To find the standard Gibbs free energy change ([tex]\(\Delta G^{\circ}\)[/tex]) for the reaction:
[tex]\[ 2 Cr^{3+}(aq) + 6 Hg(l) \rightarrow 2 Cr(s) + 3 Hg_2^{2+}(aq) \][/tex]
we can use the relationship between the standard Gibbs free energy change [tex]\(\Delta G^{\circ}\)[/tex] and the standard cell potential [tex]\(E^{\circ}\)[/tex]:
[tex]\[ \Delta G^{\circ} = -nFE^{\circ} \][/tex]
where:
- [tex]\( \Delta G^{\circ} \)[/tex] is the standard Gibbs free energy change,
- [tex]\( n \)[/tex] is the number of moles of electrons transferred in the reaction,
- [tex]\( F \)[/tex] is Faraday's constant ([tex]\( F = 96485 \, \text{C/mol} \)[/tex]),
- [tex]\( E^{\circ} \)[/tex] is the standard cell potential.
### Step-by-Step Solution:
1. Determine the number of electrons transferred ([tex]\( n \)[/tex]):
In the reaction, the change in oxidation state for chromium and mercury indicates the number of electrons transferred.
[tex]\[
2 Cr^{3+} + 6 e^- \rightarrow 2 Cr \quad \text{(each Cr^{3+} gains 3 electrons, 2 Cr^{3+} gain 6 electrons)}
\][/tex]
Therefore, [tex]\( n = 6 \)[/tex] electrons are transferred.
2. Given Data:
- Standard cell potential, [tex]\(E^{\circ}\)[/tex] = 1.59 V
- Faraday's constant, [tex]\(F = 96485 \, \text{C/mol}\)[/tex]
3. Substitute the values into the formula:
[tex]\[
\Delta G^{\circ} = -n \times F \times E^{\circ}
\][/tex]
[tex]\[
\Delta G^{\circ} = -6 \times 96485 \, \text{C/mol} \times 1.59 \, \text{V}
\][/tex]
4. Calculate the change in Gibbs free energy ([tex]\(\Delta G^{\circ}\)[/tex]):
[tex]\[
\Delta G^{\circ} = -6 \times 96485 \times 1.59
\][/tex]
[tex]\[
\Delta G^{\circ} = -920466.9 \, \text{J/mol}
\][/tex]
5. Convert Joules to Kilojoules (since 1 kJ = 1000 J):
[tex]\[
\Delta G^{\circ} = \frac{-920466.9}{1000} \, \text{kJ/mol}
\][/tex]
[tex]\[
\Delta G^{\circ} = -920.4669 \, \text{kJ/mol}
\][/tex]
6. Conclusion:
The closest value to [tex]\(\Delta G^{\circ} = -920.4669 \, \text{kJ/mol}\)[/tex] from the given options is:
[tex]\[
\boxed{-921 \, \text{kJ}}
\][/tex]
