Question

Given the standard reduction potentials \((E^\Theta)\) for the half-cell reactions below, the standard Gibbs free energy of the dissolution of silver chloride in water, at 298 K, is ---- J mol\(^{-1}\) (rounded off to the nearest integer).

\[ \text{(Given: Faraday constant, } F = 96500 \text{ C mol}^{-1}; \quad J = C \times V) \]

The standard reduction half-cell reactions are:

\[ \text{AgCl(s) + e}^- \rightarrow \text{Ag(s) + Cl}^-(\text{aq}); \quad E^\Theta = 0.22 \text{ V at 298 K} \] \[ \text{Ag}^+(\text{aq}) + e^- \rightarrow \text{Ag(s)}; \quad E^\Theta = 0.80 \text{ V at 298 K} \]

Hint:

To solve problems involving Gibbs free energy and electrochemical cells: - Use the formula (Delta G^Theta = -nFE^Theta_{{cell}}). - Identify the number of electrons transferred ((n)). - Ensure the correct subtraction of standard reduction potentials to find (E^Theta_{{cell}}).

Answer 1

Step 1: Determine the overall cell reaction and potential \( (E^\Theta_{\text{cell}}) \).

The overall reaction for the dissolution of silver chloride is:

\[ \text{AgCl(s)} \rightarrow \text{Ag}^+(\text{aq}) + \text{Cl}^-(\text{aq}) \]

Using the given half-reactions, the overall cell potential \( (E^\Theta_{\text{cell}}) \) can be calculated as:

\[ E^\Theta_{\text{cell}} = E^\Theta_{\text{Ag}^+/\text{Ag}} - E^\Theta_{\text{AgCl}/\text{Ag}} \]

Substituting the given values:

\[ E^\Theta_{\text{cell}} = 0.80 \, \text{V} - 0.22 \, \text{V} = 0.58 \, \text{V}. \] Step 2: Calculate the Gibbs free energy \( (\Delta G^\Theta) \).

The relation between \( \Delta G^\Theta \) and \( E^\Theta_{\text{cell}} \) is:

\[ \Delta G^\Theta = -n F E^\Theta_{\text{cell}} \]

Here, \( n = 1 \) (number of electrons transferred), \( F = 96500 \, \text{C mol}^{-1} \), and \( E^\Theta_{\text{cell}} = 0.58 \, \text{V} \). Substituting the values:

\[ \Delta G^\Theta = -(1)(96500)(0.58) = -55970 \, \text{J mol}^{-1}. \] Step 3: Round the value to the nearest integer.

The Gibbs free energy is:

\[ \Delta G^\Theta = -55970 \, \text{J mol}^{-1}. \]