Oxidation-Reduction (Redox) Reactions 1 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 2 Balancing Redox Reactions • Reaction needs to be charge balanced; eneed 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 3 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 4 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°) 5 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 6 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 7 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 8 Standard Potentials Written as reductions Strong Reducing Agents Half-Reaction Li+(aq) + e- → Li(s) K+(aq) + e- → K(s) Ba2+(aq) + 2 e- → Ba(s) Sr2+(aq) + 2 e- → Sr(s) Ca2+(aq) + 2 e- → Ca(s) Na+(aq) + e- → Na(s) Mg2+(aq) + 2 e- → Mg(s) Be2+(aq) + 2 e- → Be(s) Al3+(aq) + 3 e- → Al(s) Mn2+(aq) + 2 e- → Mn(s) 2 H2O + 2 e- → H2(g) + 2 OH-(aq) Zn2+(aq) + 2 e- → Zn(s) Cr3+(aq) + 3 e- → Cr(s) Fe2+(aq) + 2 e- → Fe(s) Cd2+(aq) + 2 e- → Cd(s) PbSO4(s) + 2 e- → Pb(s) + SO42-(aq) Co2+(aq) + 2 e- → Co(s) Ni2+(aq) + 2 e- → Ni(s) Sn2+(aq) + 2 e- → Sn(s) Pb2+(aq) + 2 e- → Pb(s) 2 H+(aq) + 2 e- → H2(g) E0 (V) -3.05 -2.93 -2.90 -2.89 -2.87 -2.71 -2.37 -1.85 -1.66 -1.18 -0.83 -0.76 -0.74 -0.44 -0.40 -0.31 -0.28 -0.25 -0.14 -0.13 0 The greater the E°, the more easily the substance reduced Half-Reaction E0 (V) 2 H+(aq) + 2 e- → H2(g) 0 4+ 2+ Sn (aq) + 2 e → Sn (aq) 0.13 2+ + Cu (aq) + e → Cu (aq) 0.13 2+ SO4 (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 2+ Hg2 (aq) + 2 e → 2 Hg(l) 0.85 2+ 2+ 2 Hg (aq) + 2 e → Hg2 (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 + 2+ MnO2(s) + 4 H (aq) + 2 e → Mn (aq) + 2 H2O 1.23 2+ 3+ Cr2O7 (aq) + 14 H (aq) + 6 e → 2 Cr (aq) + 7 H2O 1.33 Cl2(g) + 2 e → 2 Cl (aq) 1.36 Au3+(aq) + 3 e- → Au(s) 1.50 + 2+ MnO4 (aq) + 8 H (aq) + 5 e → Mn (aq) + 4 H2O 1.51 4+ 3+ Ce (aq) + e → Ce (aq) 1.61 + 2PbO2(s) + 4H (aq) + SO4 (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 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 10 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 11 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 12 Electrochemical Cell Negative Electrode (e- removed) Positive Electrode (e- added) Zn(s) Zn2+ + 2e- Cu2+ + 2e- Cu(s) Zn(s) + Cu2+ Zn2+ + Cu(s) 13 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+ 14 Redox Cell Pt wire electrode Salt bridge 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 15 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 16 Redox Cell using Platinum • Fe3+ + e- ↔ Fe2+ – – • [e-] proportional to the ratio of activities of the reduced to the oxidized species 17 Redox Cell Pt wire electrode Salt bridge H2 gas (1 atm) Fe2+ and Fe3+ [H+] = 1 H+ + e- ↔ ½ H2(g) 18 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 19 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) 20 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 21 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 22 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 – – 23 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 24 Eh as Master Variable • • Rewrite to put oxidized species on top 25 Eh (example) • SO42- + 8e- + 10H+ ↔ H2S + 4 H2O – 8 electrons transferred – • • E° = ( -1 / (8 x 96.5)) x (-231.82) = +0.30 V 26