[PDF] Oxidation/Reduction Limits for H2O

Oxidation/Reduction Limits for H2O

Oxidation/Reduction Limits for H2O Consider the Oxidation of H2O to yield O2(g), the half reaction can be written as; 2 H2O === O2(g) + 4 H + + 4 e-Eo = -1 23 V (from tables)

Les réactions d’oxydoréduction

l’oxydant de l’autre couple gagne des électrons - La réduction est la demi équation relative à un couple : elle correspond à un gain d’électrons L’oxydation est la demi équation relative à l’autre couple : elle correspond à une perte d’électrons - Si la réaction Ox 1 Red 2 o Red 1 Ox 2 est spontanée, la réaction Ox 2 Red

Méthode - pagesperso-orangefr

et l’on constate que le couple H2O/H2 n’est en définitive pas autre chose que le couple H+/H 2 Potentiel redox Rappels Le potentiel rédox est une grandeur, mesurée en Volts, associée à chaque couple oxydant/réducteur Il permet de comparer les pouvoirs oxydant et réducteur de différents couples : plus le potentiel redox

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H2O/H2 Cd2+/Cd Al3+/Al Highest Electronegative Redox Couple E0 = -3 045 V /NHE High Specific Capacity : 3860 mAh/g Li is the only alkaline metal that can be handle easily (small precautions) but Lithium density: 0 541 Why Lithium Chemistry is such Interesting for Electrochemical Energy Storage Standard Electrode Potential / V

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Tout couple redox oxyde un couple redox de potentiel inférieur Exemple : Que se passe-t-il si on plonge une lame de cadmium métallique dans une solution de sulfate de cuivre ? E Volt et E Volt Cu 2 Cu Cd 2 Cd 0 0 34 0 0 40 + = + = − / / • Le couple du cuivre est oxydant vis à vis de celui du cadmium La réaction susceptible de se

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There are two redox systems e g For 2H 2O →2H 2 + O 2 (the most important) one reacting with the holes at the surface of the semiconductor electrode ( at the anode) 4OH-+ 4h+→O 2 + 2H 2O , E o O2/OH-= 0 401 V NHE the second reacting with the electrons entering the counter-electrode (at the cathode) 4H 2O+ 4e-→2H 2 + 4OH-, Eo H2O/H2 = -0

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Ecrire la demi-´equation redox reliant les deux esp`eces du´ couple (Cu(CN)2− 3 /Cu) 2) On cherche a d´eterminer le potentiel standard E 2 du couple (Cu(CN)2− 3 /Cu) 2 a) Etablir la relation de´ Nernst pour le couple (Cu+/Cu) de potentiel standard E 1 = 0,52 V 2 b) Etablir la relation de´ Nernst pour le couple (Cu(CN)2−

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Oxidation/Reduction Limits for H

2 O

Consider the Oxidation of H

2

O to yield O

2 (g), the half reaction can be written as; 2 H 2

O === O

2 (g) + 4 H + 4 e E o = -1.23 V (from tables) Re-writing this as a reduction (by convention) and dividing by 4 (for convenience) yields;

¼ O

2 (g) + H + e ==== ½ H 2 O E o = 1.23 V (note the sign change in E o , but the magnitude remains unchanged) Writing an equilibrium expression for the half reaction yields; 1--1-1/4- O2 -1/4 O2eq }{e }{H P }{e }{H P1 K

Isolating {e

-1 yields; {e -1 = K eq P O2

1/4 {H

and taking log of both sides yields; log {e -1 = log K eq + log P O2 1/4 + log {H

Defining p

e = -log {e } and pH = -log {H } yields; pe = log K eq + log P O2 -1/4 - pH

At the boundary of chemical oxidation of H

2

O, the P

O2 = 1 atm and so; p e = log K eq - pH

Since,

RTG eq o K 3.2 10 RTGK o eq

3.2log

and G o = - nFE o , then

RTnFEK

o eq

3.2log

and pHRTnFEpe o 3.2

In the present case, where n = 1 and Eo

= 1.23 V, the dependence of pe on pH is given by; pe = 20.8 - pH for the boundary for the oxidation of H 2

O to O

2 where F = 96,485 C/mol, R = 8.314 J/mol K, T = 298 K and 1 CV = J/mol.

When H

is at standard state (i.e., 1 M, pH = 0) at 25 o

C, then

8.203.2RTnFEpepe

o o(for the oxidation of H 2

O to O

2

In general;

0591.03.2log

oo eq o E

RTGKpe

, and

0.0591E pe

for any one electron process at 25 o C. water redox boundaries.doc

Consider the Reduction of H

2

O to yield H

2 (g), the half reaction can be written as; 2 H 2

O + 2

e === H 2 (g) + 2 OH E o = - 0.827 V (from tables)

Re-writing this reduction dividing

by 2 (for convenience) yields; 1/2 H 2 O + e === ½ H 2 (g) + OH E o = - 0.827 V (note the magnitude of E o remains unchanged) Writing an equilibrium expression for the half reaction yields; K eq = P H2 1/2 {OH e -1

Isolating {e

-1 and taking log of both sides yields; log { e -1 = log K eq + log P H2 -1/2 + log {OH -1

Defining p

e = -log {e } and pOH = -log {OH } yields; p e = log K eq + log P O2 -1/2 + pOH

At the boundary of chemical reduction of H

2

O, the P

H2 = 1 atm and so; p e = log K eq + pOH

Since pK

w = pOH + pH, we can substitute pOH = pK wquotesdbs_dbs11.pdfusesText_17