How is the pH of a solution related to redox potential?

Question

Why would the pH i.e the hydrogen ion concentration of a solution affect its redox potential? Similarly why would the compound concentration (the concentration of the chemical species itself) affect the redox potential?

Answer 1

The standard redox potential is not affected by pH or concentration, because it is defined under standard conditions.

If you have a redox reaction which does not involve hydrogen ions or hydroxide ions, the redox potential of the solution does not depend on pH. Example,

I2 + 2e- = 2I-

If the redox half-equation does involves these, the potential will depend on pH. Example,

H2O2 + 2H+ + 2e- = 2 H2O

As to the effect of compound concentration, this does not affect the standard potential, but does affect the potential under the particular circumstances that you are investigating. The higher the concentration of the material, the greater its tendency to react, so that a more concentrated solution of an oxidising agent will be more powerfully oxidising than it would if the same solution were diluted.

The difference between redox potential under actual conditions, and standard redox potential, is very like the difference between free energy of reaction under actual conditions and standard free energy.

Hope this helps!

Answer 2

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You're almost there. The missing information is that the potential of a redox reaction depends on the concentration of a solution. As the solution becomes more dilute, the reduction potential decreases. For a common example, think of a battery. As the concentration of the two halves of the cell decreases, the battery loses juice. Eventually it dies.

Answer 3

There is a famous Nernst equation which relates concentrations of reagents and products of a redox half reaction to its electrode potential:

E=Eo-RT/(nF)Ln([Red]/[Ox]), where Eo-standard electrode ponetial for a given half reaction, n - number of electrons transfered during the reaction, F - Faraday constant, [Red] concentrations of species produced after reduction, [Ox] - concentration of species participated in oxidation.

Example:
MnO4(-) + 8H(+) +5e --> Mn(2+) + 4H2O
Eo(H2O2/H2O)=1.51eV

E=Eo-(RT/5F)*Ln ([Mn(2+)]/[MnO4(-)]]/[H+]^8)
(Water does not count in aqueous solutions. )

from here it follows:

E=Eo-(RT/5F)*Ln ([Mn(2+)]/[MnO4-]) + (RT/5F)*8*Ln [H+]

But Ln[H+] = 2.303Log[H+] = -2.303*pH

Consequently,

E=Eo-(RT/5F)*Ln ([Mn2+]/[MnO4-])-(8RT/5F)*pH

You can see that now you can actually calculate standard electrode potential of the half reaction at different pH values. Because of strong dependence of electrode potential on pH, MnO4- is very strong oxidant in acidic medium, but mild in basic solutions.