REDOX REACTIONS
What happens upon diping a needle in a CuSO 4 solution!? Fe + Cu 2+ SO 4 2- Fe 2+ SO 4 2- + Cu
In a redox reaction, one (or more) electrons are donated by a reductant to an oxidant
Multiple e- transfer (ET- chain) NADH, FADH 2 Respiratory chain Reducing Substraes O 2 H 2 O
1 e- Reductant 1 + Oxidant 2 Oxidised 1 + Reduced 2 A redox reaction to occur needs: at least 2 reagents a different tendency to attract electrons to become in physical contact (directly or indirectly)
Reduction of TMPD by ascorbate (Vit C) 1 e- R red + O ox R ox.ed + O rid.ed Ascorbate (-) + TMPD (+) ascorbate (+) + TMPD (-)
Reduction of TMPD by ascorbate (Vit C) VitC rid + TMPD ox VitC ox + TMPD rid 2 e - + TMPD + + TMPD - + 2 H + + + blue colourless
Oxidation number (N ) variation, to probe a redox-reaction (N ) It is the formal charge that an atom in a molecule would acquire if all electrons would be attributed to the most electronegative element in the structure. Es. NO 2 N = + 4 N = 2 x (-2) = - 4
Formal roules, to attribute N All atoms in the elementary state, N = 0 e.g. all diatomic molecules (N 2, Cl 2, O 2, H 2 etc.) Oxygen in all compounds N = 2 (exception H 2 O 2, Li 2 O 2 etc. N = - 1) H, in all compounds N = +1 (exceptions hydrures, LiH; AlH 3, N = -1)
metal oxidation by O 2 2 Mg s + O 2g 2MgO N 0 0 +2-2 2 Zn s + O 2g 2ZnO N 0 0 +2-2 metal oxidation by other e - - acceptors (Cl 2, Cu ++, H + ) Mg s + Cl 2g MgCl 2 N 0 0 +2 2 x (-1) Mg s + Cu ++ Mg ++ + Cu N 0 + 2 + 2 0 Mg s + 2H + Mg ++ + H 2g N 0 2 x (+1) +2 0 And equilibria!?
Single e - exchange 1 e- Reductant 1 + Oxidant 2 Oxidised 1 + Reduced 2
Zn + CuSO 4 ZnSO 4 + Cu Electronegativity Zn = 1.6 Cu = 1.9 Surface stabilization electrons d + d - e - Fe + CuSO 4 FeSO 4 + Cu (dipping a paper clip!)
Zn bar in H 2 O What happens!? H 2 O H 2 O H 2 O H 2 O H 2 O H 2 O - Zn ++ sol H 2 O - metal atom ion hydrated ion reticular energy ionization energy hydration energy Zn Zn ++ + 2e - [Zn ++ ] [e - ] 2 Keq = ---------------- E (Zn electrode potential) [Zn]
Electrochemical potential gradient DE Zn ++ Zn ++ Zn ++ Zn ++ - - - - Zn bar Using DE is just matter of technicality!! De - DE Cu ++ - - Cu bar
Cell 1 Cell 2 - - - - - - Zn bar Cu bar Zn Zn ++ + 2e - Cu Cu ++ + 2e - K Zn 4 x K Cu Zn electrode is electron-loaded 4 times more than Cu bar Shall we short circuit the two cells?! What do we expect!?
Whether & how an ET can make a work! electrochemical cell (battery) Chemical Energy Electrical energy electrolysis Zn + CuSO 4 ZnSO 4 + Cu (T = 25 C) Let s dip solid Zn in CuSO 4 a) Zn bars are consumed b) Metallic copper accumulates c) T rises (DH?)
D on + metallic Cu accumulates electrons loading D off - + hydrated Zn ++ accumulates electrons deprivation
Electrochemical cell (two half-cells) Where : oxidation, half reaction (anode, 1) reduction, half reaction (catode, 2) Zn + CuSO 4 (1) (2) 2 e- / event ZnSO 4 + Cu 2Li + ZnSO 4 2 e- / event Li 2 SO 4 + Zn (1) (2)
Daniel cells (1836)
dry wet - + + - Anode Catode dry convention - + wet convention + -
Cu e Zn : 2 active electrodes
1 active electrode (Zn) & 1 passive (inert) electrode (Pt) Zn + CuSO 4 (1) 2 e- / event ZnSO 4 + Cu (2)
Inert (highly conductive) electrodes Ni Ni 2+ + 2 e - Cl 2 + 2e - 2Cl - e - Pt e - Pt Fe 3+ Fe 2+ Fe 2+ Fe 3+ Half-reaction of Fe 3+ reduction Half-reaction of Fe 2+ oxidation Fe 3+ Cl - 3 + 1 e - Fe 2+ Cl - 2 + HCl Who/what is going to tell us whether a redox-active compound will act as reductant of Fe 3+ or oxidant of Fe 2+?
