# What is the equation that connects pH and its effect on electric potential of an electrochemical cell?

##### 1 Answer

Well, first off, you'll need to begin with the Nernst equation:

#E_(cell) = E_(cell)^@ - (RT)/(nF) lnQ# where:

#E_(cell)# is the overall cell potential.#""^@# indicates#"1 atm"# and#25^@ "C"# .#R# and#T# are known from the ideal gas law.#n# is the mols of electrons reportedly transferred in the redox reaction.#F = "96485 C/mol e"^(-)# is Faraday's constant.

Note that

**FROM THE PERSPECTIVE OF POURBAIX DIAGRAMS**

The

This can be seen in **Pourbaix diagrams**:

These are

- Diagonal lines indicate an
#E_(cell)^@# that is#"pH"# -dependent. - Vertical lines indicate
#100%# #"pH"# dependence. - Horizontal lines indicate
#0%# #"pH"# dependence.

As a cursory example, **for a given** *half-reaction* equilibrium in basic media is:

#"Mn"_3"O"_4(s) + 4"H"_2"O"(l) + 2e^(-) rightleftharpoons 3"Mn"("OH")_2(s) + 2"OH"^(-)(aq)#

It shifts towards

**FROM THE PERSPECTIVE OF THE NERNST EQUATION**

Or, using the Nernst equation, we can investigate the **reaction quotient**

#Q = \frac(prod_i^(N_P) [P_i]^(nu_i))(prod_j^(N_R) [R_j]^(nu_j))# ,where:

#prod_k^N# indicates a multiplication of terms#1, 2, . . . , k, . . . , N# .#i# is the index of individual product concentrations#[P_i]# being multiplied together.#j# is the index of individual reactant concentrations#[R_j]# being multiplied together.#nu_k# is the stoichiometric coefficient of the#k# th reactant or product. Note that in general,#i ne j# , meaning the number of products and reactants may differ.

So, we have the following conditions:

As the

#"pH"# decreases, the solution is more acidic, so#10^(-"pH") = ["H"^(+)]# increasesand#10^("pH" - 14) = ["OH"^(-)]# decreases.

- If
#"H"^(+)# is a**product**,#Q# therefore*increases*, and the nonstandard cell potential**decreases**. - If
#"H"^(+)# is a**reactant**,#Q# therefore*decreases*, and the nonstandard cell potential**increases**. - If
#"OH"^(-)# is a**product**,#Q# therefore*decreases*, and the nonstandard cell potential**increases**. - If
#"OH"^(-)# is a**reactant**,#Q# therefore*increases*, and the nonstandard cell potential**decreases**.

This is very situational, so it's not exactly possible to write one single equation for this, unless we make some assumptions.

**PLAUSIBLE EQUATION**

** If** the reaction does have an

#E_(cell) = E_(cell)^@ - (RT)/(nF) ln (Q^"*"cdot["H"^(+)]^(pmnu_("H"^(+))))# where:

#Q^"*"# is the#"pH"# -independent reaction quotient with#["H"^(+)]# separated out.#nu_("H"^(+))# is the stoichiometric coefficient of#"H"^(+)# in the reaction.- In the
#pm# for#nu_("H"^(+))# , the#(+)# indicates#"H"^(+)# product, while the#(-)# indicates#"H"^(+)# reactant.

We could write

#E_(cell) = E_(cell)^@ - (RT)/(nF) ln Q^"*" - (RT)/(nF) ln (["H"^(+)]^(pmnu_("H"^(+))))#

#= E_(cell)^@ - (RT)/(nF) ln Q^"*" - (pmnu_("H"^(+))RT)/(nF) ln (10^(-"pH"))#

#= E_(cell)^@ - (RT)/(nF) ln Q^"*" + "pH" cdot (pmnu_("H"^(+))RT)/(nF) ln 10#

Rearrange this further to get:

#E_(cell) - E_(cell)^@ + (RT)/(nF) ln Q^"*" = "pH" cdot (pmnu_("H"^(+))RT)/(nF) ln 10#

#"pH" = (E_(cell) - E_(cell)^@ + (RT)/(nF) ln Q^"*")/((pmnu_("H"^(+))RT)/(nF) ln 10)#

#= ((nF)/(RT)(E_(cell) - E_(cell)^@) + ln Q^"*")/(pmnu_("H"^(+))ln 10)#

Since

#color(blue)(barul|stackrel(" ")(" ""pH" ~~ ((nF)/(2.303RT)(E_(cell) - E_(cell)^@) + log Q^"*")/(pmnu_("H"^(+)))" ")|)# keeping in mind that

#Q^"*"# neglects the#["H"^(+)]# that would have been in the mass action expression, and that whether it was a product or reactant changes the sign of#nu_("H"^(+))# .