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To determine the standard Gibbs free energy change ([tex]\( \Delta G^\circ \)[/tex]) for the given reaction:
[tex]\[ \text{Zn (s) + Cu}^{2+} \text{(aq)} \rightarrow \text{Zn}^{2+} \text{(aq) + Cu (s)} \][/tex]
we proceed through the following steps:
### Step 1: Identify Standard Reduction Potentials
The standard reduction potentials ([tex]\( E^\circ \)[/tex]) for the half-reactions are:
- [tex]\( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)[/tex] [tex]\( E^\circ = 0.34 \, \text{V} \)[/tex]
- [tex]\( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \)[/tex] [tex]\( E^\circ = -0.76 \, \text{V} \)[/tex]
### Step 2: Determine the Overall Cell Potential
The overall cell potential [tex]\( E_{\text{cell}}^\circ \)[/tex] can be calculated using the standard reduction potentials of the two half-reactions. For the given reaction:
[tex]\[ \text{Zn (s) + Cu}^{2+} \text{(aq)} \rightarrow \text{Zn}^{2+} \text{(aq) + Cu (s)} \][/tex]
the cell potential is given by the difference:
[tex]\[ E_{\text{cell}}^\circ = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} \][/tex]
Plugging in the given standard reduction potentials:
[tex]\[ E_{\text{cell}}^\circ = 0.34 \, \text{V} - (-0.76 \, \text{V}) \][/tex]
[tex]\[ E_{\text{cell}}^\circ = 0.34 \, \text{V} + 0.76 \, \text{V} \][/tex]
[tex]\[ E_{\text{cell}}^\circ = 1.10 \, \text{V} \][/tex]
### Step 3: Use Faraday's Constant
Faraday's constant ([tex]\( F \)[/tex]) is:
[tex]\[ F = 96485 \, \text{C/mol e}^- \][/tex]
### Step 4: Determine Number of Electrons Transferred
In the given electrochemical reaction, 2 moles of electrons are transferred ([tex]\( n = 2 \)[/tex]).
### Step 5: Calculate Standard Gibbs Free Energy Change
The standard Gibbs free energy change ([tex]\( \Delta G^\circ \)[/tex]) is related to the standard cell potential by the equation:
[tex]\[ \Delta G^\circ = -nFE_{\text{cell}}^\circ \][/tex]
Plugging in the values:
[tex]\[ \Delta G^\circ = - (2 \, \text{mol e}^-) \times (96485 \, \text{C/mol e}^-) \times (1.10 \, \text{V}) \][/tex]
Perform the multiplication:
[tex]\[ \Delta G^\circ = -2 \times 96485 \times 1.10 \][/tex]
[tex]\[ \Delta G^\circ = -212267 \, \text{J} \][/tex]
Hence, the standard Gibbs free energy change for the reaction is:
[tex]\[ \Delta G^\circ = -212267 \, \text{J} \][/tex]
Or equivalently:
[tex]\[ \Delta G^\circ = -212.267 \, \text{kJ} \][/tex]
Therefore, the standard Gibbs free energy change for the given reaction is [tex]\(-212267 \, \text{J}\)[/tex] (or [tex]\(-212.267 \, \text{kJ}\)[/tex]).
[tex]\[ \text{Zn (s) + Cu}^{2+} \text{(aq)} \rightarrow \text{Zn}^{2+} \text{(aq) + Cu (s)} \][/tex]
we proceed through the following steps:
### Step 1: Identify Standard Reduction Potentials
The standard reduction potentials ([tex]\( E^\circ \)[/tex]) for the half-reactions are:
- [tex]\( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)[/tex] [tex]\( E^\circ = 0.34 \, \text{V} \)[/tex]
- [tex]\( \text{Zn}^{2+} + 2e^- \rightarrow \text{Zn} \)[/tex] [tex]\( E^\circ = -0.76 \, \text{V} \)[/tex]
### Step 2: Determine the Overall Cell Potential
The overall cell potential [tex]\( E_{\text{cell}}^\circ \)[/tex] can be calculated using the standard reduction potentials of the two half-reactions. For the given reaction:
[tex]\[ \text{Zn (s) + Cu}^{2+} \text{(aq)} \rightarrow \text{Zn}^{2+} \text{(aq) + Cu (s)} \][/tex]
the cell potential is given by the difference:
[tex]\[ E_{\text{cell}}^\circ = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - E^\circ_{\text{Zn}^{2+}/\text{Zn}} \][/tex]
Plugging in the given standard reduction potentials:
[tex]\[ E_{\text{cell}}^\circ = 0.34 \, \text{V} - (-0.76 \, \text{V}) \][/tex]
[tex]\[ E_{\text{cell}}^\circ = 0.34 \, \text{V} + 0.76 \, \text{V} \][/tex]
[tex]\[ E_{\text{cell}}^\circ = 1.10 \, \text{V} \][/tex]
### Step 3: Use Faraday's Constant
Faraday's constant ([tex]\( F \)[/tex]) is:
[tex]\[ F = 96485 \, \text{C/mol e}^- \][/tex]
### Step 4: Determine Number of Electrons Transferred
In the given electrochemical reaction, 2 moles of electrons are transferred ([tex]\( n = 2 \)[/tex]).
### Step 5: Calculate Standard Gibbs Free Energy Change
The standard Gibbs free energy change ([tex]\( \Delta G^\circ \)[/tex]) is related to the standard cell potential by the equation:
[tex]\[ \Delta G^\circ = -nFE_{\text{cell}}^\circ \][/tex]
Plugging in the values:
[tex]\[ \Delta G^\circ = - (2 \, \text{mol e}^-) \times (96485 \, \text{C/mol e}^-) \times (1.10 \, \text{V}) \][/tex]
Perform the multiplication:
[tex]\[ \Delta G^\circ = -2 \times 96485 \times 1.10 \][/tex]
[tex]\[ \Delta G^\circ = -212267 \, \text{J} \][/tex]
Hence, the standard Gibbs free energy change for the reaction is:
[tex]\[ \Delta G^\circ = -212267 \, \text{J} \][/tex]
Or equivalently:
[tex]\[ \Delta G^\circ = -212.267 \, \text{kJ} \][/tex]
Therefore, the standard Gibbs free energy change for the given reaction is [tex]\(-212267 \, \text{J}\)[/tex] (or [tex]\(-212.267 \, \text{kJ}\)[/tex]).
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