Question

Calculate standard Gibbs energy change at 25°C for the cell reaction \[ \text{Cd(s)} + \text{Sn}^{2+}_{(aq)} \rightleftharpoons \text{Cd}^{2+}_{(aq)} + \text{Sn(s)} \] Given: \[ E^\circ_{\text{cd}} = -0.403V, \quad E^\circ_{\text{sn}} = -0.136V \]

Hint:

To calculate Gibbs free energy change from cell potential, use the relation \( \Delta G^\circ = -n F E^\circ_{\text{cell}} \).

Answer 1

Step 1: The standard Gibbs free energy change \( \Delta G^\circ \) is related to the standard cell potential \( E^\circ_{\text{cell}} \) by the equation: \[ \Delta G^\circ = -n F E^\circ_{\text{cell}} \] Where: - \( n \) is the number of moles of electrons transferred. - \( F \) is the Faraday constant (\( 96485 \, \text{C/mol} \)). - \( E^\circ_{\text{cell}} \) is the standard cell potential. Step 2: Calculate \( E^\circ_{\text{cell}} \). For the cell reaction: \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] Here: - The cathode is where reduction occurs (Cd\(^{2+}\) to Cd). - The anode is where oxidation occurs (Sn to Sn\(^{2+}\)). \[ E^\circ_{\text{cell}} = E^\circ_{\text{Cd}} - E^\circ_{\text{Sn}} = (-0.403 \, \text{V}) - (-0.136 \, \text{V}) = -0.403 + 0.136 = -0.267 \, \text{V} \] Step 3: For this reaction, 2 electrons are transferred (\( n = 2 \)). Step 4: Now calculate the Gibbs free energy change: \[ \Delta G^\circ = -2 \times 96485 \times (-0.267) \, \text{J/mol} \] \[ \Delta G^\circ = 2 \times 96485 \times 0.267 = 51571.5 \, \text{J/mol} = 51.57 \, \text{kJ/mol} \] Thus, the standard Gibbs energy change is: \[ \boxed{\Delta G^\circ = 51.57 \, \text{kJ/mol}} \]