Galvanic (or Voltaic) Cells

Electrochemistry = the interchange of chemical and electrical energy

= used constantly in batteries, chemical instruments, etc…

I. Galvanic CellsA. Definitions

1) Redox Reaction = oxidation/reduction reaction = chemical reaction in which electrons are transferred from a reducing agent (which gets oxidized) to an oxidizing agent (which gets reduced)

2) Oxidation = loss of electron(s) to become more positively charged

3) Reduction = gain of electron(s) to become more negatively charged

B. Using Redox Reactions to generate electric current (moving electrons)

1) 8H+(aq) + MnO4-(aq) + 5Fe2+(aq) Mn2+(aq) + 5Fe3+(aq) + 4H2O(l)

a) Fe2+ is oxidized and MnO4- is reduced

b) Half Reaction = oxidation or reduction process only

Reduction: 8H+ + MnO4- + 5e- Mn2+ + 4H2O

Oxidation: 5(Fe2+ Fe3+ + e-)

Sum = Redox Rxn

2) In solution:

a) Fe2+ and MnO4- collide and electrons are transferred

b) No work can be obtained; only heat is generated

3) In separate compartments, electrons must go through a wire = Galvanic Cell

a) Generates a current = moving electrons from Fe2+ side to MnO4- side

b) Current can produce work in a motor

c) Salt Bridge = allows ion flow without mixing solutions (Jello-like matrix)

d) Chemical reactions occur at the Electrodes = conducting solid dipped into the solution

i) Anode = electrode where oxidation occurs (production of e-)

ii) Cathode = electrode where reduction occurs (using up e-)

C. Cell Potential

1) Think of the Galvanic Cell as an oxidizing agent “pulling” electrons off of the reducing agent. The “pull” = Cell Potential

a) cell = Cell Potential = Electromotive Force = emf

b) Units for cell = Volt = V 1 V = 1 Joule/1 Coulomb

2) Voltmeter = instrument drawing current through a known resistance to find V

Potentiometer = voltmeter that doesn’t effect V by measuring it

II. Standard Reduction PotentialsA. Standard Hydrogen Electrode

1) When measuring a value, you must have a standard to compare it to

2) Cathode = Pt electrode in 1 M H+ and 1 atm of H2(g)

Half Reaction: 2H+ + 2e- H2(g) 1/2 = 0

3) We will use this cathode to find cell of other Half Reactions

4) Standard Reduction Potentials can be found in your text appendicesa) Always given as a reduction processb) All solutes are 1M, gases = 1 atm

5) Combining Half Reactions to find Cell Potentials

a) Reverse one of the half reactions to an oxidation; this reverses the sign of 1/2 b) Don’t need to multiply for coefficients = Intensive Property (color, flavor)c) Example: 2Fe3+(aq) + Cuo 2Fe2+(aq) + Cu2+(aq)

i. Fe3+ + e- Fe2+ 1/2 = +0.77 V

ii. Cu2+ + 2e- Cuo 1/2 = +0.34 V

iii. Reverse of (ii) added to (i) = -0.34 V + +0.77 V = +0.43 V = cell

B. Direction of electron flow in a cell

1) Cell always runs in a direction to produce a positive cell

2) Fe2+ + 2e- Feo 1/2 = -0.44 V

MnO4- + 5e- + 8H+ Mn2+ + 4H2O 1/2 = +1.51 V

3) We put the cell together to get a positive potential

a) 5(Feo Fe2+ + 2e-) 1/2 = +0.44 V

b) 2(MnO4- + 5e- + 8H+ Mn2+ + 4H2O)1/2 = +1.51 V

16H+(aq) + 2MnO4-(aq) + 5Feo(s) 2Mn2+(aq) + 5Fe2+(aq) + 8H2O(l) cell = 1.95V

C. The complete description of a Galvanic Cell

1) Items to include in the description

a) Cell potential (always +) and the balanced overall reaction

b) Direction of electron flow

c) Designate the Anode and the Cathode

d) Identity of the electrode materials and the ions present with concentration

2) Example: Completely describe the Galvanic Cell based on these reactions

Ag+ + e- Ag o = +0.80 V

Fe3+ + e- Fe2+ o = +0.77 V

Ag+(aq) + Fe2+(aq) Ago(s) + Fe3+(aq) ocell = +0.03 V

Electrochemical Potential, Work, and Energy

III. Potential, Work, and EnergyA. Units

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

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

3) = electrical potential ()

