Courtnee A. answered • 04/04/24
Chemistry Expert with Bachelor's Degree
To calculate the total charge (in Coulombs) produced by the cell before reaching equilibrium, we first need to determine the moles of electrons transferred during the redox reaction and then use Faraday's constant to convert moles of electrons to Coulombs.
The balanced redox reaction involved in this cell is the reduction of copper ions (\( \text{Cu}^{2+} \)) to copper metal (\( \text{Cu} \)):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]
The number of moles of electrons transferred can be determined from the change in concentration of copper ions in each compartment. The Nernst equation is used to calculate the cell potential (\( E_{\text{cell}} \)):
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{n} \log\frac{[\text{Cu}^{2+}]_{\text{cathode}}}{[\text{Cu}^{2+}]_{\text{anode}}} \]
where:
- \( E_{\text{cell}} \) is the cell potential (V)
- \( E^\circ_{\text{cell}} \) is the standard cell potential (V)
- \( n \) is the number of moles of electrons transferred (moles of electrons in the balanced redox reaction)
- \( [\text{Cu}^{2+}]_{\text{cathode}} \) is the concentration of copper ions in the cathode compartment (M)
- \( [\text{Cu}^{2+}]_{\text{anode}} \) is the concentration of copper ions in the anode compartment (M)
Given data:
- Volume of cathode compartment (\( V_{\text{cathode}} \)) = 0.500 L
- Concentration of copper ions in cathode compartment (\( [\text{Cu}^{2+}]_{\text{cathode}} \)) = 1.00 M
- Volume of anode compartment (\( V_{\text{anode}} \)) = 0.825 L
- Concentration of copper ions in anode compartment (\( [\text{Cu}^{2+}]_{\text{anode}} \)) = 0.0500 M
- Standard cell potential (\( E^\circ_{\text{cell}} \)) for the reduction of copper ions = +0.34 V
First, calculate the cell potential using the Nernst equation:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0592}{2} \log\frac{1.00}{0.0500} \]
\[ E_{\text{cell}} = 0.34 - 0.0296 \log(20) \]
\[ E_{\text{cell}} = 0.34 - 0.0296 \times 1.301 \]
\[ E_{\text{cell}} = 0.34 - 0.0385 \]
\[ E_{\text{cell}} = 0.3015 \, \text{V} \]
Next, calculate the number of moles of electrons transferred (\( n \)) in the balanced redox reaction:
Since 1 mole of copper ions (\( \text{Cu}^{2+} \)) gains 2 moles of electrons to form 1 mole of copper (\( \text{Cu} \)), \( n = 2 \).
Now, we can use Faraday's constant (\( F \)) to convert moles of electrons to Coulombs:
\[ F = 96485 \, \text{C/mol} \]
The total charge (\( Q \)) produced by the cell is given by:
\[ Q = n \times F \times E_{\text{cell}} \]
\[ Q = 2 \times 96485 \times 0.3015 \]
\[ Q = 58206.27 \, \text{Coulombs} \]
Therefore, the cell can produce approximately 58206.27 Coulombs of electric charge before the compartments reach equilibrium at 25.0 °C.
