QuestionJuly 30, 2025

Calculate the value of E^circ and Delta G^circ for the following reaction. 2Fe^2+(aq)+Br_(2)(aq)leftharpoons 2Fe^3+(aq)+2Br^-(aq) K=4.85times 10^10

Calculate the value of E^circ and Delta G^circ for the following reaction. 2Fe^2+(aq)+Br_(2)(aq)leftharpoons 2Fe^3+(aq)+2Br^-(aq) K=4.85times 10^10
Calculate the value of E^circ and Delta G^circ for the following reaction. 2Fe^2+(aq)+Br_(2)(aq)leftharpoons 2Fe^3+(aq)+2Br^-(aq) K=4.85times 10^10

Solution
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Answer

E^{\circ} \approx 0.312 \, \text{V}; \Delta G^{\circ} \approx -60200 \, \text{J/mol} Explanation 1. Calculate E^{\circ} using the Nernst Equation Use the formula **E^{\circ} = \frac{RT}{nF} \ln K**. For standard conditions, R = 8.314 \, \text{J/mol K}, T = 298 \, \text{K}, F = 96485 \, \text{C/mol}, and n = 2. Substitute K = 4.85 \times 10^{10}. \[ E^{\circ} = \frac{(8.314)(298)}{(2)(96485)} \ln(4.85 \times 10^{10}) \] \[ E^{\circ} = \frac{2478.572}{192970} \times 24.401 \] \[ E^{\circ} \approx 0.312 \, \text{V} \] 2. Calculate \Delta G^{\circ} using the relation with E^{\circ} Use the formula **\Delta G^{\circ} = -nFE^{\circ}**. \[ \Delta G^{\circ} = -(2)(96485)(0.312) \] \[ \Delta G^{\circ} \approx -60200 \, \text{J/mol} \]

Explanation

1. Calculate $E^{\circ}$ using the Nernst Equation<br /> Use the formula **$E^{\circ} = \frac{RT}{nF} \ln K$**. For standard conditions, $R = 8.314 \, \text{J/mol K}$, $T = 298 \, \text{K}$, $F = 96485 \, \text{C/mol}$, and $n = 2$. Substitute $K = 4.85 \times 10^{10}$.<br /><br />\[ E^{\circ} = \frac{(8.314)(298)}{(2)(96485)} \ln(4.85 \times 10^{10}) \]<br /><br />\[ E^{\circ} = \frac{2478.572}{192970} \times 24.401 \]<br /><br />\[ E^{\circ} \approx 0.312 \, \text{V} \]<br /><br />2. Calculate $\Delta G^{\circ}$ using the relation with $E^{\circ}$<br /> Use the formula **$\Delta G^{\circ} = -nFE^{\circ}$**.<br /><br />\[ \Delta G^{\circ} = -(2)(96485)(0.312) \]<br /><br />\[ \Delta G^{\circ} \approx -60200 \, \text{J/mol} \]
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