Con el objetivo de hacer una representación gráfica del $\log C_i$ vs. $E$:
$\ce{a A_{ox} + $n$_{\ce{A}} $e$ <=> b A_{red}}$
$ E_{\ce{A}} = \rlap{\hphantom{E}°}E_{\lower 2mu {\ce{A}}} + \dfrac{0{,}059}{n_{\ce{A}}} \log \dfrac{[\ce{A_{ox}}]^a}{[\ce{A_{red}}]^b} $
$ \begin{array}{l} \! \underset{ \begin{subarray}{c} \displaystyle \uparrow \\[0.5ex] \displaystyle \ce{p$e$} \\[1ex] \displaystyle \Downarrow \end{subarray} } {\boxed{\dfrac{E_{\ce{A}}}{0{,}059}}} = \dfrac{\rlap{\hphantom{E}°}E_{\lower 2mu {\ce{A}}}}{0{,}059} + \dfrac{1}{n_{\ce{A}}} \log \dfrac{[\ce{A_{ox}}]^a}{[\ce{A_{red}}]^b} \\ \ce{p$e$_A} = \ce{p$\rlap{\hphantom{e}°}e_{\lower 2mu {\ce{A}}}$} + \dfrac{1}{n_{\ce{A}}} \log \dfrac{[\ce{A_{ox}}]^a}{[\ce{A_{red}}]^b} \end{array} $
$ \ce{p$e$_A} = \ce{p$\rlap{\hphantom{e}°}e_{\lower 2mu {\ce{A}}}$} + \dfrac{a}{n_{\ce{A}}} \log [\ce{A_{ox}}] - \dfrac{b}{n_{\ce{A}}} \log [\ce{A_{red}}] $
$\Rightarrow \log C_i = f(\ce{p$e$_{\ce{A}}}) \quad i = \ce{A_{ox}} \text{ ó } \ce{A_{red}}$
Puede observarse, por ejemplo, que para el caso más sencillo de $a = b = 1$:
$\ce{p$e$_A} < \ce{p$\rlap{\hphantom{e}°}e_{\lower 2mu \ce{A}}$} \rightarrow [\ce{A_{red}}] > [\ce{A_{ox}}]$
$\ce{p$e$_A} = \ce{p$\rlap{\hphantom{e}°}e_{\lower 2mu \ce{A}}$} \rightarrow [\ce{A_{red}}] = [\ce{A_{ox}}]$
$\ce{p$e$_A} > \ce{p$\rlap{\hphantom{e}°}e_{\lower 2mu \ce{A}}$} \rightarrow [\ce{A_{ox}}] > [\ce{A_{red}}]$
Ejemplo:
$ \ce{ MnO4- + 8H+ + 5 $e^-$ <=> Mn^2+ + 4 H2O } \qquad E° = \pu{1,51 v} $
$[\ce{H+}] = 1$
$ E = E° + \dfrac{0{,}059}{5} \log \dfrac{[\ce{MnO4-}]}{[\ce{Mn^2+}]} $
$ \dfrac{E}{0{,059}} = \dfrac{E°}{0{,}059} + \dfrac{1}{5} \log [\ce{MnO4-}] - \dfrac{1}{5} \log [\ce{Mn^2+}] $
$\ce{p$e$} = 25{,}6 + \dfrac{1}{5} \log [\ce{MnO4-}] - \dfrac{1}{5} \log [\ce{Mn^2+}]$
1ª región
$ \begin{align} [\ce{Mn^2+}] \gg [\ce{MnO4-}] \rightarrow C_{\ce{A}} = [\ce{Mn^2+}] &+ \! \cancel{[\ce{MnO4-}]} \Rightarrow [\ce{Mn^2+}] = C_{\ce{A}} \\ &\uparrow \\ [\ce{Mn^2+}] &\geq 100 [\ce{MnO4-}] \quad (\text{error} \leq \pu{1\%}) \end{align} $
$ \ce{p$e$} = \ce{p$e$°} + \dfrac{1}{5} \log \underbrace{ \boxed{\dfrac{[\ce{MnO4-}]}{[\ce{Mn^2+}]}} }_{\large \leq \frac{1}{100}} \Rightarrow \begin{aligned}[t] &\ce{p$e$} \leq \ce{p$e$°} + \dfrac{1}{5} \log \dfrac{1}{100} = \ce{p$e$°} - \dfrac{2}{5} \\[1ex] &\ce{pe} \leq 25{,}6 - 0{,}4 = 25{,}2 \end{aligned} $
$ \ce{p$e$} = \ce{p$e$°} + \dfrac{1}{5} \log \dfrac{[\ce{MnO4-}]}{C_{\ce{A}}} $
$ \log [\ce{MnO4-}] = 5 (\ce{p$e$} - \ce{p$e$°}) + \log C_{\ce{A}} \Rightarrow \log [\ce{MnO4-}] = \text{cte.} + 5 \ce{p$e$} $
2ª región
$ [\ce{Mn^2+}] \ll [\ce{MnO4-}] \rightarrow \begin{array}[t]{l} C_{\ce{A}} = \cancel{[\ce{Mn^2+}]} \! + [\ce{MnO4-}] \\[1ex] [\ce{MnO4-}] \geq 100 [\ce{Mn^2+}] \quad (\text{error} \leq \pu{1 \%}) \\[1ex] \ce{p$e$} \geq \ce{p$e$°} + 2/5 \quad \text{y} \quad \log [\ce{Mn^2+}] = \text{cte.}{}' - 5 \ce{p$e$} \end{array} $
3º, punto del sistema
$ \begin{aligned} \ce{p$e$} = \ce{p$e$°} &\Rightarrow [\ce{MnO4-}] = [\ce{Mn^2+}] \\[1em] &\Rightarrow [\ce{MnO4-}] = [\ce{Mn^2+}] = \dfrac{C_{\ce{A}}}{2} \\[1ex] &\hphantom{\Rightarrow {}} \log [\ce{MnO4-}] = \log [\ce{Mn^2+}] = \log C_{\ce{A}} - 0{,}3 \end{aligned} $
Con todo esto, la gráfica, para $C_{\ce{A}} = \pu{0,1M}$, queda así:
