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10.5: Células de concentración

Última actualización

30 oct 2022
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- 10.4: Entropía de Celdas Electroquímicas
- 10.E: Electroquímica (Ejercicios)

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La generación de una diferencia de potencial electrostático depende de la creación de una diferencia en el potencial químico entre dos medias celdas. Una manera importante en la que esto se puede crear es creando una diferencia de concentración. Usando la ecuación de Nernst, se puede expresar la diferencia de potencial para una célula de concentración (una en la que ambas medias células implican la misma semirreacción)

$E = -\dfrac{RT}{nF} \ln \dfrac{[\text{oxdizing}]}{[\text{reducing}]}$

Ejemplo $\PageIndex{1}$

Calcular el potencial celular (a 25 °C) para la celda de concentración definida por

$Cu(s) | Cu^{2+} (aq,\, 0.00100 \,M) || Cu^{2+} (aq,\, 0.100 \,M) | Cu(s) \nonumber$

Solución:

Dado que las medias reacciones de oxidación y reducción son las mismas,

$E_{cell}^o =0\,V \nonumber$

El potencial celular a 25 °C se calcula usando la ecuación de Nernst:

$E = -\dfrac{RT}{nF} \ln Q \nonumber$

Sustituyendo los valores del problema:

$\begin{align*} E_{cell} &= - \dfrac{(8.314 \,J/(mol\,K) (298\,K) }{2(96484\,C)} \ln \left( \dfrac{0.00100\,M}{0.100\,M} \right) \\[4pt] &= 0.059\,V \end{align*}$

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