Electrochemical cells are devices that convert chemical energy into electrical energy through redox reactions. One key concept in understanding these cells is cell potential (E), which indicates the voltage produced by an electrochemical reaction under standard conditions. Calculating cell potential is crucial for predicting the feasibility and efficiency of various electrochemical processes. Below are three diverse examples that illustrate the calculation of cell potential in electrochemical cells.
In this example, we will calculate the cell potential for a galvanic cell consisting of a standard hydrogen electrode (SHE) and a copper(II) ion solution. This scenario is common in electrochemistry and helps establish a baseline for measuring electrode potentials.
The half-reaction for the reduction of copper ions is:
\[ Cu^{2+} + 2e^{-} \rightarrow Cu \quad E^{\circ} = +0.34 V \]
The half-reaction for the oxidation of hydrogen is:
\[ 2H^{+} + 2e^{-} \rightarrow H_{2} \quad E^{\circ} = 0.00 V \]
To calculate the overall cell potential, we use the formula:
\[ E_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
Here, the copper half-reaction serves as the cathode and the hydrogen half-reaction is the anode. Plugging in the values:
\[ E_{cell} = 0.34 V - 0.00 V = 0.34 V \]
This positive cell potential indicates that the reaction is spontaneous, making it a practical example of a standard electrochemical cell.
Notes: The standard conditions for this experiment include a temperature of 25°C and 1 M concentrations of the reacting species.
This example explores the galvanic cell formed by zinc and silver nitrate. This setup is commonly used in laboratory demonstrations of electrochemical principles.
The half-reaction for the reduction of silver ions is:
\[ Ag^{+} + e^{-} \rightarrow Ag \quad E^{\circ} = +0.80 V \]
The half-reaction for the oxidation of zinc is:
\[ Zn \rightarrow Zn^{2+} + 2e^{-} \quad E^{\circ} = -0.76 V \]
Using the cell potential formula:
\[ E_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
In this case, silver acts as the cathode and zinc as the anode. Therefore:
\[ E_{cell} = 0.80 V - (-0.76 V) \] \[ E_{cell} = 0.80 V + 0.76 V = 1.56 V \]
This high cell potential indicates a highly favorable reaction, demonstrating the effectiveness of this electrochemical setup.
Notes: The zinc electrode will corrode over time, producing zinc ions in solution, while silver will deposit on the cathode.
In this example, we will calculate the cell potential for a lead-acid battery, which is widely used in automotive applications. This battery involves lead and lead dioxide electrodes immersed in sulfuric acid.
The half-reaction for the reduction of lead dioxide is:
\[ PbO_{2} + 3H^{+} + 2e^{-} \rightarrow PbSO_{4} + 2H_{2}O \quad E^{\circ} = +1.685 V \]
The half-reaction for the oxidation of lead is:
\[ Pb + SO_{4}^{2-} \rightarrow PbSO_{4} + 2e^{-} \quad E^{\circ} = -0.356 V \]
Calculating the cell potential:
\[ E_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode} \]
Here, lead dioxide acts as the cathode and lead as the anode:
\[ E_{cell} = 1.685 V - (-0.356 V) \] \[ E_{cell} = 1.685 V + 0.356 V = 2.041 V \]
This substantial cell potential indicates that the lead-acid battery can deliver a significant amount of energy, making it an effective power source.
Notes: The lead-acid battery operates efficiently within specific conditions, and the potential can vary based on the state of charge and temperature.