Diagram for the Electrochemical Stability of Water

Abstract

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 differenceΔpH =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 SHEHydroxide/Oxide Pb/Pb2+ -0.126 Pb(OH)2 Cu/Cu 2+ +0.34Cu(OH)2 Co/Co 3+ +0.4 (+0.46)Co(OH)3 Ru/Ru 2+ +0.45Ru(OH)2 Cu/Cu+ +0.52Cu2OTe/Te 4+ +0.568TeO2 Hg/Hg2 2+ +0.789Hg2OAg/Ag+ +0.799Ag2O 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, presented613.6 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).Reference1.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