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: \[ E^\circ_{\text{Ag}^+/\text{Ag}} = 0.80 \text{ V} \] \[ E^\circ_{\text{Zn}^{2+}/\text{Zn}} = -0.76 \text{ V} \] \[ 1F = 96500 \text{ C mol}^{-1} \] $\Delta G^\circ$ for 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:

Use $\Delta G^\circ = -nFE^\circ_{\text{cell}}$. If $E^\circ_{\text{cell}}$ is positive, $\Delta G^\circ$ is negative, indicating a spontaneous reaction.

Answer 1

The Correct Option is A

Concept: The relationship between standard Gibbs free energy and cell potential is: \[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \] where:

• $n$ = number of electrons transferred
• $F$ = Faraday constant (96500 C mol$^{-1}$)
• $E^\circ_{\text{cell}}$ = standard cell potential

Step 1: Identify oxidation and reduction reactions. Zinc is oxidized: \[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad (\text{Anode}) \] Silver ion (from Ag$_2$O) is reduced: \[ \text{Ag}^+ + e^- \rightarrow \text{Ag} \quad (\text{Cathode}) \] 
Step 2: Calculate standard cell potential. \[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \] \[ E^\circ_{\text{cell}} = 0.80 - (-0.76) = 1.56 \text{ V} \] 
Step 3: Find number of electrons transferred. From Zn $\rightarrow$ Zn$^{2+}$, 2 electrons are transferred. So, $n = 2$. 
Step 4: Calculate $\Delta G^\circ$. \[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \] \[ \Delta G^\circ = -(2)(96500)(1.56) \] \[ \Delta G^\circ = -301080 \text{ J mol}^{-1} \] \[ \Delta G^\circ = -301.080 \text{ kJ mol}^{-1} \]