1. Oxidation-Reduction (Redox) Reactions

2. Redox: Introduction
Electrons (e-) are transferred from one compound to another
e--donors (lose electrons)
e--acceptors (gain electrons)
Loss of e- by e--donor = oxidation
Gain of e- by e--acceptor = reduction
Oxidation and reduction always accompany one another
Electrons cannot exist freely in solution

3. Balancing Redox Reactions
Reaction needs to be charge balanced; e- need to be removed
To balance charge, cross multiply reactants
Redox reactions can also be written as half reactions, with e- included
Multiply so that e- are equal (LCM) and combine half reactions

4. Redox Reactions
Ability of elements to act as e--donors or acceptors arises from extent to which orbitals are filled with electrons
Property depends on decrease in energy of atoms resulting from having only incompletely filled orbitals
Some similarities to acid-base reactions
Transferring electrons (e-) instead of H+
Come in pairs (oxidation-reduction)
In acid-base reactions, we measure changes in [H+] (pH); in redox reactions, we measure voltage changes

5. Measuring voltage
Standard potential tables have been created for how much voltage (potential) a reaction is capable of producing or consuming
Standard conditions: P =1 atm, T = 298°K, concentration of 1.0 M for each product
Defined as Standard Potential (E°)

6. Non-Standard Redox Conditions (real life)
For non-standard conditions, E° needs to be corrected
Temperature
Number of electrons transferred
Concentrations of redox reactants and products

7. Nernst Equation Nernst Equation
E = electrical potential of a redox reaction
R = gas constant
T = temperature (K)
n = number of electrons transferred
F = Faraday constant = 9.65 x 104 J / V∙mole
Q = ion activity product

8. Nernst Equation (example)
Fe(s) + Cd2+  Fe2+ + Cd(s)
Fe oxidized, Cd reduced
Need to find standard potentials of half reactions
Fe2+ + 2e-  Fe(s)
Cd2+ + 2e-  Cd(s)
Look up in Table

9. Standard Potentials Written as reductions Strong Reducing Agents
Half-Reaction
E0 (V)
2 H+(aq) + 2 e- → H2(g)
Sn4+(aq) + 2 e- → Sn2+(aq)
0.13
Cu2+(aq) + e- → Cu+(aq)
SO42-(aq) + 4 H+(aq) + 2 e- → SO2(g) + 2 H2O
0.20
AgCl(s) + e- → Ag(s) + Cl-(aq)
0.22
Cu2+(aq) + 2 e- → Cu(s)
0.34
O2(g) + 2 H2 + 4 e- → 4 OH-(aq)
0.40
I2(s) + 2 e- → 2 I-(aq)
0.53
MnO4-(aq) + 2 H2O + 3 e- → MnO2(s) + 4 OH-(aq)
0.59
O2(g) + 2 H+(aq) + 2 e- → H2O2(aq)
0.68
Fe3+(aq) + e- → Fe2+(aq)
0.77
Ag+(aq) + e- → Ag(s)
0.80
Hg22+(aq) + 2 e- → 2 Hg(l)
0.85
2 Hg2+(aq) + 2 e- → Hg22+(aq)
0.92
NO3-(aq) + 4 H+(aq) + 3 e- → NO(g) + 2 H2O
0.96
Br2(l) + 2 e- → 2 Br-(aq)
1.07
O2(g) + 4 H+(aq) + 4 e- → 2 H2O
1.23
MnO2(s) + 4 H+(aq) + 2 e- → Mn2+(aq) + 2 H2O
Cr2O72-(aq) + 14 H+(aq) + 6 e- → 2 Cr3+(aq) + 7 H2O
1.33
Cl2(g) + 2 e- → 2 Cl-(aq)
1.36
Au3+(aq) + 3 e- → Au(s)
1.50
MnO4-(aq) + 8 H+(aq) + 5 e- → Mn2+(aq) + 4 H2O
1.51
Ce4+(aq) + e- → Ce3+(aq)
1.61
PbO2(s) + 4H+(aq) + SO42-(aq) + 2e- → PbSO4(s) + 2H2O
1.70
H2O2(aq) + 2 H+(aq) + 2 e- → 2 H2O
1.77
Co3+(aq) + e- → Co2+(aq)
1.82
O3(g) + 2 H+(aq) + 2 e- → O2(g) + H2O
2.07
F2(g) + 2 e- -----> F-(aq)
2.87
Written as reductions
Strong Reducing Agents
Half-Reaction
E0 (V)
Li+(aq) + e- → Li(s)
-3.05
K+(aq) + e- → K(s)
-2.93
Ba2+(aq) + 2 e- → Ba(s)
-2.90
Sr2+(aq) + 2 e- → Sr(s)
-2.89
Ca2+(aq) + 2 e- → Ca(s)
-2.87
Na+(aq) + e- → Na(s)
-2.71
Mg2+(aq) + 2 e- → Mg(s)
-2.37
Be2+(aq) + 2 e- → Be(s)
-1.85
Al3+(aq) + 3 e- → Al(s)
-1.66
Mn2+(aq) + 2 e- → Mn(s)
-1.18
2 H2O + 2 e- → H2(g) + 2 OH-(aq)
-0.83
Zn2+(aq) + 2 e- → Zn(s)
-0.76
Cr3+(aq) + 3 e- → Cr(s)
-0.74
Fe2+(aq) + 2 e- → Fe(s)
-0.44
Cd2+(aq) + 2 e- → Cd(s)
-0.40
PbSO4(s) + 2 e- → Pb(s) + SO42-(aq)
-0.31
Co2+(aq) + 2 e- → Co(s)
-0.28
Ni2+(aq) + 2 e- → Ni(s)
-0.25
Sn2+(aq) + 2 e- → Sn(s)
-0.14
Pb2+(aq) + 2 e- → Pb(s)
-0.13
2 H+(aq) + 2 e- → H2(g)
The greater the E°, the more
easily the substance reduced
Strong Oxidizing Agents

