Question

The emf of a standard cadmium cell is 1.02 V at 300 K. The temperature coefficient of the cell is \( -5.0 \times 10^{-5} \, \text{V K}^{-1} \). The value of \( \Delta H^\circ \) for the cell is ............ kJ mol\(^{-1}\) (rounded up to two decimal places). 

[1 F = 96500 \(C mol^{–1}\)]
 

Hint:

For electrochemical cells, the temperature dependence of the emf can be used to calculate the enthalpy change using the appropriate equation.

Answer 1

Correct Answer: -198 - -201

This is a thermodynamics problem involving the relationship between the electromotive force ($\text{emf}$) of an electrochemical cell, the temperature coefficient, and the standard enthalpy change ($\Delta H^\circ$) of the cell reaction.

The relationship between the standard Gibbs free energy change ($\Delta G^\circ$), the standard enthalpy change ($\Delta H^\circ$), the temperature ($T$), and the temperature coefficient of the $\text{emf}$ is given by the Gibbs-Helmholtz equation for electrochemical cells:

$$\Delta G^\circ = \Delta H^\circ + T \left(\frac{\partial \Delta G^\circ}{\partial T}\right)_P$$

Also, $\Delta G^\circ$ is related to the $\text{emf}$ ($E$) by:

$$\Delta G^\circ = -nFE$$

where $n$ is the number of moles of electrons transferred and $F$ is the Faraday constant.

Taking the derivative of $\Delta G^\circ$ with respect to temperature gives:

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

Substituting these relations back into the Gibbs-Helmholtz equation:

$$-nFE = \Delta H^\circ + T \left(-nF \frac{dE}{dT}\right)$$

$$\Delta H^\circ = -nFE + nFT \frac{dE}{dT}$$

$$\Delta H^\circ = nF \left(T \frac{dE}{dT} - E\right)$$

In this equation:

$E$ is the $\text{emf}$ of the cell.

$\frac{dE}{dT}$ is the temperature coefficient of the cell.

1. Identify Given Values

$\text{emf}$ ($E$): $1.02 \text{ V}$

Temperature ($T$): $300 \text{ K}$

Temperature coefficient ($\frac{dE}{dT}$): $-5.0 \times 10^{-5} \text{ V K}^{-1}$

Faraday constant ($F$): $96500 \text{ C mol}^{-1}$

Number of electrons ($n$): The standard cadmium cell (Weston cell) reaction involves the transfer of $n=2$ electrons:

$$\text{Cd}(s) + \text{Hg}_2\text{SO}_4(s) + \frac{8}{3}\text{H}_2\text{O}(l) \rightleftharpoons \text{CdSO}_4 \cdot \frac{8}{3}\text{H}_2\text{O}(s) + 2\text{Hg}(l)$$

2. Calculate $\Delta H^\circ$ (in $\text{J mol}^{-1}$)

Substitute the values into the $\Delta H^\circ$ equation:

$$\Delta H^\circ = nF \left(T \frac{dE}{dT} - E\right)$$

$$\Delta H^\circ = (2 \text{ mol}^{-1}) (96500 \text{ C mol}^{-1}) \left[(300 \text{ K}) (-5.0 \times 10^{-5} \text{ V K}^{-1}) - (1.02 \text{ V})\right]$$

Calculate the term inside the brackets:

$$T \frac{dE}{dT} = 300 \times (-5.0 \times 10^{-5}) = -1500 \times 10^{-5} = -0.015 \text{ V}$$

$$\left(T \frac{dE}{dT} - E\right) = (-0.015 \text{ V}) - (1.02 \text{ V}) = -1.035 \text{ V}$$

Now, calculate $\Delta H^\circ$:

$$\Delta H^\circ = (2 \times 96500) \text{ C mol}^{-1} \times (-1.035 \text{ V})$$

Since $1 \text{ J} = 1 \text{ C} \cdot 1 \text{ V}$:

$$\Delta H^\circ = 193000 \times (-1.035) \text{ J mol}^{-1}$$

$$\Delta H^\circ = -199755 \text{ J mol}^{-1}$$

3. Convert to $\text{kJ mol}^{-1}$

To convert from Joules to kilojoules, divide by $1000$:

$$\Delta H^\circ = \frac{-199755}{1000} \text{ kJ mol}^{-1}$$

$$\Delta H^\circ = -199.755 \text{ kJ mol}^{-1}$$

4. Round to Two Decimal Places

Rounding $\Delta H^\circ = -199.755 \text{ kJ mol}^{-1}$ to two decimal places:

$$\Delta H^\circ \approx -199.76 \text{ kJ mol}^{-1}$$