Otro ejemplo, con un sistema con más de dos estados de oxidación:
$ \begin{align} &\ce{Fe^3+ + $e^-$ <=> Fe^2+} \quad \rlap{\hphantom{E}°}E_{\ce{Fe^3+/\mspace{1mu}Fe^2+}} = \pu{0,77v} && C_{\ce{A}} = \pu{0,1M} \\[1ex] &\ce{Fe^2+ + 2 $e^-$ <=> Fe} \quad \rlap{\hphantom{E}°}E_{\ce{Fe^2+/\mspace{1mu}Fe}} = \pu{-0,44v} &&[\ce{Fe^3+}] + [\ce{Fe^2+}] = \pu{0,1 M} \\[1ex] &\ce{p$\rlap{\hphantom{e}°}e_{\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} = 13{,}05 \quad \ce{p$e°_{\!\!\ce{Fe^2+/\mspace{1mu}Fe}}$} = -7{,}46 && \end{align} $
Zonas y puntos del diagrama:
$\ce{p$e$} > \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$}$
Se considera que:
$ \dfrac{[\ce{Fe^3+}]}{[\ce{Fe^2+}]} \geq 100 \Rightarrow \ce{p$e$} - \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} \geq 2 $
$ \dfrac{[\ce{Fe^3+}]}{[\ce{Fe^2+}]} \geq 100 \Rightarrow [\ce{Fe^3+}] + \! \cancel{[\ce{Fe^2+}]} \! = \pu{0,1 M} = C_{\ce{A}} $
$ \ce{p$e$} = \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + \log C_{\ce{A}} - \log [\ce{Fe^2+}] \Rightarrow \log [\ce{Fe^2+}] = \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + \log C_{\ce{A}} - \ce{p$e$} $
$\ce{p$e$} = \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$}$
$[\ce{Fe^3+}] = [\ce{Fe^2+}] = C_{\ce{A}} /\mspace{1mu} 2$
$\log [\ce{Fe^3+}] = \log [\ce{Fe^2+}] = \log C_{\ce{A}} - 0{,}3$
$\ce{p$e$} < \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$}$
Se considera que:
$ \begin{array}{l} \dfrac{[\ce{Fe^3+}]}{[\ce{Fe^2+}]} \leq 0{,}01 \Rightarrow \cancel{[\ce{Fe^3+}]} \! + [\ce{Fe^2+}] = \pu{0,1 M} = C_{\ce{A}} \\[1ex] \hphantom{\dfrac{[\ce{Fe^3+}]}{[\ce{Fe^2+}]}} \Downarrow \\[1ex] \ce{p$e$} - \ce{p$e°_{\!\!\ce{Fe^3+}/\mspace{1mu}\ce{Fe^2+}}$} \leq \log \dfrac{1}{100} = -2 \\[1ex] \ce{p$e$} = \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + \log [\ce{Fe^3+}] - \log C_{\ce{A}} \Rightarrow \log[\ce{Fe^3+}] = \ce{p$e$} - \underbrace{ \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + \log C_{\ce{A}} }_{\text{cte.}} \end{array} $
Comenzará a aparecer $\ce{Fe}$ metálico cuando:
$ \ce{p$e$} = \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + \dfrac{1}{2} \log [\ce{Fe^2+}] \quad \begin{array}[t]{l} \smash[b]{\text{La especie $\ce{Fe}$ es un sólido y no aparece en la}} \\ \smash[t]{\text{expresión de $\ce{p$e$}$.}} \end{array} $
$\ce{p$e$} = \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + \dfrac{1}{2} \log C_{\ce{A}}$
$\ce{p$e$} = -7{,}46 + \dfrac{1}{2} \log 0{,}1 = -7{,}96$
Por tanto a $\ce{p$e$} < −7{,}96$ predominio de $\ce{Fe}$.
$\ce{p$e$} \lt \ce{p$e°_{\!\!\ce{Fe^2+/\mspace{1mu}Fe}}$}$
$\log\{\ce{Fe}\} = 0$
$\log [\ce{Fe^2+}] = 2 \ce{p$e$} - 2 \ce{p$e°_{\!\!\ce{Fe^2+/\mspace{1mu}Fe}}$}$
$ \begin{align} \ce{p$e$} &= \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + \log [\ce{Fe^3+}] - \log [\ce{Fe^2+}] = \\[1ex] &= \ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + 2 \ce{p$e°_{\!\!\ce{Fe^2+/\mspace{1mu}Fe}}$} - 2 \ce{p$e$} + \log [\ce{Fe^3+}] \end{align} $
$ \Rightarrow \log [\ce{Fe^3+}] = -(\ce{p$e°_{\!\!\ce{Fe^3+/\mspace{1mu}Fe^2+}}$} + 2 \ce{p$e°_{\!\!\ce{Fe^2+/\mspace{1mu}Fe}}$}) + 3 \ce{p$e$} $
La gráfica, teniendo en cuenta lo visto, queda así: