Explanation:
To calculate the concentration of Zn^2+(aq) ion at the cathode, we can use the Nernst equation, which relates the cell potential to the concentration of the species involved in the cell reaction. The Nernst equation is given as:
E = E° - (RT/nF) * ln(Q)
Where:
E = Cell potential
E° = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
n = Number of moles of electrons transferred in the balanced cell reaction
F = Faraday's constant (96485 C/mol)
Q = Reaction quotient (concentration of products divided by concentration of reactants)
Given:
E = 16.0 mV = 0.016 V (since 1 V = 1000 mV)
E° = 0 V (as the cell is at standard conditions)
T = 25 °C = 298 K
n = 2 (since 2 moles of electrons are transferred in the balanced cell reaction)
We need to find the concentration of Zn^2+(aq) at the cathode, which corresponds to the product concentration in the reaction quotient (Q). Let's assume the concentration of Zn^+2(aq) at the anode is x M.
The cell reaction is:
Zn(s) + Zn^2+(aq, 0.100 M) → Zn^2+(aq, x M) + Zn(s)
The reaction quotient (Q) is then given by:
Q = [Zn^2+(aq, x M)] / [Zn^2+(aq, 0.100 M)]
Using the Nernst equation, we can rearrange it to solve for x:
E = E° - (RT/nF) * ln(Q)
0.016 = 0 - (8.314 * 298 / (2 * 96485)) * ln([Zn^2+(aq, x M)] / [Zn^2+(aq, 0.100 M)])
Simplifying the equation:
0.016 = -0.000108 * ln([Zn^2+(aq, x M)] / 0.100)
Now, let's solve for x by rearranging the equation and solving for ln([Zn^2+(aq, x M)] / 0.100):
ln([Zn^2+(aq, x M)] / 0.100) = -0.016 / -0.000108
ln([Zn^2+(aq, x M)] / 0.100) = 148.14814814814814
Taking the exponential of both sides:
[Zn^2+(aq, x M)] / 0.100 = e^(148.14814814814814)
Now, let's solve for [Zn^2+(aq, x M)]:
[Zn^2+(aq, x M)] = 0.100 * e^(148.14814814814814)
Calculating the concentration:
[Zn^2+(aq, x M)] ≈ 4.79 x 10^64 M
Therefore, the concentration of Zn^2+(aq) ion at the cathode is approximately 4.79 x 10^64 M.