Oxidation-Reduction (Redox) Reactions

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

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

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

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°)

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

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

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

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

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

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

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

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)

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+

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

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

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

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)

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

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)

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

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 = 0.059 pe

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

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

Eh as Master Variable Rewrite to put oxidized species on top

Eh (example) SO42- + 8e- + 10H+ ↔ H2S + 4 H2O 8 electrons transferred E° = ( -1 / (8 x 96.5)) x (-231.82) = +0.30 V