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# Electrochemical Potential, Work, and Energy - PowerPoint PPT Presentation

Electrochemical Potential, Work, and Energy. Potential, Work, and Energy Units Joule (J) = unit of energy, heat, or work (w) = kg • m 2 /s 2 Coulomb (C) = unit of electrical charge (q). 1 e - = 1.6 x 10 -19 C = electrical potential ( e )

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Presentation Transcript

• Potential, Work, and Energy

• Units

• Joule (J) = unit of energy, heat, or work (w) = kg•m2/s2

• Coulomb (C) = unit of electrical charge (q). 1 e- = 1.6 x 10-19 C

• = electrical potential (e)

• 1 J of work is produced when 1 C of charge is transferred between two points differing by 1 V of electrical potential

• Work flowing out of a system (Galvanic Cell) is taken to be negative work

• Cell Potential is always positive

• From last chapter, wmax = DG

• Electrochemical Problems

• When current flows, we always waste some of the energy as heat instead of work

w < wmax

• We can, however, measure emax with a potentiometer, so we can find the hypothetical value of wmax

• Example: eocell = 2.50 V 1.33 mole e- pass through the wire. eactual = 2.10 V

• 1 Faraday (F) = the charge on 1 mole of electrons = 96,485 C

(6.022 x 1023 e-/mol)(1.6 x 10-19 C/e-) = 96,485 C/mol

• w = -qe = -(1.33 mol e-)(96,485 C/mole e-)(2.10 J/C) = -2.69 x 105 J

• wmax = -qemax = -(1.33 mol e-)(96,485 C/mole e-)(2.50 J/C) = -3.21 x 105 J

• Efficiency = w/wmax = -2.69 x10-5 J/-3.21 x 105 J = 0.838 or 83.8%

• Free Energy (DG)

• q = nF where n = number of moles, F = 96,485 C/mole

• DG = -nFe (assuming the maximum e)

• Maximum cell potential is directly related to DG between reactants and products in the Galvanic Cell (This lets us directly measure DG)

• Example: Calculate DGo for the reaction

Cu2+(aq) + Fe(s) Cu(s) + Fe2+(aq)

• Half Reactions: Cu2+ + 2e- Cuoeo = 0.34 V

Feo Fe2+ + 2e- eo = 0.44 V

b) DGo = -nFeo = -(2 mol e-)(96,485 C/mol e-)(0.78 J/C) = -1.5 x 105 J

• Example: Will 1 M HNO3 dissolve metallic gold to make 1 M Au3+?

• Half Reaction: NO3- + 4H+ + 3e- NO + 2H2O eo = +0.96 V

Auo Au3+ + 3e-eo = -1.50 V

Au(s) + NO3-(aq) + 4H+(aq) Au3+(aq) + NO(g) + 2H2O(l) eocell = -0.54V

• Since e is negative (DG = +) the reaction will not occur spontaneously

• Cell Potential and Concentration

• Concentration Cells

• Up until now, concentration for all Galvanic solutions = 1 M (Gives eo)

• What happens if we change these concentrations?

• Eocell = +0.78 V

• Cu(s) + 2Ce4+(aq) Cu2+(aq) + 2Ce3+(aq) eocell = 1.36 V

• Increase Ce4+ concentration, (e > eo)

• Increase Cu2+ concentration, (e < eo)

d) Example

• Concentration Cell = Galvanic Cell driven by the fact that concentrations of the same reactants are different on the two sides of the cell.

• Example: Ag+ + e- Agoeo1/2 = +0.80 V

• If both sides had [Ag+] = 1 M, then eocell = +0.80 V + (-0.80 V) = 0.00 V

• If [Ag+]right = 1 M and [Ag+]left = 0.1 M then we should have a potential

• Diffusion would try to equalize Ag+ on the right side and the left side

(Entropy favors even distribution, like gas particles in two chambers)

• Electrons would flow from left to right to even out [Ag+]

• A very small voltage would be generated

• Example

• The Nernst Equation

• Derivation

• DG = DGo + RTlnQ = -nFe

• DGo = -nFeo

• -nFe = -nFeo + RTlnQ

• At 25 oC, this simplifies to

• Example: 2Al(s) + 3Mn2+(aq) 2Al3+(aq) + 3Mn(s) eocell = 0.48 V

• Oxidation: 2Al(s) 2Al3+(aq) + 6e-

• Reduction: 3Mn2+(aq) + 6e- 3Mn(s)

• [Mn2+] = 0.5 M, [Al3+] = 1.5 M

• Q = [Al3+]2 / [Mn2+]3 = (1.5)2 / (0.5)3 = 18

• As the reaction proceeds, ecell 0 (Q K) = Dead Battery!

• Calculating K:

VO2+ + 2H+ + e- VO2+ + H2O eo = 1.00 V

Zn2+ + 2e- Zn eo = -0.76 V

Find Ecell

• Example: [VO2+] = 2M, [H+] = 0.5M, [VO2+] = 0.01M, [Zn2+] = 0.1M

• Ion-Selective Electrodes

• Cell potential depends on concentration of an ion

• pH meter

• Standard electrode of known potential

• Glass electrode filled with known [HCl] whose potential changes based on external [H+]

• Potentiometer measures the potential difference

• You can make similar Na+, K+, or NH4+, Cl-, F-, etc…selective electrodes

• Glass “senses” the presence of H+ in open sites (pH meter)

• Change the type of glass for sensing other ions

• Line Notation for a typical pH electrode:

Ag | AgCl | Cl- || H+ outside | H+ inside, Cl- | AgCl | Ag

Outer ref elec.

sample

Known H+

Inner ref elec.

H+ sensing glass membrane

• Batteries

• Battery Basics

• Battery = galvanic cells used as a portable source of electrical potential

• Batteries are a source of direct current only; not suitable for providing alternating current like permanent outlets do

• Highly rechargeable, durable batteries that can operate between –30 and 120 oF

Anode: Pb + H2SO4 PbSO4 + H+ + 2e-

Cathode: PbO2 + HSO4- + 3H+ + 2e- PbSO4 + 2H2O

Cell: Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4-(aq) 2PbSO4(s) + 2H2O eo = 2.0V

• For cars: 6 of these cells in series with grid electrodes provides 12 V (2 V each)

• Sulfuric Acid is consumed; so density of the acid drops over its life

• Water is also consumed; can “top off” the battery with water. New Ca/Pb electrodes no longer use up water (sealed batteries)

• Alternator recharges battery by forcing current in opposite direction

• Physical Damage, not chemical depletion, usually “kills” the battery

• Other Batteries

• Dry Cell Batteries = calculators, watches, etc…

• Acid Version: Zn anode, C cathode, MnO2/NH4Cl/C paste as electrolyte 1.5V

Anode: Zn Zn2+ + 2e-

Cathode: 2NH4+ + 2MnO2 + 2e- Mn2O3 + 2NH3 + H2O

• Alkaline Version has KOH or NaOH as electrolyte

Anode: Zn + 2OH- ZnO + H2O + 2e-

Cathode: 2MnO2 + H2O + 2e- Mn2O3 + 2OH-

Anode: Cd + 2OH- Cd(OH)2 + 2e-

Cathode: NiO2 + 2H2O + 2e- Ni(OH)2 + 2OH-

d) Nickel-Metal Hydride (NiMH) Batteries

Anode: M∙H + OH- M + H2O + e-

Cathode: NiO2 + 2H2O + 2e- Ni(OH)2 + 2OH-

e) Lithium Ion Batteries: flow of Li+ inside battery matched by e- in wire

• Fuel cells = galvanic cell with continuous source of reactants

• The Hydrogen—Oxygen Fuel Cell is used for NASA spaceflights

• The reactant gases can be stored as liquids in tanks

• Anode: 2H2 + 4OH- 4H2O + 4e-e1/2 = 0.83V

• Cathode: 4e- + O2 + 2H2O 4OH-e1/2 = 1.20V

• Overall: 2H2(g) + O2(g) + catalyst 2H2O(l) eo = 2.03V