Oxidation & Reduction II. Suggested reading: Chapter 5

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1 Lecture 1 Oxidation & Reduction II Suggested reading: Chapter 5

2 Recall from Last time: Redox Potentials The Nernst equation: E cell E 0 RT F ln Q

3 Cell Potential and ph For the H + /H couple at 1 bar and 5 o C: E cell RT ln10 F ph mV ph In general, for aox+ν - + e e +ν H H a Red ared E cell E ' H RT F e ln 10 ph E ' H e (0.059V) 059V) ph E ' E 0 RT [Red] ln F [Ox] e The potential decreases (becoming more negative) as the ph increases and the solution becomes more basic a' a

4 Redox stabilities When assessing the thermodynamic stability of a species in solution, we must consider all possible reactants: the species itself the solvent other solutes dissolved O

5 Redox Chemistry of Water What is in pure water? H ( aq), OH ( aq), H O( l) Water can act as an oxidizing agent, when it is reduced to H : 1 HO(l) e H (g) OH (aq) For the equivalent reduction of hydronium ions in water: 1 H (aq) e H(g) E H (59mV)pH Water s oxidation power increases with decreasing ph. / H

6 Redox Chemistry of Water Water can also act as a reducing agent, when it is oxidized to O : O (g) 4H (aq) 4e H O(l) When the partial pressure of O is 1 bar, the Nernst equation gives: E E 0 RT 4FF ln 1 [H ] 4 1.3V 0.059V ph

7 Oxidation by Water For metals with large, negative standard potentials, reaction with aqueous acid will occur Oxidation of the metal by water or hydrogen ions The overall reaction occurs via one of the following two processes: 1 M(s) HO(l) M (aq) H(g) OH 1 M(s) H (aq) M (aq) H(g) (aq) These reactions are thermodynamically favorable when M is an s-block metal of a 3d series metal (i.e., Ti, Cr, Al, Ni, etc) 3 Sc(s) 6H (aq) M (aq) 3H (g)

8 Oxidation by atmospheric oxygen Reactions of Al, Ti metal with moist air are spontaneous, but they can be used for many years in the presence of water and oxygen. Why?: h Their surfaces become passivated by an oxide film ie: aluminum oxide, titanium oxide, also oxides of Cu, Fe, Zn Copper carbonate on the roofs of the Chateau Frontenac in Quebec

9 Cu Oxidation Rd Redox half reactions: O (g) - 4H (aq) (q) 4e HO(l) () E 1.3V 0.059V ph Cu (aq) e - Cu(s) E 0.34V Full Reaction: Cu(s) O(g) 4H (aq) Cu (aq) HO(l) E 0.89V 0.059V ph Since E>0 in neutral and acidic environments atmospheric oxidation Since E>0 in neutral and acidic environments, atmospheric oxidation is spontaneous (E=0.48V at ph=7)

10 Reduction by water The positive potential of the O, H + /H couple is large: E 1.3V 0.059V ph Acidified water is a poor reducing agent (i.e., a poor electron supplier) except toward even stronger oxidizing i agents 3 4Co (aq) HO(l) 4Co O(g) 4H E 1.9V V (Lowering acidity favors the reaction) (aq)

11 Standard potentials at 98K Only a few redox couples have standard potentials greater than greater than 1.3V.

12 Water splitting & Photosynthesis HO(l) e H(g) O(g) Easy with metals HO(l) e H(g) OH (aq) 4OH O(g) H O(l) 4e Hard in general: barrier is kinetic transfer 4 electrons, form O=O

13 Photosynthesis Nature uses a metal-oxide cluster containing 4 Mn atoms and 1 Ca atom that is located in photosystem II for O evolution The Porphyrin Porphine Chlorophyll-a (located in giant proteins known as light harvesting antennae

Photosynthesis A light harvesting complex hv P* CO

+ P e - P+ Charged reaction center chlorophyll,

14 Photosynthesis A light harvesting complex hv P* CO CH O e - (in 1 ps) Reduction of CO to carbohydrates via reduction of NADP+ to NADPH In PSII, this electron comes from oxidation of water into O and H + P e - P+ Charged reaction center chlorophyll, reduced back to it s ground state by accepting an electron

