The voltage generated by the zinc concentration cell described by,

Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)

is 11.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.

Ecell = Eocell - (0.0592/2)log(x)/(Zn^2+)

0.011 = 0 -(0.0592/2)log(0.1/x)
Solve for x.

To calculate the concentration of the Zn2+ ion at the cathode, we need to use the Nernst equation, which relates the voltage of an electrochemical cell to the concentration of the ions involved.

The Nernst equation is given by:

E = E° - (RT / nF) * ln(Q)

Where:
E: Voltage of the cell
E°: Standard cell potential
R: Gas constant (8.314 J/(mol∙K))
T: Temperature in Kelvin
n: Number of electrons transferred in the balanced equation for the cell reaction
F: Faraday constant (96485 C/mol)
ln: Natural logarithm
Q: Reaction quotient

In this case, we have a concentration cell, which means that the zinc electrode on the left is the anode and the zinc electrode on the right is the cathode. Since they are both pure zinc electrodes, their concentrations are 1 M.

The balanced equation for the cell reaction is:

Zn(s) → Zn2+(aq) + 2e-

The number of electrons transferred (n) is 2.

Substituting the given information into the Nernst equation:

11.0 mV = E° - (8.314 J/(mol∙K)) * (298 K) / (2 * (96485 C/mol)) * ln(Q)

Simplifying the equation:

11.0 * 10^(-3) V = E° - (2.303 / 0.059) * ln(Q)

Since we don't have the value for E°, we cannot directly solve the equation. However, we can use the given information that the Zn2+ concentration at the anode is 0.100 M to find it.

Using the Nernst equation with the known concentration:

11.0 * 10^(-3) V = E° - (2.303 / 0.059) * ln(0.100)

Now we can solve for E° by rearranging the equation:

E° = 11.0 * 10^(-3) V + (2.303 / 0.059) * ln(0.100)

Calculating E°:

E° = -0.439 V

Now that we have the value for E°, we can go back to the original Nernst equation and solve for Q, the reaction quotient:

11.0 * 10^(-3) V = -0.439 V - (2.303 / 0.059) * ln(Q)

Simplifying the equation:

2.303 / 0.059 * ln(Q) = 0.428

ln(Q) = 0.428 * 0.059 / 2.303

ln(Q) = 0.01098

Taking the inverse of the natural logarithm (exponential function), we get:

Q = e^(0.01098)

Q = 1.011

Now we have the value for Q. We can use this value to find the concentration of the Zn2+ ion at the cathode using the equation for Q:

Q = [Zn2+ (cathode)] / [Zn2+ (anode)]

1.011 = [Zn2+ (cathode)] / (0.100 M)

Simplifying the equation:

[Zn2+ (cathode)] = 1.011 * 0.100 M

[Zn2+ (cathode)] = 0.1011 M

Therefore, the concentration of the Zn2+ ion at the cathode is approximately 0.1011 M.