Question

Consider the following reaction :
\( \text{Zn}_{(s)} + \text{Ag}_2\text{O}_{(s)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2\text{Ag}_{(s)} + 2\text{OH}^-_{(aq)} \)
Given : \( E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \, \text{V} \), \( E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \, \text{V} \), \( 1 \, \text{F} = 96500 \, \text{C mol}^{-1} \).
\( \Delta_r G^\circ \) for the above reaction is :

• A. \( -301.080 \, \text{kJ mol}^{-1} \)
• B. \( +310.080 \, \text{kJ mol}^{-1} \)
• C. \( -326.070 \, \text{kJ mol}^{-1} \)
• D. \( -375.060 \, \text{kJ mol}^{-1} \)

Hint:

The unit of \( \Delta G \) calculated from \( nFE \) is Joules (J). Always remember to convert it to kiloJoules (kJ) as most multiple-choice options are provided in kJ.

Answer 1

The Correct Option is A

Concept: The standard Gibbs energy change (\( \Delta_r G^\circ \)) is the maximum work that can be obtained from a chemical reaction. In electrochemistry, it is calculated using the formula: \[ \Delta_r G^\circ = -nFE^\circ_{\text{cell}} \] where \( n \) is the number of electrons transferred, \( F \) is Faraday's constant, and \( E^\circ_{\text{cell}} \) is the standard cell potential. Step 1: Determining the standard cell potential (\( E^\circ_{\text{cell}} \)).
First, identify the half-cells:

• Anode (Oxidation): \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \); \( E^\circ_{\text{anode}} = -0.76 \, \text{V} \)

• Cathode (Reduction): Silver ions are reduced; \( E^\circ_{\text{cathode}} = 0.80 \, \text{V} \)
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = 0.80 \, \text{V} - (-0.76 \, \text{V}) = 1.56 \, \text{V} \]

Step 2: Identifying the value of \( n \).
The balanced equation shows that Zinc loses 2 electrons and two Silver atoms are formed (each Ag\(^+\) in the oxide gaining 1 electron, total 2 electrons). Thus, \( n = 2 \).

Step 3: Calculating \( \Delta_r G^\circ \) and converting units.
Using the formula: \[ \Delta_r G^\circ = -2 \times 96500 \, \text{C/mol} \times 1.56 \, \text{V} \] \[ \Delta_r G^\circ = -301080 \, \text{J/mol} \] Convert Joules to kiloJoules (kJ) by dividing by 1000: \[ \Delta_r G^\circ = -301.080 \, \text{kJ/mol} \] Final Answer: \( \Delta_r G^\circ \) for the reaction is -301.080 kJ mol\(^{-1}\).