Thermodynamic Functions of a Galvanic Cell
A galvanic cell converts chemical energy into electrical energy. The electrical work done by the cell comes at the expense of the Gibbs Free Energy ($G$) of the cell reaction. By measuring the Electromotive Force (EMF) of the cell and its variation with temperature, we can calculate fundamental thermodynamic parameters like $\Delta G$, $\Delta S$, and $\Delta H$.
1. Electrical Work and Gibbs Free Energy ($\Delta G$)
The electrical work done by a cell in one second is equal to the product of the potential difference (EMF) and the total charge passed.
If $n$ moles of electrons are transferred in the balanced cell reaction, the total charge is $nF$ (where $F$ is Faraday's constant $\approx 96500 \ C/mol$).
According to thermodynamics, the maximum useful work done by a system (at constant T and P) is equal to the decrease in Gibbs Free Energy ($-\Delta G$).
Conditions for Spontaneity:
- For a spontaneous reaction, $\Delta G$ must be Negative ($\Delta G < 0$).
- Therefore, $E_{\text{cell}}$ must be Positive ($E_{\text{cell}} > 0$).
Standard Gibbs Free Energy ($\Delta G^\circ$)
At standard conditions (1 M concentration, 1 bar pressure, 298 K):
$$ \Delta G^\circ = -nF E^\circ_{\text{cell}} $$Relation with Equilibrium Constant ($K_{eq}$)
At equilibrium, $E_{\text{cell}} = 0$, so $\Delta G = 0$. Using the relationship $\Delta G^\circ = -RT \ln K_{eq}$:
2. Entropy Change ($\Delta S$)
From the fundamental thermodynamic equation $dG = VdP - SdT$, at constant pressure ($dP=0$), we have:
$$ \left(\frac{\partial G}{\partial T}\right)_P = -S \quad \implies \quad \left(\frac{\partial (\Delta G)}{\partial T}\right)_P = -\Delta S $$Substituting $\Delta G = -nFE$:
$$ \frac{\partial}{\partial T}(-nFE) = -\Delta S $$Here, $\left(\frac{\partial E}{\partial T}\right)_P$ is called the Temperature Coefficient of the cell EMF.
- If $\frac{\partial E}{\partial T} > 0$, then $\Delta S$ is positive (Disorder increases).
- If $\frac{\partial E}{\partial T} < 0$, then $\Delta S$ is negative (Disorder decreases).
3. Enthalpy Change ($\Delta H$)
From the Gibbs-Helmholtz equation:
$$ \Delta G = \Delta H - T\Delta S $$Rearranging for Enthalpy:
$$ \Delta H = \Delta G + T\Delta S $$Substituting the values derived above:
Heat of Reaction: If the cell operates reversibly, the heat absorbed or evolved ($q_{rev}$) is given by $q = T\Delta S = nFT (\frac{\partial E}{\partial T})_P$.
4. Summary of Formulas
| Parameter | Formula |
|---|---|
| Gibbs Energy ($\Delta G$) | $-nFE_{\text{cell}}$ |
| Entropy ($\Delta S$) | $nF (\frac{\partial E}{\partial T})_P$ |
| Enthalpy ($\Delta H$) | $-nFE + nFT (\frac{\partial E}{\partial T})_P$ |
| Equilibrium Constant | $\Delta G^\circ = -RT \ln K_{eq}$ |
5. Example Calculation
Problem: For a cell with $n=2$, EMF is $1.10 \ V$ at $298 \ K$, and the temperature coefficient is $-0.0005 \ V/K$. Calculate $\Delta G$ and $\Delta S$. ($F = 96500 \ C/mol$)
Solution:
1. $\Delta G = -nFE = -2 \times 96500 \times 1.10 = -212,300 \ J/mol = -212.3 \ kJ/mol$.
2. $\Delta S = nF (\frac{\partial E}{\partial T}) = 2 \times 96500 \times (-0.0005) = -96.5 \ J/K \cdot mol$.
Thermodynamics Quiz
Test your concepts on Cell Potential and Energy. 10 MCQs with explanations.
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