15 O evolution catalyst Various hypotheses for O evolution: 1) As Mn sites are progressively oxidized, d coordinated d H O molecules l become increasingly polarized and lose protons H OOH - O - ) Since Mn(IV) and Mn(V)=O are equivalent to Mn(II)-[O] and Mn(III)-[O], if two [O] atoms are close together, and O molecule may form 3) Presence of Ca + is essential: remains in + state and provides a fast binding site for H O

16 Dis & co-proportionation Disproportionation A redox reaction in which the oxidation number of element is simultaneously raised and lowered. Example: Cu+ a) b) : Cu a ) : Cu b) : Cu E 0 ( aq) ( aq ) e ( aq) e Cu V ( aq) Cu( s) Cu ( s ), 0.5 V Cu ( aq), 0.16V Coproportionation The reverse of disproportionation. Example Fe 3+ and Fe 3 Fe ( aq ) Fe ( s ) 3 Fe ( aq ) Fe Fe E 0 3 ( aq) e ( aq ) e Fe ( aq), 0.77V Fe ( s ), 0.45 V 0.77 ( 0.45) 1.V

17 Redox Reactions in Biology: Neuroscience

18 The Soma Membrane Cl -

19 Re-deriving the Nernst Equation The Nernst equation specifies the voltage that will result from maintenance of a given concentration difference.

20 Flux due to drift and diffusion Diffusion i flux: The net number of charges (ions) crossing some position per unit time per unit area, Γ is: J diff DD dc( x) dx Fick s First Law D = diffusion coefficient of charges = l /τ dc/dx = charge concentration gradient Drift flux: Ions will drift with a velocity proportional to their mobility mu, charge, and the strength of the E field: vdrift qe x J drift qc( x) E x (Assuming single charged ions, so coefficient of q=1)

21 Total Flux J tot qc( x) E x D dc ( x ) dx D kt Einstein relation In equilibrium, J=0 kt E xdx dc C V out E x in dx

22 Membrane Potential V out E in dx x kt q [ln( C( in)) ln( C( out))] V kt q ln C ( in ) C ( in ) 5mV ln C( out) C( out) (Recall: R=N a k, F=qN a ) [K + + in ]:[K out ]=10:1 ]10:1 ΔV=-58mV resting potential of neuron

When a nerve or muscle cell is stimulated by synaptic transmission, the mobility (or channel conductance, or permeability) for sodium

23 Action potentials In the resting state of a nerve cell, sodium mobility across the membrane is much lower than potassium, and the cell maintains a negative voltage. When a nerve or muscle cell is stimulated by synaptic transmission, the mobility (or channel conductance, or permeability) for sodium transiently increases to a value greater than that for potassium and the cell internal voltage "spikes" above zero volts for about a millisecond. tures/electron.htm Clark, Nature Neuroscience 8 (005)

24 Electrode : Solution Interactions Outer Helmholtz plane Cations surrounded by solvent particles Specifically Primary solvent (solvated adsorbed Electrolytic layer (acting as a cations) anions solution dielectric) Metal plane Inner Helmhol l tz plane Metal (with negative charge)

25 Electrical Double Layer Negatively charged particle To derive the potential ψ, start with Poisson s equation: / ρ: charge density = ρ(x) Boundary conditions: slipping plane ψ Potential, ψ(0)=ψ o ψ( )=0

26 Electrical Double Layer Assume the ions obey a Boltzmann distribution ib i (probability bili an ion having a local potential ψ ): #ions/vol n i z q / kt n i e i valence i z i qn i q i d q ziq / zinie dx kt Poisson-Boltzman equation i

27 Electrical Double Layer: Huckel approximation Poisson-Boltzman equation has no general solution. kt z i Assume: q kt ~ 5.7mV T=300K q (Low surface potential approximation) Expand exponential into a power series d q dx i z i n i (1 ziq ) kt At infinity, electrical neutrality all charges cancel0

28 Electrical Double Layer: Huckel approximation d dx d dx 1 i kt z i q n kt q i z i n i i e o x e κ: Double layer thickness (typically y 1-3 nm) Beyond the Huckle approximation: Gouy-Chapman-Stern: o 8kTn i zq sinh kt

Ch 20 Electrochemistry: the study of the relationships between electricity and chemical reactions.

Ch 20 Electrochemistry: the study of the relationships between electricity and chemical reactions. In electrochemical reactions, electrons are transferred from one species to another. Learning goals and