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Temperature Coefficient of EMF & Thermodynamics of Electrochemical Cells

Temperature Coefficient of EMF & Thermodynamics of Electrochemical Cells | Chemca

Exhaustive Guide: Temperature Coefficient of EMF & Thermodynamics of Electrochemical Cells

Master the derivations linking Cell Potential ($E_{\text{cell}}$) with Gibbs Free Energy ($\Delta G$), Entropy ($\Delta S$), and Enthalpy ($\Delta H$) for CBSE, JEE Advanced, and NEET.

1. Introduction to Electrochemical Thermodynamics

Welcome back to Chemca.in. While the Nernst Equation elegantly describes how cell potential changes with concentration, it is only one piece of the puzzle. Physical chemistry demands a complete thermodynamic profile of chemical reactions. How much absolute heat does a battery release? Does the disorder of the system increase or decrease as it discharges? How does the voltage of a battery change on a freezing winter day compared to a scorching summer afternoon?

To answer these questions, we must bridge the gap between Electrochemistry and classical Thermodynamics. Measuring the Electromotive Force (EMF) of a reversible cell is one of the most precise experimental methods available to physical chemists for determining extensive thermodynamic properties like Gibbs Free Energy ($\Delta G$), Entropy ($\Delta S$), and Enthalpy ($\Delta H$).

In this exhaustive treatise, we will introduce the concept of the Temperature Coefficient of EMF, utilize the Gibbs-Helmholtz equation, and derive all interconnected formulas. This guide is specifically tailored for students aiming to dominate the advanced conceptual questions often found in JEE Advanced and NEET examinations.

2. The Foundation: EMF and Gibbs Free Energy ($\Delta G$)

Before we introduce temperature variations, we must firmly establish the anchor of electrochemical thermodynamics: Gibbs Free Energy. As we learned in our previous notes, in a reversible Galvanic cell operating at constant temperature and pressure, the maximum useful non-expansion electrical work done by the system is equal to the decrease in Gibbs Free Energy.

If an electrochemical cell transfers $n$ moles of electrons through an external circuit across a potential difference of $E_{\text{cell}}$, the total charge transferred is $q = nF$ (where $F$ is Faraday's constant, $\approx 96485 \text{ C mol}^{-1}$).

The electrical work $W_{\text{elec}} = nFE_{\text{cell}}$. Therefore, the relationship is:

Fundamental Equation of Electrochemistry: $$ \Delta G = -nFE_{\text{cell}} $$

Under standard conditions ($1 \text{ M}$ concentration, $1 \text{ bar}$ pressure, pure solids/liquids), this becomes:

$$ \Delta G^{\circ} = -nFE^{\circ}_{\text{cell}} $$

Significance: This equation acts as a spontaneity check. For any cell to be a viable battery (spontaneous reaction), $\Delta G$ must be negative. Because $n$ and $F$ are always positive constants, $E_{\text{cell}}$ must be positive. A negative voltage indicates a non-spontaneous reaction that requires an external power source to proceed (electrolysis).

3. The Temperature Coefficient of EMF

Batteries are sensitive to temperature. If you measure the voltage of a standard lead-acid car battery at $25^{\circ}\text{C}$ and then again at $0^{\circ}\text{C}$, the readings will be slightly different. The rate at which the Electromotive Force (EMF) of a cell changes with a change in absolute temperature (at constant pressure) is called the Temperature Coefficient of EMF.

Mathematically, it is expressed as a partial derivative:

$$ \text{Temperature Coefficient} = \left( \frac{\partial E}{\partial T} \right)_P $$

Where:

  • $\partial E$ = Infinitesimal change in cell potential (Volts)
  • $\partial T$ = Infinitesimal change in temperature (Kelvin)
  • Subscript $P$ indicates that the pressure is kept constant.

The unit of the temperature coefficient is Volts per Kelvin ($V K^{-1}$). For most typical electrochemical cells, the value of $\left( \frac{\partial E}{\partial T} \right)_P$ is quite small, typically on the order of $10^{-4}$ to $10^{-5} \text{ V K}^{-1}$. It can be positive, negative, or even exactly zero depending on the specific redox reaction.

