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Use standard reduction potentials to calculate the equilibrium constant for the reaction:

[tex]\[
\begin{array}{r}
Fe^{3+}(aq) + Cr^{2+}(aq) \rightarrow Fe^{2+}(aq) + Cr^{3+}(aq) \\
\begin{array}{|l|l|}
\hline
Fe^{3+}(aq) + e^{-} \rightarrow Fe^{2+}(aq) & E_{red}^{\circ} = 0.771\, V \\
\hline
Cr^{3+}(aq) + e^{-} \rightarrow Cr^{2+}(aq) & E_{red}^{\circ} = -0.410\, V \\
\hline
\end{array}
\end{array}
\][/tex]

Hint: Carry at least 5 significant figures during intermediate calculations to avoid round-off error when taking the antilogarithm.

Equilibrium constant at 298 K: [tex]$\square$[/tex]

[tex]$\Delta G^{\circ}$[/tex] for this reaction would be [tex]$\square$[/tex] than zero.

Sagot :

GinnyAnsweridn
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.
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