4) 1 J of work is produced when 1 C of charge is transferred between two points differing by 1 V of electrical potential

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

6) Cell Potential is always positive

7) From last chapter, wmax = G

(C) Charge

(J)Work (V)Volt

qε w-or q

w- ε

maxmax qε- wG

B. Electrochemical Problems

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

w < wmax

2) We can, however, measure max with a potentiometer, so we can find the hypothetical value of wmax

3) Example: ocell = 2.50 V 1.33 mole e- pass through the wire. actual = 2.10 V

a) 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

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

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

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

4) Free Energy (G)

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

b) G = -nF (assuming the maximum )

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

maxmaxmax nFεqε- wG

Process sSpontaneou G- ε

5) Example: Calculate Go for the reaction

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

a) Half Reactions: Cu2+ + 2e- Cuo o = 0.34 V

Feo Fe2+ + 2e- o = 0.44 V

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

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

a) Half Reaction: NO3- + 4H+ + 3e- NO + 2H2O o = +0.96 V

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

Au(s) + NO3-(aq) + 4H+(aq) Au3+(aq) + NO(g) + 2H2O(l) o

cell = -0.54V

b) Since is negative (G = +) the reaction will not occur spontaneously

Eocell = +0.78 V

C. The Nernst Equation

1) Derivation

a) G = Go + RTlnQ = -nF

b) Go = -nFo

c) -nF = -nFo + RTlnQ

2) At 25 oC, this simplifies to

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

a) Oxidation: 2Al(s) 2Al3+(aq) + 6e-

b) Reduction: 3Mn2+(aq) + 6e- 3Mn(s)

c) [Mn2+] = 0.5 M, [Al3+] = 1.5 M

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

e) As the reaction proceeds, cell 0 (Q K) = Dead Battery!

f) Calculating K:

lnQnF

RT - ε ε o

logQn

0.0592 - ε ε o

V 47.0log186

0.0592 - V 0.48 εcell

0.0592

nεlogKlogK

n

0.0592 - ε 0

oo

IV. Electrolysis = using electric energy to produce chemical change (opposite of cell)A. Example

1) Consider the Cu/Zn Galvanic Cell

a) Anode: Zn Zn2+ + 2e-

b) Cathode: Cu2+ + 2e- Cu ocell = +1.10 V

2) If we attach a power source of o > +1.10 V, we can force e- to go the other way

a) Anode: Cu Cu2+ + 2e-

b) Cathode: Zn2+ + 2e- Zn

c) Called an Electrolytic Cell

B. Calculations with Electrolytic Cells1) How much Chemical Change? Is usually the question.2) Find mass Cuo plated out passing 10 amps (10 C/s) through Cu2+ solution 30 min.

a) Cu2+ + 2e- Cuo(s) b) Steps: current/time, charge (C), moles e-, moles Cu, grams Cu

3) Example: How long must a current of 5.00 amps be applied to a Ag+ solution to produce 10.5 g of silver metal?

C. Electrolysis of Water

1) Galvanic: 2H2 + O2 2H2O (Fuel Cell)2) Electrolytic Cell:

a) Anode: 2H2O O2 + 4H+ + 4e- -o = -1.23 V

b) Cathode: 4H2O + 4e- 2H2 + 4OH- o = -0.83 V

c) Overall: 6H2O 2H2 + O2 + 4(H+ + OH-)

2H2O 2H2 + O2 ocell = -2.06 V

3) We must add a salt to increase the conductance of pure water [H+] = [OH-] = 10-7

Cu g 5.94 g/mol) Cu)(63.546 mol 0935.0(

Cu mol 0935.0-e mol 2

Cu mol 1e-) mol (0.187

-e mol 0.187 C 96,485

-e mol 1C) (18,000

C 18,000 s/min) min)(60 C/s)(30 10(

D. Electrolysis of Mixtures

1) Mixture of Cu2+, Ag+, Zn2+; What is the order of plating out?

a) Ag+ + e- Ag o1/2 = +0.80 V

b) Cu2+ + 2e- Cu o1/2 = +0.34 V

c) Zn2+ + 2e- Zn o1/2 = -0.76 V

2) Reduction of Ag+ is easiest (o = most positive) followed by Cu, then Zn

3) Example: Ce4+ (o1/2 = +1.70 V), VO2+ (o

1/2 = +1.00 V), Fe3+ (o1/2 = +0.77 V)