E o = 0.852 V

14-7 Biochemists use E

o

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 Cd 2+ .

Solution:

Cd 2+ + 2e  Cd(s)  0 = -0.403

0.0591

log 2 1  Cd 2+ ε = -0.403 0.0591

log 2 1 0.0150

  0.457

V

Example: Calculate the potential for a platinum electrode immersed in a solution prepared by saturating a 0.015 M solution of KBr with Br 2 Br 2 (

l

) + 2e

2Br

-

E 0 = 1.065 V Solution:

Br

 2 E= 1.065 0.0591

2 log 1.0

E = 1.065 0.0591

2  2  1.173

V

Important notes 1. When you multiply a reaction by a coefficient, the potential of such a reaction does not change!

2. When you subtract or add reactions, the Gibbs energy of those reactions can always be directly added or subtracted, not their potentials!