# Full steps of getting Nernst's equation?

May 9, 2017

We can begin from:

$\Delta G = - n F {E}_{\text{cell}}$ $\text{ "" "" "" } \boldsymbol{\left(1\right)}$

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$ $\text{ "" } \boldsymbol{\left(2\right)}$

where $n$ is the mols of electrons transferred, $F = {\text{96485 C/mol e}}^{-}$ is Faraday's constant, and ${E}_{\text{cell}}$ is the cell potential.

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

$\Delta G$ is the change in Gibbs' free energy in units of $\text{J/mol}$.

Actually, we could begin further back if you wish to derive $\boldsymbol{\left(2\right)}$, but it's up to your professor where you could start. I'll start with $\boldsymbol{\left(2\right)}$ already known, and provide its derivation at the end.

From $\boldsymbol{\left(1\right)}$ modified for standard and nonstandard conditions, we have, with $\boldsymbol{\left(2\right)}$:

$- n F {E}_{\text{cell" = -nFE_"cell}}^{\circ} + R T \ln Q$

Divide through by $- n F$:

${E}_{\text{cell" = E_"cell}}^{\circ} - \frac{R T}{n F} \ln Q$

A neat trick is that $\ln x = {2.302}_{59 \cdots} \log x \approx 2.303 \log x$, for any $x$. Therefore, for general chemistry, the Nernst equation could be written as:

${E}_{\text{cell" = E_"cell}}^{\circ} - \frac{2.303 R T}{n F} \log Q$

Using a standard temperature of $\text{298.15 K}$ (allowing concentrations to vary), we can further condense down the coefficient in front of $\log Q$. Since a $\text{J/C}$ is a $\text{V}$,

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

$= {0.0591}_{596 \cdots} \approx \text{0.0592 V"cdot"mol e"^(-)"/mol}$

since $\text{mols substance}$ is not necessarily equal to ${\text{mol e}}^{-}$.

Therefore, since the units of $n$ are ${\text{mol e}}^{-}$, we now have:

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

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

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

## APPENDIX

Derivation of $\boldsymbol{\Delta G = \Delta {G}^{\circ} + R T \ln Q}$

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

Consider the chemical potential $\mu \equiv \overline{G} = \frac{G}{n}$, where $\frac{G}{n}$ is the molar Gibbs' free energy.

In a non-equilibrium situation for the general reaction

${\nu}_{A} A + {\nu}_{B} B r i g h t \le f t h a r p \infty n s {\nu}_{C} C + {\nu}_{D} D$,

we can define:

${\sum}_{j} {\nu}_{j} {\mu}_{j} = \left({\nu}_{C} {\mu}_{C} + {\nu}_{D} {\mu}_{D}\right) - \left({\nu}_{A} {\mu}_{A} + {\nu}_{B} {\mu}_{B}\right)$ $\text{ "" "" } \boldsymbol{\left(1\right)}$

$\equiv \Delta \mu = \Delta \overline{G}$

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} \left(T , P\right) = {\mu}_{j}^{\circ} \left(T\right) + R T \ln \left(\frac{\left[j\right]}{{\left[j\right]}^{\circ}}\right)$$\text{ "" "" "" "" } \boldsymbol{\left(2\right)}$

where $\left[j\right]$ is the concentration of component $j$ in $\text{M}$, and ${\left[j\right]}^{\circ}$ is the standard concentration of $j$, $\text{1 M}$.

We can multiply through by the stoichiometric coefficient for a particular component $j$ to get:

${\nu}_{j} {\mu}_{j} \left(T , P\right) = {\nu}_{j} {\mu}_{j}^{\circ} \left(T\right) + R T {\nu}_{j} \ln \left(\frac{\left[j\right]}{{\left[j\right]}^{\circ}}\right)$$\text{ "" "" } \boldsymbol{\left(3\right)}$

Into $\boldsymbol{\left(1\right)}$, if we then plug the definition, $\boldsymbol{\left(3\right)}$, for each ${\nu}_{j} {\mu}_{j}$, we get:

