# Full steps of getting Nernst's equation?

##### 1 Answer

We can begin from:

#DeltaG = -nFE_"cell"# #" "" "" "" "bb((1))#

#DeltaG = DeltaG^@ + RTlnQ# #" "" "bb((2))# where

#n# is the mols of electrons transferred,#F = "96485 C/mol e"^(-)# is Faraday's constant, and#E_"cell"# is the cell potential.

#R# and#T# are known from the ideal gas law, and#Q# is the reaction quotient (not-yet-equilibrium).

#DeltaG# is the change in Gibbs' free energy in units of#"J/mol"# .

Actually, we could begin further back if you wish to derive

From

#-nFE_"cell" = -nFE_"cell"^@ + RTlnQ#

Divide through by

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

A neat trick is that

#E_"cell" = E_"cell"^@ - (2.303RT)/(nF)logQ#

Using a standard temperature of

#(2.303("8.314472 J/mol"cdotcancel"K")(298.15 cancel"K"))/("96485 C/mol e"^(-))#

#= 0.0591_(596cdots) ~~ "0.0592 V"cdot"mol e"^(-)"/mol"# since

#"mols substance"# is not necessarily equal to#"mol e"^(-)# .

Therefore, since the units of

#E_"cell" = E_"cell"^@ - ("0.0592 V/mol")/(n)logQ#

Since

#color(blue)(E_"cell" = E_"cell"^@ - ("0.0592 V")/(n)logQ)#

**APPENDIX**

**Derivation of**

You may or may not need to know this, but the full derivation is below.

Consider the chemical potential

In a *non-equilibrium* situation for the general reaction

#nu_A A + nu_B B rightleftharpoons nu_C C + nu_D D# ,

we can define:

#sum_j nu_jmu_j = (nu_C mu_C + nu_D mu_D) - (nu_A mu_A + nu_B mu_B)# #" "" "" "bb((1))#

#-= Deltamu = DeltabarG#

In general, for a one-component system, the chemical potential can be written in reference to some standard state as a function of temperature:

#mu_j(T,P) = mu_j^@(T) + RTln(([j])/([j]^@))# #" "" "" "" "" "bb((2))# where

#[j]# is the concentration of component#j# in#"M"# , and#[j]^@# is the standard concentration of#j# ,#"1 M"# .

We can multiply through by the stoichiometric coefficient for a particular component

#nu_j mu_j(T,P) = nu_j mu_j^@(T) + RTnu_j ln(([j])/([j]^@))# #" "" "" "bb((3))#

Into

#Deltamu = nu_Cmu_C^@ + RTnu_Cln(([C])/([C]^@)) + nu_Dmu_D^@ + RTnu_Dln(([D])/([D]^@)) - nu_Amu_A^@ - RTnu_Aln(([A])/([A]^@)) - nu_Bmu_B^@ - RTnu_Bln(([B])/([B]^@))#

Gather your terms together to get:

#Deltamu = (nu_Cmu_C^@ + nu_Dmu_D^@) - (nu_Amu_A^@ + nu_Bmu_B^@) + RTnu_Cln(([C])/([C]^@)) + RTnu_Dln(([D])/([D]^@)) - RTnu_Aln(([A])/([A]^@)) - RTnu_Bln(([B])/([B]^@))#

But recall from

#Deltamu = Deltamu^@ + RTnu_Cln(([C])/([C]^@)) + RTnu_Dln(([D])/([D]^@)) - RTnu_Aln(([A])/([A]^@)) - RTnu_Bln(([B])/([B]^@))#

Then, to simplify this, we note that

Therefore, after grouping terms:

#Deltamu = Deltamu^@ + RT(nu_Cln[C] + nu_Dln[D]) - RT(nu_Aln[A] + nu_Bln[B])#

Next, use the property that

#Deltamu = Deltamu^@ + RT(ln[C]^(nu_C) + ln[D]^(nu_D)) - RT(ln[A]^(nu_A) + ln[B]^(nu_B))#

Then, use the property that

#Deltamu = Deltamu^@ + RTln(([C]^(nu_C)[D]^(nu_D))/([A]^(nu_A)[B]^(nu_B)))#

We recognize that the

#Deltamu = Deltamu^@ + RTlnQ#

Or, in terms of the Gibbs' free energy,

#color(blue)(DeltabarG = DeltabarG^@ + RTlnQ)#

In general chemistry, I have not seen the usage of

#DeltaG = DeltaG^@ + RTlnQ#