what's the cell potential of a copper electrolytic cell if both electrodes are copper? (the e0 value of the copper reaction is 0.34V)
It depends upon the concentration of the electrolyte in each half cell. If each electrode is dipping into a 1 M solution of Cu^2+ then the cell potential will be zero. Look up cell potential in a "concentration cell" on Google or in your text book/notes for a full description.
To calculate the cell potential of a copper electrolytic cell where both electrodes are copper, you can use the Nernst equation. The Nernst equation is a mathematical relationship that relates the standard cell potential to the cell potential under non-standard conditions.
The equation is:
Ecell = Eºcell - (0.0592/n) * log(Q)
Where:
Ecell is the cell potential under non-standard conditions
Eºcell is the standard cell potential
0.0592 is a constant at 25°C (known as the "Nernst constant")
n is the number of electrons transferred in the balanced redox equation
Q is the reaction quotient
In this case, since both electrodes are copper, the redox reaction is:
Cu(s) → Cu2+(aq) + 2e-
From this equation, we can determine that n = 2 (two electrons are transferred).
The reaction quotient, Q, can be determined by calculating the concentration ratio of the products (Cu2+) to the reactant (Cu(s)).
Since both electrodes are made of copper, their concentrations are the same, so the ratio is 1.
Therefore, Q = 1.
Now, substituting the values into the Nernst equation:
Ecell = Eºcell - (0.0592/2) * log(1)
Assuming the Eº value of the copper reaction is 0.34V:
Ecell = 0.34V - (0.0592/2) * log(1)
Ecell = 0.34V - (0.0296) * log(1)
Ecell ≈ 0.34V
Therefore, the cell potential of a copper electrolytic cell with both electrodes made of copper is approximately 0.34V.
To determine the cell potential of a copper electrolytic cell with both electrodes made of copper, you can use the Nernst equation. The Nernst equation relates the cell potential to the standard cell potential and the concentrations of the reactants and products.
Before applying the Nernst equation, we need to identify the half-reactions taking place at both electrodes:
1. At the anode (oxidation): Cu(s) → Cu2+(aq) + 2e^-
2. At the cathode (reduction): Cu2+(aq) + 2e^- → Cu(s)
Given that both electrodes are made of copper, the concentration of Cu2+ ions in the electrolyte is essentially zero. Therefore, we can assume the concentration of Cu2+ (aq) to be 1M at the anode and 0.1M at the cathode (since copper sulfate solutions are commonly used in electrolytic cells).
Now, let's apply the Nernst equation:
Ecell = E°cell - (RT / (nF)) * ln(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential (0.34V)
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (usually 298 K)
- n is the number of electrons transferred in the balanced half-reaction (2 in this case)
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, which can be calculated as Q = [Cu2+]/[Cu] = 0.1/1 = 0.1
Plugging in the given values:
Ecell = 0.34V - (8.314 J/(mol·K) * 298 K / (2 * 96,485 C/mol)) * ln(0.1)
Calculating this equation will give you the cell potential of the copper electrolytic cell with both electrodes made of copper.