$\Delta \mu = {\nu}_{C} {\mu}_{C}^{\circ} + R T {\nu}_{C} \ln \left(\frac{\left[C\right]}{{\left[C\right]}^{\circ}}\right) + {\nu}_{D} {\mu}_{D}^{\circ} + R T {\nu}_{D} \ln \left(\frac{\left[D\right]}{{\left[D\right]}^{\circ}}\right) - {\nu}_{A} {\mu}_{A}^{\circ} - R T {\nu}_{A} \ln \left(\frac{\left[A\right]}{{\left[A\right]}^{\circ}}\right) - {\nu}_{B} {\mu}_{B}^{\circ} - R T {\nu}_{B} \ln \left(\frac{\left[B\right]}{{\left[B\right]}^{\circ}}\right)$

Gather your terms together to get:

$\Delta \mu = \left({\nu}_{C} {\mu}_{C}^{\circ} + {\nu}_{D} {\mu}_{D}^{\circ}\right) - \left({\nu}_{A} {\mu}_{A}^{\circ} + {\nu}_{B} {\mu}_{B}^{\circ}\right) + R T {\nu}_{C} \ln \left(\frac{\left[C\right]}{{\left[C\right]}^{\circ}}\right) + R T {\nu}_{D} \ln \left(\frac{\left[D\right]}{{\left[D\right]}^{\circ}}\right) - R T {\nu}_{A} \ln \left(\frac{\left[A\right]}{{\left[A\right]}^{\circ}}\right) - R T {\nu}_{B} \ln \left(\frac{\left[B\right]}{{\left[B\right]}^{\circ}}\right)$

But recall from $\boldsymbol{\left(1\right)}$ that we defined $\Delta \mu$. Apply that for the standard state to get:

$\Delta \mu = \Delta {\mu}^{\circ} + R T {\nu}_{C} \ln \left(\frac{\left[C\right]}{{\left[C\right]}^{\circ}}\right) + R T {\nu}_{D} \ln \left(\frac{\left[D\right]}{{\left[D\right]}^{\circ}}\right) - R T {\nu}_{A} \ln \left(\frac{\left[A\right]}{{\left[A\right]}^{\circ}}\right) - R T {\nu}_{B} \ln \left(\frac{\left[B\right]}{{\left[B\right]}^{\circ}}\right)$

Then, to simplify this, we note that ${\left[C\right]}^{\circ} = {\left[D\right]}^{\circ} = {\left[A\right]}^{\circ} = {\left[B\right]}^{\circ} = \text{1 M}$, and can be considered a means to eliminate units inside a logarithm (the argument of a logarithm must be unitless).

Therefore, after grouping terms:

$\Delta \mu = \Delta {\mu}^{\circ} + R T \left({\nu}_{C} \ln \left[C\right] + {\nu}_{D} \ln \left[D\right]\right) - R T \left({\nu}_{A} \ln \left[A\right] + {\nu}_{B} \ln \left[B\right]\right)$

Next, use the property that $b \ln a = \ln {a}^{b}$ to get:

$\Delta \mu = \Delta {\mu}^{\circ} + R T \left(\ln {\left[C\right]}^{{\nu}_{C}} + \ln {\left[D\right]}^{{\nu}_{D}}\right) - R T \left(\ln {\left[A\right]}^{{\nu}_{A}} + \ln {\left[B\right]}^{{\nu}_{B}}\right)$

Then, use the property that $\ln a + \ln b = \ln \left(a b\right)$ and $\ln a - \ln b = \ln \left(\frac{a}{b}\right)$:

$\Delta \mu = \Delta {\mu}^{\circ} + R T \ln \left(\frac{{\left[C\right]}^{{\nu}_{C}} {\left[D\right]}^{{\nu}_{D}}}{{\left[A\right]}^{{\nu}_{A}} {\left[B\right]}^{{\nu}_{B}}}\right)$

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

$\Delta \mu = \Delta {\mu}^{\circ} + R T \ln Q$

Or, in terms of the Gibbs' free energy,

$\textcolor{b l u e}{\Delta \overline{G} = \Delta {\overline{G}}^{\circ} + R T \ln Q}$

In general chemistry, I have not seen the usage of $\Delta \overline{G}$ as $\frac{\Delta G}{n}$, and the bar is just dropped. The units of $\Delta G$ are implied to be $\text{J/mol}$ or #"kJ/mol", and for general chemistry we write:

$\Delta G = \Delta {G}^{\circ} + R T \ln Q$