Question

Full steps of getting Nernst's equation?

Answer 1

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)`

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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`