consider the following electrochemical cell:

  • silver (I) oxide is anode, platinum electrode is cathode, 6M KOH is electrolyte

what voltage is needed to be supplied to the cell OR what voltage is generated by this electrolytic cell, with the reaction being:

$\ce{2 Ag2O -> 4 Ag + O2}$

where half reactions are:

$\ce{2 Ag2O + 2H2O + 4e- -> 4Ag + 4OH-}$

$\ce{4OH- -> 2 H2O + 4e- + O2}$

Is the reverse reaction spontaneous in alkaline electrolyte with inert electrode like platinum:

$\ce{4 Ag + 4OH- -> 2 Ag2O + 2 H2O + 4e-}$

$\ce{O2 + 2H2O + 4e- -> 4OH-}$

(the platinum electrode is for oxygen reduction reaction and oxygen evolution reaction)

I am a computer science engineer trying to learn chemistry, excuse me if my equations are not precise


1 Answer 1


You need to look up the standard electrode potentials of the materials involved. While wikipedia provides an answer for $\ce{AgO}$ in acidic media, the abstract of this primary publication mentions a value of $(0.599 \pm 0.001)$ V of the $\ce{AgO / Ag2O / OH-}$ electrode, as determined against the H electrode at 25 C in KOH solution. Note the authors find the potential dependent on the grid material their AgO-electrode was made; beside the expected-to-happen changes in function of temperature of the solution and concentration of $\ce{KOH}$ present.

A generally useful secondary literature reference for physico-chemical data you (still) may carry with you is the CRC Handbook of Chemistry and Physics by CRC Press, including tables like this, too.

  • $\begingroup$ thanks! I will lookup the resources.. if the standard potential of AgO electrode is (+0.6V), and the electrode potential for oxygen evolution reaction at the platinum electrode is (-0.4V), can the reaction be spontaneous because the overall cell potential is (+0.2V)? $\endgroup$
    – bsaaa
    Mar 11, 2017 at 15:57
  • $\begingroup$ The standard potentials are determined at 1M concentrations and 1 bar pressures. Some adjustment will need to be accounted for the higher contraction of hydroxide. $\endgroup$
    – Ben Norris
    Mar 12, 2017 at 1:31

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