Experimental Determination: In a laboratory, this coefficient is found by measuring the cell's EMF at various temperatures, plotting a graph of $E$ versus $T$, and finding the slope of the line (or tangent to the curve) at a specific temperature.

4. Derivation of Entropy Change ($\Delta S$) from EMF

Now, let us connect the macroscopic voltage measurements to the microscopic disorder (Entropy) of the chemical system. From classical chemical thermodynamics, we know the fundamental Maxwell relation that links Gibbs Free Energy, Entropy, and Temperature at constant pressure:

$$ \left( \frac{\partial G}{\partial T} \right)_P = -S $$

If we apply this to a chemical reaction (products minus reactants), we get the relation for the change in these properties:

$$ \left( \frac{\partial (\Delta G)}{\partial T} \right)_P = -\Delta S $$

Now, let us substitute our fundamental electrochemistry equation ($\Delta G = -nFE$) into this thermodynamic identity:

$$ \left( \frac{\partial (-nFE)}{\partial T} \right)_P = -\Delta S $$

Since $n$ (number of electrons) and $F$ (Faraday's constant) are independent of temperature, they can be pulled out of the derivative:

$$ -nF \left( \frac{\partial E}{\partial T} \right)_P = -\Delta S $$

Canceling the negative signs on both sides, we arrive at the beautiful equation for Entropy:

Equation for Entropy Change: $$ \Delta S = nF \left( \frac{\partial E}{\partial T} \right)_P $$

What this means: The entropy change of a redox reaction is directly proportional to the temperature coefficient of the cell. If the voltage of a battery increases as it gets hotter (positive coefficient), the chemical reaction inside has a positive entropy change (products are more disordered than reactants).

5. Derivation of Enthalpy Change ($\Delta H$) from EMF

Enthalpy ($\Delta H$) represents the total heat content of the system. To find it, we use the famous Gibbs-Helmholtz Equation, which relates Free Energy, Enthalpy, and Entropy at a constant temperature $T$:

$$ \Delta G = \Delta H - T\Delta S $$

We want to solve for Enthalpy, so we rearrange the equation:

$$ \Delta H = \Delta G + T\Delta S $$

Now, we simply substitute the electrochemical expressions we have already derived for $\Delta G$ and $\Delta S$ into this equation:

  • Substitute $\Delta G = -nFE$
  • Substitute $\Delta S = nF \left( \frac{\partial E}{\partial T} \right)_P$
$$ \Delta H = -nFE + T \left[ nF \left( \frac{\partial E}{\partial T} \right)_P \right] $$

Factoring out the common $-nF$ term yields the final, highly tested equation for Enthalpy:

Equation for Enthalpy Change (Gibbs-Helmholtz Form): $$ \Delta H = -nF \left[ E - T \left( \frac{\partial E}{\partial T} \right)_P \right] $$

This remarkable equation proves that we can determine the exact heat of reaction ($\Delta H$) purely by measuring voltages with a voltmeter at a couple of different temperatures, without ever needing an expensive, complex bomb calorimeter!

6. Derivation of Heat Capacity Change ($\Delta C_p$) [JEE Advanced Bonus]

For top-tier competitive exams, you might be asked to go one step further and find the change in heat capacity at constant pressure ($\Delta C_p$) for the cell reaction. From Kirchhoff's Law of thermodynamics, we know:

$$ \Delta C_p = \left( \frac{\partial \Delta H}{\partial T} \right)_P $$

We can also relate it to entropy: $\Delta C_p = T \left( \frac{\partial \Delta S}{\partial T} \right)_P$. Let's use the entropy route. Substitute our $\Delta S$ equation ($\Delta S = nF \left( \frac{\partial E}{\partial T} \right)_P$):

$$ \Delta C_p = T \left( \frac{\partial}{\partial T} \left[ nF \left( \frac{\partial E}{\partial T} \right)_P \right] \right)_P $$
Equation for Heat Capacity Change: $$ \Delta C_p = nFT \left( \frac{\partial^2 E}{\partial T^2} \right)_P $$

This tells us that the heat capacity change is proportional to the second derivative of the EMF with respect to temperature (the curvature of the $E$ vs. $T$ graph).

