Diagram for the Electrochemical Stability of Water
Authors: Chernomorskii Alexandr I. 
Published: September 2023

According to the electrochemical stability diagram of water (Fig. 1), Ox/Red pairs displace hydrogen and oxygen respectively, when equilibrium potentials
E
Ox/Red <
E
H+/H2 (line
a
) and
E
Ox/Red >
E
O2/OH- (line
b
). However, if this is theoretically possible, the fact is that only pairs with
E
Ox/Red > ~ +0.5 V SHE (
E

o
I2/I- = + 0.535 V SHE;
E
o
NiO2/Ni(OH)2,OH- = + 0.49 V SHE) displace oxygen from alkaline solutions and only pairs with
E
Ox/Red < ~ +0.5 V SHE (
E
o
Cu/Cu2+ = +0.34 V SHE;
E
o
Co/Co3+
= +0.4 V SHE) can displace hydrogen from acid solutions. There will be no displacement of oxygen by ions Cu2+ (Ox-form) and hydrogen by ions I- (Red-form) at the appropriate pH (Fig.1) [1,2].
The reason for this is that Red-forms (I-ions) with
E
Red/Ox > ~+0.5 V SHE and Ox-forms (Cu2+, Co3+ ions) with
E
Red/Ox < ~+0.5 V SHE is already thermodynamically stable in solutions, their transitions to conjugated forms (I2, Cu) are not spontaneous. With the theoretical possibility of H2 displacement, ions I-will form acids (HI) with H+ ions. With the theoretical possibility of O2 displacement, ions Cu2+, Co3+ will form hydroxides (Cu(OH)2) with OH- ions. These reactions are more thermodynamically probable. Accordingly, electrodes displace either hydrogen or oxygen from water.
Between pairs with low oxidizing and reducing activities, there should be a pair of zero activity, in which Ox + ne and Red-forms are chemically equilibrium (line
c
, Fig.1) [1-3]. Its average potential is
E
Red/Ox ~ +0.44 V SHE between
E
o
I2/I- = + 0.535 V SHE and
E
o
Cu/Cu2+ = +0.34 V SHE and
E
Red/Ox ~ +0.47 V SHE between nearest
E
o
NiO2/Ni(OH)2,OH- = +0.49 V SHE and
E
o
Ru/Ru3+ = +0.45 V SHE determined opposite activities [1,2].
The intersection of line
c
with lines
a
(pH = -8; point
d
) and
b
(pH=12.8; point e) should also correspond to zero activity of H+/H2 - and O2/OH- - electrodes (Fig.1). The difference
ΔpH =12.8 – (-8) = 20.8 determines emf of H+/H2//O2/OH- element (20.8 x 0.0 59 = 1.227 V) [1].
The instability of metal ions reflects itself in the decomposition of hydroxides 2CuOH→ Cu2O + H2O; 2AuOH → Au2O +H2O. Unstable metal ions (Cu+, Te4+, Ag+, Au+) can polarize hydroxyl ions without the formation of the pair OH-/O2[2]. Indeed, the boundary at
E

Red/Ox ~+0.47 V SHE separates stable and unstable hydroxides (Table).

Me/Men+

E
o, V SHE
Hydroxide/Oxide

Pb/Pb2+

-0.126

Pb(OH)2

Cu/Cu
2+

+0.34
Cu(OH)2

Co/Co 3+

+0.4 (+0.46)
Co(OH)3

Ru/Ru
2+

+0.45
Ru(OH)2

Cu/Cu+

+0.52
Cu2O

Te/Te
4+

+0.568
TeO2

Hg/Hg2
2+

+0.789
Hg2O

Ag/Ag+

+0.799
Ag2O

Absolute values

|-E
H+/H2
| = |+E
O2/OH-
| = 1.227 V/ 2 = ~ 613.6 mV (10.4 x 59 mV) are equal at pH = 10.4 (line
f
, Fig.1), where 1.227 V is the distance between lines
a
and
b
(emf H2/H+//OH-/O2 element).
Line
f
divides the horizontal distances (as
g - i
) from pH = 0 to pH = 14 (14 x 59 mV= 826 mV) into parts 1227 mV / 2 = 613.6 mV (
g - h
) and 212.4 mV (3.6 pH x 59 mV;
h - i
). Their ratio is 613.6 mV / 212.4 mV = ~ 2.89. Thus, presented
613.6 mV = (212.4 mV + 0.89 x 212.4 mV) + 212.4 mV =
= 401.2 mV + 212.4 mV = 2.89 x
212.4 mV =
=2.89 (ψ O2/OH-, pH=14 + ψ O2/OH-,pH=10.4)
where ψ O2/OH-, pH=14 + ψ O2/OH-,pH=10.4 are the sum of external potential drops (not Galvani potentials) arising in half-reactions between oppositely charged particles in double-electrical layers on electrodes.
Both sides are equal to 212.4 mV. Therefore, dividing 212.4 mV (left) / 2.89 = 73.5 mV should give the smallest actual component ψ O2/OH-,pH=14of equal 212.4 mV (right).
Then, its second component ψ O2/OH-,pH=10.4 is 401.2 mV (left) / 2.89 =138.8 mV or 212.4 mV - 73.5 mV = 138.9 mV. Close ψ O2/OH-, pH=14 (14 - 12.8) x 59 mV = 70.8 mV and ψ O2/OH-, pH=10.4 - (12.8 - 10.4) x 59 mV =141.6 mV calculated relative to pH=12.8 (point
e
, Fig.1) [1].
The potential of pairs of zero activity (line
c
, Fig.1) is calculated as
E
O2/OH-,pH=10.4 - ψ O2/OH-,pH=10.4 = 613.6 mV SHE - 138.8 mV = + 474.8 mV SHE. The potential drops

ψ H
+/H2,pH=0 = - 474.5 mV; ψ O2/OH-,pH=0 = + 752.4 mV and ψ H
+/H2= ψ O2/OH- at pH = 2.4 are calculated (Fig.1).
Reference
1.A.I.Chernomorskii, Zh. Fiz. Khim., 51,924 (1977).
2.A.I.Chernomorskii, Dokl. Akad.Nauk Uzb.SSR, 3,37(1989).
3.A.I.Chernomorskii, Journal of The Electrochemical Society,
2021,168,116514.

Figure 1