Question

Calculate the \(\Delta G\) of the following cell:
\[ \text{Mg}(s) | \text{Mg}^{2+} (aq) || \text{Cu}^{2+} (aq) | \text{Cu}(s) \] \[ \left[ E^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37 \, \text{V}, E^\circ_{\text{Cu}^{2+}/\text{Cu}} = +0.34 \, \text{V}, F = 96500 \, \text{C mol}^{-1} \right] \]

Hint:

To calculate \(\Delta G\), use the relation \(\Delta G = -nFE_{\text{cell}}\), where \(n\) is the number of electrons, \(F\) is Faraday’s constant, and \(E_{\text{cell}}\) is the cell potential.

Answer 1

The cell potential is given by: \[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.34 - (-2.37) = 2.71 \, \text{V} \] The Gibbs free energy \(\Delta G\) is related to the cell potential \(E_{\text{cell}}\) by the equation: \[ \Delta G = -nFE_{\text{cell}} \] For this reaction, \(n = 2\) (2 electrons involved), and \(F = 96500 \, \text{C mol}^{-1}\). Thus: \[ \Delta G = -2 \times 96500 \times 2.71 = -522,870 \, \text{J/mol} \]