7. Physical Significance: Heating and Cooling of Batteries

Let's analyze the Enthalpy equation ($\Delta H = \Delta G + T\Delta S$) from a practical battery-design perspective. When a battery discharges reversibly, it does electrical work ($\Delta G$). What happens to the heat? There are three cases based on the sign of the temperature coefficient $\left( \frac{\partial E}{\partial T} \right)_P$:

Case 1: Temperature Coefficient is Positive ($> 0$)

  • $\Delta S$ is positive ($+ve$). Disorder increases during discharge.
  • The term $T\Delta S$ is positive.
  • Therefore, $\Delta H$ is less negative than $\Delta G$. The electrical work produced by the cell is actually greater than the decrease in enthalpy.
  • Physical reality: How is this possible? The cell must absorb heat from its surroundings to convert into extra electrical energy! A battery with a positive temperature coefficient will naturally cool down as it operates reversibly.

Case 2: Temperature Coefficient is Negative ($< 0$)

  • $\Delta S$ is negative ($-ve$). Disorder decreases during discharge.
  • The term $T\Delta S$ is negative.
  • Therefore, $\Delta H$ is more negative than $\Delta G$. The electrical work is less than the total enthalpy drop.
  • Physical reality: The "excess" enthalpy that wasn't converted into electrical work is released as heat. A battery with a negative temperature coefficient will heat up as it operates. (This is the most common case for commercial batteries).

Case 3: Temperature Coefficient is Zero ($= 0$)

  • $\Delta S = 0$.
  • Therefore, $\Delta H = \Delta G = -nFE$.
  • Physical reality: The cell operates with 100% thermodynamic efficiency. All the chemical enthalpy is converted entirely into electrical work. The cell will neither heat up nor cool down. (Such cells are very rare).

8. Comprehensive Solved Numericals

Problem 1 (Standard Application):
The EMF of a standard Weston Cadmium cell is $1.0183 \text{ V}$ at $293 \text{ K}$ and its temperature coefficient is $-5.0 \times 10^{-5} \text{ V K}^{-1}$. Calculate $\Delta G$, $\Delta S$, and $\Delta H$ for the cell reaction at $293 \text{ K}$. (Given: $n = 2$ for the Weston cell, $F = 96500 \text{ C mol}^{-1}$).
Solution:
Given data:
$E = 1.0183 \text{ V}$
$T = 293 \text{ K}$
$\left( \frac{\partial E}{\partial T} \right)_P = -5.0 \times 10^{-5} \text{ V K}^{-1}$
$n = 2$

Step 1: Calculate Gibbs Free Energy ($\Delta G$)
$\Delta G = -nFE$
$\Delta G = -2 \times 96500 \text{ C mol}^{-1} \times 1.0183 \text{ V}$
$\Delta G = -196531.9 \text{ Joules} = \mathbf{-196.53 \text{ kJ mol}^{-1}}$

Step 2: Calculate Entropy ($\Delta S$)
$\Delta S = nF \left( \frac{\partial E}{\partial T} \right)_P$
$\Delta S = 2 \times 96500 \times (-5.0 \times 10^{-5})$
$\Delta S = \mathbf{-9.65 \text{ J K}^{-1} \text{ mol}^{-1}}$

Step 3: Calculate Enthalpy ($\Delta H$)
We use $\Delta H = \Delta G + T\Delta S$
$\Delta H = -196531.9 \text{ J} + [293 \text{ K} \times (-9.65 \text{ J K}^{-1})]$
$\Delta H = -196531.9 - 2827.45$
$\Delta H = -199359.35 \text{ Joules} = \mathbf{-199.36 \text{ kJ mol}^{-1}}$
Problem 2 (JEE Level - Differential form):
The cell potential for a particular reversible Galvanic cell is given by the equation:
$E (\text{in Volts}) = 1.25 - 2.0 \times 10^{-4} (T - 298) - 1.5 \times 10^{-6} (T - 298)^2$
where $T$ is temperature in Kelvin. Calculate the entropy change ($\Delta S$) for the cell reaction ($n=2$) operating exactly at $298 \text{ K}$.
Solution:
Step 1: Find the temperature coefficient expression by differentiating $E$ with respect to $T$.
$E = 1.25 - 2.0 \times 10^{-4} (T - 298) - 1.5 \times 10^{-6} (T - 298)^2$
$\left( \frac{\partial E}{\partial T} \right)_P = 0 - 2.0 \times 10^{-4}(1) - 1.5 \times 10^{-6} \times 2(T - 298) \times (1)$
$\left( \frac{\partial E}{\partial T} \right)_P = -2.0 \times 10^{-4} - 3.0 \times 10^{-6} (T - 298)$

