Full steps of getting Nernst's equation?

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 #bb((2))#, but it's up to your professor where you could start. I'll start with #bb((2))# already known, and provide its derivation at the end.

From #bb((1))# modified for standard and nonstandard conditions, we have, with #bb((2))#:

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

Divide through by #-nF#:

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

A neat trick is that #lnx = 2.302_(59 cdots)logx ~~ 2.303logx#, for any #x#. Therefore, for general chemistry, the Nernst equation could be written as:

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

Using a standard temperature of #"298.15 K"# (allowing concentrations to vary), we can further condense down the coefficient in front of #logQ#. Since a #"J/C"# is a #"V"#,

#(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 #n# are #"mol e"^(-)#, we now have:

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

Since #E_"cell"# does not change value when we scale the reaction by a stoichiometric coefficient (the new factors in front of #"mol substance"# and #"mol e"^(-)# cancel each other out), we tend to just drop the "#"/mol"#" (of what is being reduced or oxidized) to avoid confusion with #"mol e"^(-)#.

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

APPENDIX

Derivation of #bb(DeltaG = DeltaG^@ + RTlnQ)#

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

Consider the chemical potential #mu -= barG = G/n#, where #G/n# is the molar Gibbs' free energy.

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 #j# to get:

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

Into #bb((1))#, if we then plug the definition, #bb((3))#, for each #nu_jmu_j#, we get:

#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 #bb((1))# that we defined #Deltamu#. Apply that for the standard state to get:

#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 #[C]^@ = [D]^@ = [A]^@ = [B]^@ = "1 M"#, and can be considered a means to eliminate units inside a logarithm (the argument of a logarithm must be unitless).

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 #blna = lna^b# to get:

#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 #lna + lnb = ln(ab)# and #lna - lnb = ln(a/b)#:

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

We recognize that the #ln# argument is just the definition of the reaction quotient, #Q#. Therefore:

#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 #DeltabarG# as #(DeltaG)/n#, and the bar is just dropped. The units of #DeltaG# are implied to be #"J/mol"# or #"kJ/mol", and for general chemistry we write:

#DeltaG = DeltaG^@ + RTlnQ#