The two reactions involving hydrogen! e- Pt e- Pt H 2 2H + H2 2H + + 2e - H 2 2 H + 2H + can be reduced to H 2 H 2 can be oxidised to 2H +
The standard potential E This half-reaction is: Easily reproducible Experimentally reliable and May procede in both directions - depending on the (second) half-reaction Hydrogen Half-cell H +, H 2 Pt 2H + + 2e - H 2 1 atm H + Cl - 1M anode cathode
Battery, electromotive force (emf) = DE = E o (more positive) - E o (more negative) e- e- anode cathode Zn Cu ZnSO 4 CuSO 4 DE 1.1 Volts = 0.34 - (- 0.76) sn Zn Zn ++ oxidation E 0 = - 0.76 V dx Cu ++ Cu reduction E 0 = 0.34 V Conventions Pt e- e- Pt anode Pt Fe 2+, Fe 3+ cathode Cl 2, Cl - Pt Fe 2+, Fe 3+ Cl 2, Cl- oxidation E 0 = + 0.77 V reduction E 0 = + 1.36 V DE 0.59 Volts = 1.36 - (+ 0.77)
DE = electromotive force (emf) of a battery DE = E (more positive) - E (less positive/more negative) Ex. Electrochemical cell copper/zinc (Cu) E 0 = +0.34 V (Zn) E 0 = - 0.76 V DE = fem = 0.34 (- 0.76) = 1.1 V
DE of some electrochemical cells (batteries) Overall reaction Mg + Cu ++ Mg ++ + Cu anode cathode Mg Mg ++ Cu ++ Cu DE = 2.7 V if [Mg ++ ] = [Cu ++ ] = 1 M (T = 25 C) standard conditions Overall reaction Zn + 2 Ag + Zn ++ + 2Ag anode cathode Zn Zn ++ Ag + Ag DE = 1.56 V If [Zn ++ ] = [Ag + ] = 1 M (T = 25 C) standard conditions
Nernst equation : Evaluating the electrode potential (half-cell) experimentally by making use of the standard hydrogen half-cell Ox + e - Red R= 8.314472 J K -1 mol -1 T = 298 K F = 9.6485309*10 4 C mol -1 ln = log 10 x 2.303 E 0 potential depends on: 1) chemical nature of molecule 2) activity(a) species concentration [ ] (Le Chatelier) 3) Temperature if [ox] = [red] = 1M, T = 298 K then E = E 0 = standard potential
Example Half cell Nernst equation Cu 2+ + Zn Cu + Zn 2+ 0.059 [Cu 2+ ] E cu = E 0 Cu + ------- log -------- 2 [Cu] 0.059 [Zn 2+ ] E Zn = E 0 Zn+ ------- log -------- 2 [Zn] Electrochemical cell Nernst equation RT [Zn] [Cu 2+ ] DE = DE 0 + ----- ln ---------------- nf [Cu] [Zn 2+ ] 0.059 [Zn] [Cu 2+ ] DE = DE 0 + ------- log ---------------- n [Cu] [Zn 2+ ] E cell = E 0 cell - (RT/nF)lnQ
Concentration cells left Cu Cu 2+ Cu 2+ Cu right [a] left >> [a] right Electrochemical concentration cell - Nernst equation 0.059 [Cu 2+ ] left DE = ------- log ------------- n [Cu 2+ ] right
Alcaline battery (common, manganese-zinc) KOH electrolyte paste anode cathode Zn Zn ++ Mn ++, Mn graphite E 0= - 0.76 V E 0= + 1.51 V DE = 1.51 - (- 0.76) = 2.27 V
Another alcaline! Zn ++ + 2 e- Zn E = - 0.76 V Hg ++ + 2 e- Hg E = + 0.85 V anode cathode Zn Zn ++ Hg ++ Hg electrolyte DE = 0.85 - (- 0.76) = 1.56 V,
Pacemaker Li/Carbon monofluoride Li/(CF) x Li (anode) (CF) x (cathode) anode cathode Li Li + (CF) x + (CF) x DE = 2 Volts (5-7 years life)
The Respiratory Electron Transport Chain NADH E h,7 = -0.28 V e - O 2 E h,7 = +0.78 V ~ 100 kj 24 Kcal (1 mol of e - transferred) DE = 0.78 - (- 0.28) = 1.06 V
nicotinammide NADH Nucleotide 1 2 e -, 2 H + Nucleotide 2 adenine
Nicotinic acid nicotinamide reduction oxidation (standard potential) E = 0.28 V (NAD + /NADH)
emf of respiratory chain (mitochondria) anode cathode E 0 = - 0.3 V Pt [NADH],[NAD + ] [H +], H 2 Pt E 0 (approx.) anode cathode E 0 = 0.8 V Pt [H 2 ], [H + ] [O 2 ], [H 2 O] Pt DE = fem = 0.8 - (- 0.3) 1.1 V
Termodynamics of a redox reaction DG 0 = -RTlnK DG = nfde DG = L DG = DG 0 + RTlnQ DG = -RTlnK + RTlnQ -RTlnK + RTlnQ = nfde solving forde RT RT DE = - ------ lnk + ------ lnq (for Q = 1) nf nf if Q K?? E 0
Free Energy and equilibrium [C] c [D] d K eq =------------ [A] a [B] b aa + bb on off cc + dd DG related to K eq (Van t Hoff equation) DG = (-RT lnk eq + RT ln Q) = RT ln Q/K eq for Q Keq..? Paolo Sarti 2003 Dip. Scienze Biochimiche La DG = 0
The Respiratory Chain Yoshida et al. (2001)