When an ion-selective electrode for X was immersed in 0.0210 M XCl, the measured potential was 0.0480 V. What is the concentration of X when the potential is 0.0630 V? Assume that the electrode follows the Nernst equation, the temperature is at 25°C, and that the activity coefficient of X is 1.
I would think you could use the Nernst equation, use the standard to calculate Eocell (they give you the concn XCl as well as Ecell). Then knowing Eocell you can calculate concn XCl when they give you Ecell for the unknown.
The Equation is going to be
E = constant + (0.05916)log[X]
find constant first
use that constant to find the new []
To determine the concentration of X when the potential is 0.0630 V, we can use the Nernst equation:
E = E° - (RT / nF) * ln(Q)
Where:
- E is the measured potential (0.0630 V)
- E° is the standard electrode potential
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (25°C = 298 K)
- n is the number of moles of electrons transferred in the electrochemical reaction (assume 1 for simplicity)
- F is the Faraday constant (96485 C/mol)
- ln() is the natural logarithm
- Q is the reaction quotient, which is equal to the concentration of X (C_x) divided by the concentration of XCl (C_xCl)
We can rearrange the Nernst equation to solve for C_x:
E = E° - (RT / nF) * ln(C_x / C_xCl)
First, we need to calculate E°. Since it is not given in the question, we'll assume it is zero (0).
E = 0 - (RT / nF) * ln(C_x / C_xCl)
Now we can substitute the given values:
0.0630 V = 0 - (8.314 J/(mol·K) * 298 K / (1 * 96485 C/mol)) * ln(C_x / 0.0210 M)
Simplifying:
0.0630 V = -(2.303 * 8.314 * 298 / 96485) * ln(C_x / 0.0210 M)
Now we can solve for ln(C_x / 0.0210 M):
ln(C_x / 0.0210 M) = -(0.002303 * 8.314 * 298 / 96485) * 0.0630
Calculating the right-hand side:
ln(C_x / 0.0210 M) ≈ -0.01024
Now, we can solve for C_x by taking the exponential of both sides:
C_x / 0.0210 M = exp(-0.01024)
Calculating the right-hand side:
exp(-0.01024) ≈ 0.98984
Finally, we can solve for C_x:
C_x ≈ 0.98984 * 0.0210 M
Calculating:
C_x ≈ 0.0207 M
Therefore, the concentration of X when the potential is 0.0630 V is approximately 0.0207 M.
To find the concentration of X when the potential is 0.0630 V, we can use the Nernst equation. The Nernst equation relates the measured potential of an ion-selective electrode to the concentration of the ion being measured.
The Nernst equation is given by:
E = E° - (RT / nF) * ln(Q)
Where:
- E is the measured potential
- E° is the standard electrode potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (25°C = 298.15 K)
- n is the number of moles of electrons transferred in the reaction
- F is the Faraday constant (96485 C/mol)
- Q is the reaction quotient, which is the ratio of the concentrations of the products and reactants raised to their respective stoichiometric coefficients
In this case, the reaction being measured by the ion-selective electrode is:
X⁺ + e⁻ → X
Since there is no information given about the stoichiometry of this reaction, we assume that only one electron is involved (n = 1).
First, we can calculate the standard electrode potential (E°) using the given concentration and measured potential:
E° = E - (RT / nF) * ln(Q)
E° = 0.0480 V - ((8.314 J/(mol·K)) * (298.15 K) / (1 * 96485 C/mol)) * ln(0.0210 M)
E° ≈ 0.0480 V - 0.001982 V * ln(0.0210 M)
Next, we can rearrange the Nernst equation to solve for the concentration of X (Q) when the potential is 0.0630 V:
Q = exp(((E° - E) * nF) / (RT))
Q = exp(((0.0480 V - 0.0630 V) * 1 * 96485 C/mol) / ((8.314 J/(mol·K)) * (298.15 K)))
Q ≈ exp(-982.560 C/(J·mol))
Finally, we can substitute the calculated value of Q into the Nernst equation to find the concentration of X:
E = E° - (RT / nF) * ln(Q)
0.0630 V = E° - ((8.314 J/(mol·K)) * (298.15 K) / (1 * 96485 C/mol)) * ln(Q)
Solve this equation to find the concentration of X when the potential is 0.0630 V.