# Why is reduction potential of water ( 2H₂O + 2e⁻ → H₂ + 2OH⁻ ) is -0.83 insted of 0, as in the case of (2H⁺ → H₂)?

I came across this value of reduction potential of water while solving a question. Also, I noticed that, for example

$$\ce{Ag+ → Ag}$$

has a fixed standard potential, and the same is used when we do it for a reaction like

$$\ce{AgCl → Ag + Cl-}$$

Then why here in the reduction of water the value differs from the $$\ce{H+ → 1/2 H2}$$ reaction?

Also, while doing electrolysis of water, which reaction actually happens at the cathode:

$$\ce{2 H2O + 2 e- → H2 + 2 OH-}\quad\text{or}\quad\ce{2 H+ → H2}~?$$

• The potential of $\ce{Ag+ -> Ag}$ is not actually fixed. It depends on the concentration of $\ce{Ag^+}$, and the form of this dependency is well known. Same thing here. – Ivan Neretin Apr 16 '19 at 5:24
• – Loong Apr 16 '19 at 9:49
• $$\ce{Ag+(aq) + e- ->Ag(s)};\ E^\circ_{red} = \pu{0.80 V}$$ $$\ce{AgCl(s) + e- ->Ag(s) + Cl-(aq)};\ E^\circ_{red} =\pu{ 0.22 V}$$ – Karsten Theis Apr 16 '19 at 13:51

Note that the actual potential for a particular redox reaction is not a fixed value, but depends on concentrations ( more exactly activities ) of reagents. The standard redox potentials are potentials with activities equal to 1,

If we consider reactions

\begin{align}\ce{ 2 H+ + 2e- &<=> H2 \\ 2 H2O + 2 e- &<=> 2 OH- + H2 \\ }\end{align}

their standard redox potentials imply $$\rm{p}H=0$$ for the former, $$\rm{p}H=14$$ for the latter.

For the former reaction and $$\rm{p}H=14$$, $$E=-0.83\ \rm{V}$$. As for $$\ce{H+/H2}$$ potential, at $$\pu{p_{\ce{H_2}} = 1 atm}$$

$$E_\ce{H2\ /H+} = E^{\circ}_\ce{H2\ /H+} + 0.059 \log [\ce{H+}] = E^{\circ} - 0.059\ \rm{p}H$$

Thermodynamically, the equations they are equivalent, but kinetically, the latter reaction has in strongly alkaline environment the advantage of many concentration orders. And vice versa.

Elements can have multiple standard redox potentials, as such a potential can be expressed for any redox reaction (even if for some only formally from the thermodynamic data).

For example, some standard redox potentials for $$\ce{Ag}$$ are:

$$\begin{array}{lcll} \ce{Ag+ + e- &<=>& Ag(s)};\ & E^\circ_{red} = \pu{0.80 V}\\ \ce{AgCl(s) + e- &<=>& Ag(s) + Cl-(aq)};\ & E^\circ_{red} =\pu{ 0.22 V} \end{array}$$

(There are more; see the Standard electrode potential data table).

The latter can be calculated from the former using $$E_{\ce{Ag/Ag+}} = E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+}$$

For the $$\ce{Ag / AgCl / Cl-}$$ system, we have to include into potential calculations the $$\ce{AgCl}$$ solubility constant.

\begin{align} K_{\rm s, AgCl} &= c_{\ce{Ag+}}\cdot c_{\ce{Cl-}} \\[1em] E_{\ce{Ag/AgCl/Cl-}} &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log c_\ce{Ag+} =\\ &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} - 0.059 \log {c_\ce{Cl-}}= \\ &= E^{\circ}_{\ce{Ag/AgCl/Cl-}} - 0.059 \log {c_\ce{Cl-}} \\ E^{\circ}_{\ce{Ag/AgCl/Cl-}} &= E^{\circ}_{\ce{Ag/Ag+}} + 0.059 \log {K_{\rm s, AgCl}} =\\ &= 0.80 + 0.059\log(1.77\cdot10^{−10}) = \\ & = 0.22 \end{align}

Note that the $$\ce{Ag/AgCl/Cl-}$$ electrode is often used as the secondary standard of the electrode potential, aside of the calomel electrode $$\ce{Hg/Hg2Cl2/Cl-}$$, because maintaining of the primary potential standard electrode $$\ce{Pt-H2/H+}$$ is not convenient. The $$\ce{Ag/AgCl/Cl-}$$ electrode is used also as the electrode to measure $$c_{\ce{Cl-}}$$.

• I'd write the two different half-cell reactions as reductions since that is the standard for tabular values. -- LOL, Now I'm the Standard Police. – MaxW Apr 16 '19 at 5:34
• Equilibrium reactions are neither reductions neither oxidations. They are both. The orientation is purely formal. Similarly, the electrode potentials do not flip signs.:-) – Poutnik Apr 16 '19 at 5:37
• Btw, about flipping potential signs, should not it to be rather flipping equation signs ? As the flipped potential does not have meaning of a measurable potential. – Poutnik Apr 16 '19 at 6:03
• When I switch sides of the reaction. :-) Additionally, the potential of the electrode does not flip the sign with the flipping the multimeter orientation. – Poutnik Apr 16 '19 at 6:47
• I'd love to see your answer stressing even more prominently that -0.83 can be simply calculated via concentration using Nernst equation, and that H+'s half-reaction is no different from any other half reaction in this regard, and that it is OK for an element to have multiple half-reactions (say, there is one for Ag+ and another for AgCl, and they are related in the same manner). Oh, and please include the link: en.wikipedia.org/wiki/Standard_electrode_potential_(data_page) – Ivan Neretin Apr 16 '19 at 7:58

$$\ce{2 H+ + 2 e- <=> H2}\label{rxn:1}\tag{1}$$
$$\ce{2 H2O + 2 e- <=> H2 + 2 OH-}\label{rxn:2}\tag{2}$$
are two different experimental procedures. Hence these are two distinct half cell reactions. It happens that you get $$\ce{H2}$$ as a product in $$\eqref{rxn:1}$$ and $$\eqref{rxn:2}$$.
In $$\eqref{rxn:1}$$ you are reducing hydrogen ions and the experimental set-up is very specific. You have $$\ce{Pt}$$ electrode (or $$\ce{Pt}$$ black) and you bubble hydrogen over it in $$\pu{1 M}$$ $$\ce{H+}$$ ion. It is impossible to determine its absolute potential and we set this half cell to $$\pu{0 V}$$ (by definition). The electrode is being exposed to $$\ce{H+}$$ and $$\ce{H2}$$. This system is called NHE and SHE (normal hydrogen electrode and standard hydrogen electrode, respectively).
In $$\eqref{rxn:2}$$ instead you have $$\ce{H2O}$$ which is being reduced here in alkaline medium in the presence of $$\ce{H2}$$. This reaction is not $$\pu{0 V}$$ with reference to SHE.