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

To find the standard Gibbs free energy change, \(\Delta G^\circ\), for the given reaction, we can use the Nernst equation in its standard form:

\(\Delta G^\circ = -nFE^\circ_{\text{cell}}\) 

where:

• \(n\) is the number of moles of electrons exchanged (for the given reaction, \(n = 2\)),
• \(F\) is Faraday's constant, given as \(96500 \text{ C mol}^{-1}\),
• \(E^\circ_{\text{cell}}\) is the standard cell potential.

Step 1: Calculate \(E^\circ_{\text{cell}}\):

The overall cell potential is calculated using the standard electrode potentials of the individual half-reactions.

Reduction Half-Cell	Electrode Potential
\(\text{Ag}^+ + e^- \rightarrow \text{Ag} \quad E^\circ = +0.80\, \text{V}\)	Reduction
\(\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \quad E^\circ = -0.76\, \text{V}\)	Oxidation (reverse of reduction)

The reaction given is:

\(\text{Zn}_{(s)} + \text{Ag}_2\text{O}_{(s)} + \text{H}_2\text{O}_{(l)} \rightarrow \text{Zn}^{2+}_{(aq)} + 2\text{Ag}_{(s)} + 2\text{OH}^-_{(aq)}\)

First, break it down in terms of half-reactions to get the reduction and oxidation potentials:

The complete cell reaction is:

1. \(\text{Ag}_2\text{O} + \text{H}_2\text{O} + 2e^- \rightarrow 2\text{Ag} + 2\text{OH}^- \quad E^\circ = 0.80\, \text{V}\)
2. \(\text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \quad E^\circ = -0.76\, \text{V} \, (\text{oxidation})\)

Therefore, the overall standard cell potential is:

\(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = 0.80 \, \text{V} - (-0.76 \, \text{V}) = 1.56 \, \text{V}\)

Step 2: Calculate \(\Delta G^\circ\):

\(\Delta G^\circ = -nFE^\circ_{\text{cell}} = -(2 \times 96500 \, \text{C mol}^{-1} \times 1.56\, \text{V})\)

This becomes:

\(\Delta G^\circ = -301080 \, \text{J mol}^{-1} = -301.080 \, \text{kJ mol}^{-1}\)

Thus, the standard Gibbs free energy change for the reaction is -301.080 kJ/mol.

Therefore, the correct answer is: \(-301.080 \, \text{kJ mol}^{-1}\).