1. Electrochemistry
Applications of Redox

2. Cell Potential Oxidizing agent pull the electron.
Reducing agent push the electron.
The push or pull (“driving force”) is called the cell potential Ecell
Also called the electromotive force (emf)
Unit is the volt(V) = 1 joule of work/coulomb of charge
Measured with a voltmeter
Measures the ability for a ½ cell to attract electrons ie: to be reduced.
If ½ cell undergoes reduction Ecell > 0
If ½ cell undergoes oxidation Ecell < 0

3. Standard Hydrogen Electrode
This is the reference all other oxidations are compared to
Eº = 0
º indicates standard states of 25ºC, 1 atm, 1 M solutions.
H2 in
H+ Cl-
1 M HCl

4. 0.76
H2 in
Cathode
Anode
H+ Cl-
Zn+2 SO4-2
1 M ZnSO4
1 M HCl

5. Cell Potential Zn(s) + Cu+2 (aq) ® Zn+2(aq) + Cu(s)
The total cell potential is the sum of the potential at each electrode.
Eº cell = EºZn® Zn+2 + Eº Cu+2 ® Cu
We can look up reduction potentials in a table.
One of the reactions must be reversed, so change it sign.

6. Cell Potential
Determine the cell potential for a galvanic cell based on the redox reaction.
Cu(s) + Fe+3(aq) ® Cu+2(aq) + Fe+2(aq)
1. Fe+3(aq) + e-® Fe+2(aq) Eº = 0.77 V
2. Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V
Since the iron electrode has a higher Eº value, the forward reaction is observed. Reaction 2 must be written in reverse and the sign changed.
Cu(s) ® Cu+2(aq)+2e Eº = V
Balance the charges for each ½ cell.
2Fe+3(aq) + 2e-® 2Fe+2(aq) Cu(s) ® Cu+2(aq)+2e-
Add the ½ cells and the E º
2Fe+3(aq) + Cu(s) ® 2Fe+2(aq) + Cu+2(aq)+
Eº = V
= 0.43V

7. Line Notation solid½Aqueous½½Aqueous½solid
Anode on the left½½Cathode on the right
Single line different phases.
Double line porous disk or salt bridge.
If all the substances on one side are aqueous, a platinum electrode is indicated.
For the last reaction
Cu(s)½Cu+2(aq)½½Fe+2(aq),Fe+3(aq)½Pt(s)

8. Galvanic Cell
The reaction always runs spontaneously in the direction that produced a positive cell potential.
Four things for a complete description.
Cell Potential
Direction of flow
Designation of anode and cathode
Nature of all the components- electrodes and ions

9. Practice
Completely describe the galvanic cell based on the following half-reactions under standard conditions.
MnO H+ +5e- ® Mn+2 + 4H2O Eº=1.51
Fe+3 +3e- ® Fe(s) Eº=0.036V
p. 708 # p.709 #1-3, 7, 8

10. Batteries are Galvanic Cells
Car batteries are lead storage batteries.
Pb +PbO2 +H2SO4 ®PbSO4(s) +H2O
Dry Cell Zn + NH4+ +MnO2 ® Zn+2 + NH3 + H2O
Alkaline Zn +MnO2 ® ZnO+ Mn2O3 (in base)
NiCad
NiO2 + Cd + 2H2O ® Cd(OH)2 +Ni(OH)2

11. Corrosion Rusting - spontaneous oxidation.
Most structural metals have reduction potentials that are less positive than O2 .
Fe ® Fe+2 +2e- Eº= 0.44 V
O2 + 2H2O + 4e- ® 4OH- Eº= 0.40 V
Fe+2 + O2 + H2O ® Fe2 O3 + H+
Reaction happens in two places.

12. Salt speeds up process by increasing conductivity
Water
Rust e-
Iron Dissolves- Fe ® Fe+2

13. Preventing Corrosion Coating to keep out air and water.
Galvanizing - Putting on a zinc coat
Has a lower reduction potential, so it is more. easily oxidized.
Alloying with metals that form oxide coats.
Cathodic Protection - Attaching large pieces of an active metal like magnesium that get oxidized instead.

14. Electrolysis Running a galvanic cell backwards.
Put a voltage bigger than the potential and reverse the direction of the redox reaction.
Used for electroplating.

15. 1.10 e- e-
Anode
Cathode Zn Cu
1.0 M Zn+2
1.0 M Cu+2

16. A battery >1.10V e- e-
Cathode
Anode Zn Cu
1.0 M Zn+2
1.0 M Cu+2

17. Other uses Electroysis of water. Separating mixtures of ions.
More positive reduction potential means the reaction proceeds forward.
We want the reverse.
Most negative reduction potential is easiest to plate out of solution.

18. Potential, Work and DG emf = potential (V) = work (J) / Charge(C)AP
Potential, Work and DG
emf = potential (V) = work (J) / Charge(C)
E = work done by system / charge
E = -w/q
Charge is measured in coulombs.
-w = qE
Faraday = 96,485 C/mol e-
q = nF = moles of e- x charge/mole e-
w = -qE = -nFE = DG

19. Potential, Work and DG DGº = -nFE ºAP
Potential, Work and DG
DGº = -nFE º
if E º < 0, then DGº > 0 spontaneous
if E º > 0, then DGº < 0 nonspontaneous
In fact, reverse is spontaneous.
Calculate DGº for the following reaction:
Cu+2(aq)+ Fe(s) ® Cu(s)+ Fe+2(aq)
Fe+2(aq) + e-® Fe(s) Eº = 0.44 V
Cu+2(aq)+2e- ® Cu(s) Eº = 0.34 V

20. Cell Potential and ConcentrationAP
Cell Potential and Concentration
Qualitatively - Can predict direction of change in E from LeChâtelier.
2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s)
Predict if Ecell will be greater or less than Eºcell if [Al+3] = 1.5 M and [Mn+2] = 1.0 M
if [Al+3] = 1.0 M and [Mn+2] = 1.5M
if [Al+3] = 1.5 M and [Mn+2] = 1.5 M

21. The Nernst Equation E = Eº - RTln(Q) nF DG = DGº +RTln(Q)AP
The Nernst Equation
DG = DGº +RTln(Q)
-nFE = -nFEº + RTln(Q)
E = Eº - RTln(Q) nF
2Al(s) + 3Mn+2(aq) ® 2Al+3(aq) + 3Mn(s) Eº = 0.48 V
Always have to figure out n by balancing.
If concentration can gives voltage, then from voltage we can tell concentration.

22. The Nernst Equation 0 = Eº - RTln(K) nF Eº = RTln(K) nFAP
The Nernst Equation
As reactions proceed concentrations of products increase and reactants decrease.
Reach equilibrium where Q = K and Ecell = 0
0 = Eº - RTln(K) nF
Eº = RTln(K) nF
nFEº = ln(K) RT

23. Calculating plating Have to count charge.AP
Calculating plating
Have to count charge.
Measure current I (in amperes)
1 amp = 1 coulomb of charge per second
q = I x t
q/nF = moles of metal
Mass of plated metal
How long must 5.00 amp current be applied to produce 15.5 g of Ag from Ag+