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)
Re-writing this as a reduction (by convention) and dividing by 4 (for convenience) yields;
Eo = 1.23 V
¼ O2(g) + H+ + e- ==== ½ H2O
(note the sign change in Eo, but the magnitude remains unchanged)
Writing an equilibrium expression for the half reaction yields;
1
-1/4
K eq = 1/4 +
= PO2 {H + }-1 {e - }-1
PO2 {H } {e }
Isolating {e-}-1 yields; {e-}-1 = Keq PO21/4 {H+}
and taking log of both sides yields; log {e-}-1 = log Keq + log PO21/4 + log {H+}
Defining pe = -log {e-} and pH = -log {H+} yields;
pe = log Keq + log PO2-1/4 – pH
At the boundary of chemical oxidation of H2O, the PO2 = 1 atm and so;
pe = log Keq – pH
Since, K eq = 10
log K eq =
pe =
−
ΔG o
2.3 RT
, log K eq =
− ΔG o
and ΔGo = - nFEo, then
2.3RT
nFE o
and
2.3RT
nFE o
− pH
2.3RT
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 H2O to O2
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 25oC, then
nFE o
pe = pe o =
= 20.8 (for the oxidation of H2O to O2)
2.3RT
− ΔG o
Eo
E
In general; pe o = log K eq =
=
, and pe =
for any one electron
2.3RT 0.0591
0.0591
process at 25oC.
water redox boundaries.doc
Consider the Reduction of H2O to yield H2(g), the half reaction can be written as;
2 H2O + 2 e- ===
H2(g) + 2 OH-
Eo = - 0.827 V (from tables)
Re-writing this reduction dividing by 2 (for convenience) yields;
1/2 H2O + e- === ½ H2(g) + OH(note the magnitude of Eoremains unchanged)
Eo = - 0.827 V
Writing an equilibrium expression for the half reaction yields;
Keq = PH21/2 {OH-} {e-}-1
Isolating {e-}-1 and taking log of both sides yields;
log {e-}-1 = log Keq + log PH2-1/2 + log {OH-}-1
Defining pe = -log {e-} and pOH = -log {OH-} yields;
pe = log Keq + log PO2-1/2 + pOH
At the boundary of chemical reduction of H2O, the PH2 = 1 atm and so;
pe = log Keq + pOH
Since pKw = pOH + pH, we can substitute pOH = pKw – pH thus;
pe = log Keq + pKw - pH
As before,
log K eq
pe =
nFE o
=
and
2.3RT
nFE o
+ pK w − pH
2.3RT
In the present case, where n = 1, Eo = - 0.827 V and pKw = 14, the dependence of pe on
pH is given by;
(1)(96485)(−0.827)
pe =
+ 14 − pH = − pH (note that the units in nFEo/2.3RT cancel)
2.3(8.314)(298)
When H+ is at standard state (i.e., 1 M, pH = 0) at 25o, then pe = peo = 0 for the reduction
of H2O to H2.
At other {H+}, the boundary between H2O and H2 is given by; pe = - pH
water redox boundaries.doc