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

Step 1: Splitting the reaction into its two half-cells.
Zinc is oxidised, so it is the anode, with $E^{\circ}_{anode} = -0.76\ V$ (for $Zn^{2+}/Zn$). Silver ions are reduced to silver metal at the cathode, with $E^{\circ}_{cathode} = 0.80\ V$.
Step 2: Getting the standard cell potential. \[ E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} = 0.80 - (-0.76) = 1.56\ V \]
Step 3: Counting the electrons transferred.
Zinc gives up two electrons as it turns into $Zn^{2+}$, and those two electrons are what reduce the two silver centres in $Ag_2O$ down to metallic silver, so $n = 2$ electrons move through the circuit per formula unit reacting.
Step 4: Plugging into the Gibbs energy relation. \[ \Delta_r G^{\circ} = -nFE^{\circ}_{cell} = -(2)(96500)(1.56) = -301080\ J/mol \] Converting to kilojoules by dividing by 1000 gives the final value. \[ \boxed{-301.080\ \text{kJ mol}^{-1}} \]