1. Oxidation Reduction reactions ARAR B ox A ox BRBR ++ Transfer of electrons Fe 2+ Cu 2+ + Fe 3+ Cu + + Transfer of H + + e - AH 2 B A ++BH 2 Addition of O 2 RCH3 + 1/2O 2 RCH 2 OH Transfer of hydride ions :H - AH 2 NAD A + +NADH + H + Dehydrogenases (see Horton: p319-323)

2. Redox potential E o’ the standard redox potential at pH 7 and 1 M concentrations measure of the affinity for electrons units of volts the greater the affinity, the larger the value of E o ’

3. Half cells or redox couples AH 2 B A ++BH 2 A + 2H + + 2 e - AH 2 Eo’Eo’ -0.12V  E o ’ = E o ’ acceptor – E o ’ donar B + 2H + + 2 e - BH 2 +0.10V = + 0.1 -(-0.12) = + 0.22V

4. V. 1 MX + 1MX - 1M H 2 + 1 atm H + Salt bridge KCl Agar bridge E o of the H half cell is defined as 0 For biochemical reactions E o ’= -0.42V

5. V. -0.197v CH 2 OH CH 2 O 2H + H2H2 V. H2H2 +0.771v Fe 3+ Fe 2+

6. ARAR B ox A ox BRBR ++ n = number of electrons transferred F – Faraday’s constant = 96.485 kJ/Vmol  E o’ = difference in redox potential of e - accepter and e - donar = E o ’ acceptor - E o ’ donar  G o’ = -nF  E o ’ For a reaction involving the transfer of 2 electrons:  G o’ = -nF  E o ’ = (-2 x 96.5)  E o ’ = - 193  E o ’

7. Example of calculation of E o ’ -0.197V +0.320 Eo’Eo’  E o ’ = E o ’ acceptor - E o ’donor = 0.320 –(-0.197) = 0.320 + 0.197 = +0.517  G o’ = -nF  E o ’ NAD + 2H + + 2 e - NADH + H + 2 half cells: CH 3 CHO + 2H + + 2 e - CH 3 CH 2 OH CH 3 CHO + NADH + H + CH 3 CH 2 OH + NAD = -2 x 96.5 x 0.517 = -99.8kJ/mol Positive  E o ’ = spontaneous rx Negative  G o ’ = spontaneous rx

8. Oxidation of NADH by oxygen NADH + H + + O 2 NAD + H 2 O NAD + 2H + + 2e - NADH + H + 1/2 O 2 + 2H + + 2e - H 2 O -0.32V +0.82V Eo’Eo’  E o ’ = E o ’ acceptor - E o ’donor = +0.82 - (-0.32) = +1.14V  G o’ = -nF  E o ’ = -2 x 96.5 x 1.14 = -220kJ/mol

9. Nernst equation (see page 333 Horton) for determination of actual  E o ’  E =  E o ’ - RT nF ln Q At 25 o c (298 o K) for 1 e transfer RT F = 0.026 and equation becomes  E =  E o ’ - 0.026 n ln Q