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To calculate the standard Gibbs free energy change (ΔrG⁰) and the equilibrium constant (log Kc) for the given cell reaction, we follow these steps:
### Step 1: Calculate ΔrG⁰
The standard Gibbs free energy change ΔrG⁰ for the cell reaction can be calculated using the following relationship:
[tex]\[ \Delta_{ r} G ^0 = -nFE_{ \text{cell} } \][/tex]
where:
- [tex]\( n \)[/tex] is the number of moles of electrons transferred in the reaction.
- [tex]\( F \)[/tex] is the Faraday constant (96500 C mol⁻¹).
- [tex]\( E_{\text{cell}} \)[/tex] is the standard cell potential in volts (1.10 V).
Let's plug in the given values:
- [tex]\( n = 2 \)[/tex] (as an example value)
- [tex]\( F = 96500 \)[/tex] C mol⁻¹
- [tex]\( E_{\text{cell}} = 1.10 \)[/tex] V
Thus,
[tex]\[ \Delta_{ r} G ^0 = - (2) (96500) (1.10) \, \text{J} \][/tex]
Calculating the value, we get:
[tex]\[ \Delta_{ r} G ^0 = -212300.00000000003 \, \text{J} \][/tex]
### Step 2: Calculate log Kc
The equilibrium constant Kc can be related to the standard cell potential through the following formula:
[tex]\[ \log K_{ c } = \frac{n FE_{ \text{cell} }}{RT \ln 10 } \][/tex]
However, for simplicity, we can directly compute the natural logarithm and convert it to log base 10:
[tex]\[ \log K_{ c } = \frac{n FE_{ \text{cell} }}{RT} \][/tex]
where:
- [tex]\( R \)[/tex] is the gas constant (8.314 J K⁻¹ mol⁻¹).
- [tex]\( T \)[/tex] is the temperature in Kelvin (298 K).
Using the same values:
- [tex]\( n = 2 \)[/tex]
- [tex]\( F = 96500 \)[/tex] C mol⁻¹
- [tex]\( E_{\text{cell}} = 1.10 \)[/tex] V
- [tex]\( R = 8.314 \)[/tex] J K⁻¹ mol⁻¹
- [tex]\( T = 298 \)[/tex] K
Let's plug in these values:
[tex]\[ \log K_{ c } = \frac{(2)(96500)(1.10)}{(8.314)(298)} \][/tex]
Calculating the value, we get:
[tex]\[ \log K_{ c } = 85.6887307412257 \][/tex]
### Conclusion
Thus, the calculated values are:
- [tex]\(\Delta_{ r} G ^0 = -212300.00000000003 \, \text{J} \)[/tex]
- [tex]\(\log K_{ c } = 85.6887307412257\)[/tex]
These are the standard Gibbs free energy change and the logarithm of the equilibrium constant for the cell reaction given the provided parameters.
### Step 1: Calculate ΔrG⁰
The standard Gibbs free energy change ΔrG⁰ for the cell reaction can be calculated using the following relationship:
[tex]\[ \Delta_{ r} G ^0 = -nFE_{ \text{cell} } \][/tex]
where:
- [tex]\( n \)[/tex] is the number of moles of electrons transferred in the reaction.
- [tex]\( F \)[/tex] is the Faraday constant (96500 C mol⁻¹).
- [tex]\( E_{\text{cell}} \)[/tex] is the standard cell potential in volts (1.10 V).
Let's plug in the given values:
- [tex]\( n = 2 \)[/tex] (as an example value)
- [tex]\( F = 96500 \)[/tex] C mol⁻¹
- [tex]\( E_{\text{cell}} = 1.10 \)[/tex] V
Thus,
[tex]\[ \Delta_{ r} G ^0 = - (2) (96500) (1.10) \, \text{J} \][/tex]
Calculating the value, we get:
[tex]\[ \Delta_{ r} G ^0 = -212300.00000000003 \, \text{J} \][/tex]
### Step 2: Calculate log Kc
The equilibrium constant Kc can be related to the standard cell potential through the following formula:
[tex]\[ \log K_{ c } = \frac{n FE_{ \text{cell} }}{RT \ln 10 } \][/tex]
However, for simplicity, we can directly compute the natural logarithm and convert it to log base 10:
[tex]\[ \log K_{ c } = \frac{n FE_{ \text{cell} }}{RT} \][/tex]
where:
- [tex]\( R \)[/tex] is the gas constant (8.314 J K⁻¹ mol⁻¹).
- [tex]\( T \)[/tex] is the temperature in Kelvin (298 K).
Using the same values:
- [tex]\( n = 2 \)[/tex]
- [tex]\( F = 96500 \)[/tex] C mol⁻¹
- [tex]\( E_{\text{cell}} = 1.10 \)[/tex] V
- [tex]\( R = 8.314 \)[/tex] J K⁻¹ mol⁻¹
- [tex]\( T = 298 \)[/tex] K
Let's plug in these values:
[tex]\[ \log K_{ c } = \frac{(2)(96500)(1.10)}{(8.314)(298)} \][/tex]
Calculating the value, we get:
[tex]\[ \log K_{ c } = 85.6887307412257 \][/tex]
### Conclusion
Thus, the calculated values are:
- [tex]\(\Delta_{ r} G ^0 = -212300.00000000003 \, \text{J} \)[/tex]
- [tex]\(\log K_{ c } = 85.6887307412257\)[/tex]
These are the standard Gibbs free energy change and the logarithm of the equilibrium constant for the cell reaction given the provided parameters.
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