10. Nernst Equation (example)
By convention, half-reactions are shown as reduction reactions
Fe(s) + Cd2+  Fe2+ + Cd(s)
Fe2+ + 2e-  Fe(s), E° = V
Cd2+ + 2e-  Cd(s), E° = V
For oxidation, use negative value (+0.44 V)
Added together, get V as standard potential for complete reaction

11. Nernst Equation (example)
Assume [Fe2+] = M and [Cd2+] = M (instead of the standard 1.0 M)
For Fe2+
E = – [(8.314)(298)]/[(2)(9.65x10-4)] ln(1/0.001) = -0.50
Fe Cd2+
E = – (0.128) ln (1/0.005) = -0.47

12. Redox and Thermodynamics
Energy released in a redox reaction can be used to perform electrical work using an electrochemical cell
A device where electron transfer is forced to take an external pathway instead of going directly between the reactants
A battery is an example

13. Electrochemical Cell Zn(s)  Zn2+ + 2e- Cu2+ + 2e-  Cu(s)
Negative
Electrode
(e- removed)
Positive
Electrode
(e- added)
Instead of placing a piece of zinc directly into a solution containing copper, we can form a cell where solid pieces of zinc and copper are placed in two different solutions such as sodium nitrate. The two solids are called electrodes. The anode is the electrode where oxidation occurs and mass is lost where as the cathode is the electrode where reduction occurs and mass is gained. The two electrodes are connected by a circuit and the two (2) solutions are connected by a "salt bridge" which allows ions to pass through. The anions are the negative ions and they move towards the anode. The cations are the positive ions and they move towards the cathode
Zn(s)  Zn2+ + 2e-
Cu2+ + 2e-  Cu(s)
Zn(s) + Cu2+  Zn2+ + Cu(s)

14. Redox Cell using Platinum
Platinum is a good inert means of transferring electrons to/from solution
Consider the half-reaction in the presence of a Pt electrode:
Fe3+ + e- ↔ Fe2+

15. Redox Cell
Salt bridge
Pt wire
electrode
H2 gas (1 atm)
Fe2+
and
Fe3+
[H+] = 1
Fe3+ + e- ↔ Fe2+
←: Pt wire removes electrons from half cell A
→: Pt wire provides electrons to the solution

16. Redox Cell using Platinum
If Pt wire not connected to source/sink of electrons, there is no net reaction
But wire acquires an electrical potential reflecting tendency of electrons to leave solution
Defined as activity of electrons [e-]
pe = -log [e-]
Can be used in equilibrium expressions

17. Redox Cell using Platinum
Fe3+ + e- ↔ Fe2+
[e-] proportional to the ratio of activities of the reduced to the oxidized species

18. Redox Cell H+ + e- ↔ ½ H2(g) Salt bridge Pt wire electrode
H2 gas (1 atm)
Fe2+
and
Fe3+
[H+] = 1
H+ + e- ↔ ½ H2(g)

19. Redox Cell using Platinum
Other half-cell (B)
H+ + e- ↔ ½ H2(g)
Can write an equilibrium expression:
SHE = standard hydrogen electrode
By convention, [e-] =1 in SHE

20. Redox Cell using Platinum
If switch is closed, electrons will move from solution with higher activity of e- to the solution with lower activity of e-
Energy is released (heat)

21. Redox Cell using Platinum
Combine half reactions:
Fe3+ + ½ H2(g) ↔ Fe2+ + H+
Direction of reaction depends on which half-cell has higher activity of electrons
Now open switch: no transfer of e-
Voltage meter registers difference in potential (E) between the 2 electrodes
Potential of SHE = 0, so E = potential of electrode in half-cell A
Defined as Eh
Measured in volts

22. Eh Eh is positive when: Eh is negative when:
[e-] in solution A less than [e-] in SHE
Eh is negative when:
[e-] in solution A greater than [e-] in SHE
At 25°C, pe = 16.9
Eh = pe

23. Eh as Master Variable Fe3+ + ½ H2(g) ↔ Fe2+ + H+
Recall: G° = -RT ln Keq
Standard state
At non-standard state: GR = G° + RT ln Keq

24. Eh as Master Variable From electrochemistry: GR = -nF Eh
n = number of electrons
By convention, sign of Eh set for half-reaction written with e- on left side of equation
Divide through by –nf

25. Eh as Master Variable
Rewrite to put oxidized species on top

26. Eh (example) SO42- + 8e- + 10H+ ↔ H2S + 4 H2O 8 electrons transferred
E° = ( -1 / (8 x 96.5)) x ( ) = V