Question

Consider the following cell reaction: \[ {Mg} + {Cd}^{2+} \rightleftharpoons {Mg}^{2+} + {Cd} \] The standard Gibbs free energy change for the reaction is _________ kJ (rounded off to an integer). Given: Standard oxidation potentials for the reactions with respect to the standard hydrogen electrode are:
Mg \( \rightleftharpoons \) Mg\(^{2+}\) + 2e\(^-\) \( E^\circ = 2.37 \, {V} \) Cd \( \rightleftharpoons \) Cd\(^{2+}\) + 2e\(^-\) \( E^\circ = 0.403 \, {V} \) Faraday’s constant = 96500 C mol\(^{-1}\)

Hint:

Remember that the standard cell potential \( E^\circ_{{cell}} \) is the difference between the reduction potentials for the cathode and anode. Also, use the equation \( \Delta G^\circ = -nFE^\circ_{{cell}} \) to calculate the Gibbs free energy change.

Answer 1

The standard Gibbs free energy change (\( \Delta G^\circ \)) for the reaction can be calculated using the equation: \[ \Delta G^\circ = -nF E^\circ \] where: - \( n \) is the number of moles of electrons involved in the reaction (for this reaction, \( n = 2 \)), - \( F \) is Faraday’s constant (96500 C mol\(^{-1}\)), - \( E^\circ \) is the cell potential. The cell potential (\( E^\circ_{{cell}} \)) is calculated by: \[ E^\circ_{{cell}} = E^\circ_{{cathode}} - E^\circ_{{anode}} \] Since the reduction reaction is at the cathode and oxidation is at the anode, we take the reduction potentials for the half-reactions. Here, Mg will be oxidized (anode) and Cd\(^{2+}\) will be reduced (cathode). Therefore: \[ E^\circ_{{cell}} = 0.403 \, {V} - 2.37 \, {V} = -1.967 \, {V} \] Now, calculating \( \Delta G^\circ \): \[ \Delta G^\circ = -2 \times 96500 \times (-1.967) = 379,745 \, {J} = 379.7 \, {kJ} \] So the Gibbs free energy change is approximately between -381 kJ to -379 kJ.