Step 2: Evaluate the coefficient at the specified temperature ($T = 298 \text{ K}$).
Plug $T = 298$ into the derivative:
$\left( \frac{\partial E}{\partial T} \right)_P = -2.0 \times 10^{-4} - 3.0 \times 10^{-6} (298 - 298)$
$\left( \frac{\partial E}{\partial T} \right)_P = -2.0 \times 10^{-4} \text{ V K}^{-1}$

Step 3: Calculate $\Delta S$.
$\Delta S = nF \left( \frac{\partial E}{\partial T} \right)_P$
$\Delta S = 2 \times 96500 \times (-2.0 \times 10^{-4})$
$\Delta S = 193000 \times -0.0002$
$\Delta S = \mathbf{-38.6 \text{ J K}^{-1} \text{ mol}^{-1}}$

9. Conclusion

Understanding the thermodynamic properties of electrochemical cells goes far beyond just calculating voltages. By utilizing the Temperature Coefficient of EMF, we unlock the ability to rigorously define the Entropy ($\Delta S$) and Enthalpy ($\Delta H$) of chemical reactions using highly precise electrical measurements.

For students traversing Class 12 chemistry and preparing for engineering/medical entrance exams, the equations $\Delta G = -nFE$, $\Delta S = nF(\frac{\partial E}{\partial T})$, and $\Delta H = -nF[E - T(\frac{\partial E}{\partial T})]$ form a holy trinity of electro-thermodynamics. Keep practicing numericals focusing on units (especially converting Joules to Kilojoules), and remember to keep this guide from Chemca.in bookmarked for your final revisions!

10. Frequently Asked Questions (FAQs)

Q1. Why is the Temperature Coefficient of EMF generally very small?
The coefficient $\left( \frac{\partial E}{\partial T} \right)_P$ is equal to $\frac{\Delta S}{nF}$. While Entropy changes ($\Delta S$) in redox reactions are typically in the range of tens to hundreds of $\text{J K}^{-1}$, the denominator involves Faraday's constant ($96485 \text{ C}$), which is a massive number. Dividing a normal-sized entropy by this huge number results in a very small voltage change per degree Kelvin, usually on the order of $10^{-4} \text{ V K}^{-1}$.
Q2. Can a battery get colder while generating electricity?
Yes, theoretically. If a cell has a positive temperature coefficient, its entropy change is positive. This means $T\Delta S$ is positive, and the cell is capable of converting both its chemical enthalpy and ambient heat absorbed from the surroundings into electrical work. Such a cell will cool down when discharged reversibly.
Q3. What is the difference between $\Delta G$ and $\Delta H$ in an electrochemical cell?
Enthalpy ($\Delta H$) is the total heat energy released or absorbed by the chemical reaction. Gibbs Free Energy ($\Delta G$) is the portion of that total energy that is actually available to do useful electrical work. The difference between them is the $T\Delta S$ term, which represents energy that is "lost" to increasing disorder (entropy) or "gained" from the surroundings.
Q4. How does the Nernst equation relate to the Temperature coefficient?
The Nernst equation ($E = E^{\circ} - \frac{RT}{nF} \ln Q$) explicitly shows that temperature $T$ is a variable in cell potential. If you differentiate the Nernst equation with respect to temperature at constant pressure and concentration, you will theoretically derive the temperature coefficient for that specific non-standard state.
Q5. Why is measuring EMF considered better than calorimetry for finding $\Delta H$?
Calorimetry (measuring heat changes directly in a calorimeter) can be difficult, requires large amounts of chemicals, and is prone to heat-loss errors. Measuring voltage (EMF) is incredibly precise, requires only tiny amounts of chemicals, and using the Gibbs-Helmholtz equation alongside the temperature coefficient gives highly accurate thermodynamic data.
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