To calculate the equilibrium constant for the given redox reaction and determine the Gibbs free energy change for the reaction, follow these steps:
1. Write down the given reduction potentials:
[tex]\[
\begin{aligned}
\text{Reduction potential for } & \text{Fe}^{3+}/\text{Fe}^{2+}: & E_{\text{red}}^\circ & = 0.771 \, \text{V} \\
\text{Reduction potential for } & \text{Cr}^{3+}/\text{Cr}^{2+}: & E_{\text{red}}^\circ & = -0.410 \, \text{V}
\end{aligned}
\][/tex]
2. Determine the standard cell potential [tex]\(E_{\text{cell}}^\circ\)[/tex]:
[tex]\[
E_{\text{cell}}^\circ = E_{\text{red}}^\circ(\text{Fe}^{3+}/\text{Fe}^{2+}) - E_{\text{red}}^\circ(\text{Cr}^{3+}/\text{Cr}^{2+})
\][/tex]
[tex]\[
E_{\text{cell}}^\circ = 0.771 \, \text{V} - (-0.410 \, \text{V}) = 0.771 \, \text{V} + 0.410 \, \text{V} = 1.181 \, \text{V}
\][/tex]
3. Calculate the standard Gibbs free energy change [tex]\(\Delta G^\circ\)[/tex]:
The equation relating [tex]\( \Delta G^\circ \)[/tex] and [tex]\( E_{\text{cell}}^\circ \)[/tex] is:
[tex]\[
\Delta G^\circ = -n F E_{\text{cell}}^\circ
\][/tex]
Where:
- [tex]\( n \)[/tex] is the number of moles of electrons transferred (in this case, [tex]\( n = 1 \)[/tex]).
- [tex]\( F \)[/tex] is the Faraday constant [tex]\( (96485.33212 \, \text{C/mol}) \)[/tex].
Substituting the values:
[tex]\[
\Delta G^\circ = -(1) \times 96485.33212 \, \text{C/mol} \times 1.181 \, \text{V}
\][/tex]
[tex]\[
\Delta G^\circ = -113949.17723372001 \, \text{J/mol}
\][/tex]
4. Calculate the equilibrium constant, [tex]\( K_{\text{eq}} \)[/tex]:
The relationship between [tex]\( \Delta G^\circ \)[/tex] and [tex]\( K_{\text{eq}} \)[/tex] is given by:
[tex]\[
\Delta G^\circ = -RT \ln K_{\text{eq}}
\][/tex]
Where:
- [tex]\( R \)[/tex] is the gas constant (8.314 J/(mol K)).
- [tex]\( T \)[/tex] is the temperature (298 K).
Rearranging the equation to solve for [tex]\( K_{\text{eq}} \)[/tex]:
[tex]\[
K_{\text{eq}} = \exp\left(\frac{-\Delta G^\circ}{RT}\right)
\][/tex]
Substituting the values:
[tex]\[
K_{\text{eq}} = \exp\left(\frac{-(-113949.17723372001 \, \text{J/mol})}{8.314 \, \text{J/(mol K)} \times 298 \, \text{K}}\right)
\][/tex]
[tex]\[
K_{\text{eq}} = \exp\left(\frac{113949.17723372001}{2477.372}\right)
\][/tex]
[tex]\[
K_{\text{eq}} = \exp(46.003858)
\][/tex]
[tex]\[
K_{\text{eq}} = 9.423061542465036 \times 10^{19}
\][/tex]
5. Interpret the Gibbs free energy change [tex]\( \Delta G^\circ \)[/tex]:
Since [tex]\( \Delta G^\circ \)[/tex] is negative [tex]\((-113949.17723372001 \, \text{J/mol})\)[/tex], it indicates that the reaction is spontaneous under standard conditions.
Thus, the equilibrium constant at 298 K is [tex]\( 9.423061542465036 \times 10^{19} \)[/tex], and [tex]\( \Delta G^\circ \)[/tex] for this reaction is less than zero, indicating a spontaneous reaction.
