Oxidation-Reduction (Redox) Reactions

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 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° = -0.44 V • Cd2+ + 2e- Cd(s),E° = -0.40 V • For oxidation, use negative value (+0.44 V) • Added together, get +0.04 V as standard potential for complete reaction

11. Nernst Equation (example) • Assume [Fe2+] = 0.0100 M and [Cd2+] = 0.005 M (instead of the standard 1.0 M) • For Fe2+ • E = -0.44 – [(8.314)(298)]/[(2)(9.65x10-4)] ln(1/0.001) = -0.50 • Fe Cd2+ • E = -0.40 – (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 Negative Electrode (e- removed) Positive Electrode (e- added) 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 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: • [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 = 0.059 pe

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

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 (-231.82) = +0.30 V