An electrochemical cell, in which the reaction $$ \mathrm{Cu}+\mathrm{Hg}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Hg}+\mathrm{CuSo}_{4} $$ takes place, is connected to a motor having a back emf only slightly smaller than the \(\mathrm{emf}\) of the cell. The emf of this cell is given by Eq. (2.14), with \(\mathcal{S}_{20}=0.3497 \mathrm{~V}\), \(\alpha=-6.35 \times 10^{-4} \mathrm{~V} / \mathrm{deg}, \beta=-2.4 \times 10^{-6} \mathrm{~V} / \mathrm{deg}^{2}\), and \(\gamma=0 .\) If the cell is kept at a constant temperature of \(25^{\circ} \mathrm{C}\) and \(0.1 \mathrm{~mol}\) of copper reacts, then how much work is done on the motor?

Electromotive force (EMF) is the voltage generated by an electrochemical cell or by changing the magnetic field. In this exercise, the EMF of the cell is calculated using a specific equation: \[ \text{emf} = S_{20} + \alpha (T - 298.15) + \beta (T - 298.15)^2 + \gamma (T - 298.15)^3 \] The constants given in the problem are crucial. Here, \(S_{20} = 0.3497\) V, \(\alpha = -6.35 \times 10^{-4}\) V/deg, \(\beta = -2.4 \times 10^{-6}\) V/deg², and \(\gamma = 0\). The temperature is 25 ºC, which needs to be converted to Kelvin for the calculations. You will notice that due to the constants and zero value for \(\gamma\), the EMF remains essentially \(0.3497\) V at 25 ºC.
This is due to the derivative terms becoming null at 25ºC.

Faraday's constant is a fundamental physical constant that represents the charge of one mole of electrons. Its value is approximately 96485 C/mol. This constant is used to convert moles of electrons transferred in a reaction to the equivalent charge (Coulombs). For example:

• Given the moles of electrons (\(n_e\)) transferred is 0.2 mol, the charge \(Q\) can be calculated as:
\[ Q = n_e \times F = 0.2 \text{ mol} \times 96485 \text{ C/mol} = 19297 \text{ C} \]

This relationship is critical for determining the work done by the electrochemical cell.

Reaction stoichiometry involves relating quantities of reactants and products in a chemical reaction. For the given reaction: \[ \mathrm{Cu} + \mathrm{Hg}_{2} \mathrm{SO}_{4} \rightarrow 2 \mathrm{Hg} + \mathrm{CuSO}_{4} \] The stoichiometry shows that 1 mole of Cu reacts to produce 2 moles of electrons. Hence, if 0.1 mol of Cu is reacted,

• The moles of electrons transferred ( \( n_e \)) is: \[ 0.1 \text{ mol Cu} \times 2 \text{ mol e}^{-}/ \text{ mol Cu} = 0.2 \text{ mol e}^{-} \]

Understanding stoichiometry helps in calculating products and reactants in balanced chemical equations accurately.

In electrochemical calculations, temperatures must often be converted from Celsius to Kelvin. The formula used is: \[ T(K) = T(ºC) + 273.15 \] Therefore, for a temperature of 25 ºC, we would have: \[ 25 + 273.15 = 298.15 \text{ K} \] Using Kelvin ensures that calculations align with the standards required by the equations for EMF and other thermodynamic properties.

Finally, work calculation in the context of an electrochemical cell involves multiplying the total charge (Coulombs) by the electromotive force (EMF). For this problem, after calculating the charge (\(Q\)) and verifying the EMF of the cell, the work done (\(W\)) is given by: \[ W = Q \times \text{emf} \]

• Here, with \(Q = 19297 \text{ C}\) and \(\text{emf} = 0.3497 \text{ V}\), the calculations are: \[ W = 19297 \text{ C} \times 0.3497 \text{ V} = 6748.01 \text{ J} \]

This approach shows how the electrochemical cell's potential is converted into useful work on the motor despite internal resistances.