To calculate the concentration of Cu²⁺ ions when the cell potential (E₀cell) is 0.070 V, we can use the Nernst equation, which relates the reduction potential to the concentrations of the reactants and products in an electrochemical cell.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
- E is the cell potential under non-standard conditions (0.070 V in this case).
- E° is the standard cell potential. For a hydrogen electrode, E° = 0 V; the standard reduction potential for Cu²⁺/Cu is approximately +0.34 V.
- R is the universal gas constant (8.314 J/(mol·K)).
- T is the temperature in Kelvin (assuming 298 K for standard temperature conditions).
- n is the number of moles of electrons transferred (2 for Cu²⁺/Cu).
- F is the Faraday’s constant (96485 C/mol).
- Q is the reaction quotient, which is the ratio of the concentrations of products to reactants.
Given that we’ll consider the standard conditions, we can plug the values into the equation:
0.070 = 0.34 - (8.314 * 298 / (2 * 96485)) * ln(Q)
First, let’s calculate the term involving constants:
(8.314 * 298) / (2 * 96485) ≈ 0.00413
Substituting this back in gives us:
0.070 = 0.34 - 0.00413 ln(Q)
Rearranging it to isolate ln(Q):
ln(Q) = (0.34 - 0.070) / 0.00413
Calculate the right side:
ln(Q) ≈ (0.27) / 0.00413 ≈ 65.5
Now, exponentiate to solve for Q:
Q = e^(65.5) ≈ 4.62 × 10^28
Since Q for the half-cell reaction is defined as:
Q = 1 / [Cu²⁺]
This means:
[Cu²⁺] = 1 / Q ≈ 1 / (4.62 × 10^28) ≈ 2.16 × 10^-29 M
Therefore, the concentration of Cu²⁺ ions when the cell potential is 0.070 V is approximately 2.16 × 10^-29 M.