The voltage generated by the zinc concentration cell described by,
Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)
is 21.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.
I worked this just a few post below but the Ecell was 11 mv. Same process though.
Ecell = Eo - (0059/2)log(0.1/x) and solve for x.
To calculate the concentration of Zn2+ ions at the cathode in a concentration cell, we need to use the Nernst equation. The Nernst equation relates the concentration of reactants and products in an electrochemical cell to the measured cell potential.
The Nernst equation is given by:
E = E° - (RT / nF) * ln(Q)
Where:
E is the measured cell potential,
E° is the standard cell potential,
R is the gas constant (8.314 J/(mol·K)),
T is the temperature in Kelvin,
n is the number of electrons transferred in the balanced cell reaction, and
F is Faraday's constant (96,485 C/mol).
In the given concentration cell, Zn2+ ions are reducing at the cathode, and the reaction is:
Zn2+(aq) + 2e- -> Zn(s)
Since the electrode potential of the Zn2+/Zn half-cell is being measured against the standard hydrogen electrode (SHE), E° for this reaction is 0.00 V.
Given data:
E = 21.0 mV = 0.021 V
T = 25 °C = 298 K
Now, let's calculate the concentration of Zn2+ ions at the cathode.
Step 1: Calculate the value of the natural logarithm term.
RT / nF = (8.314 J/(mol·K) * 298 K) / (2 * 96,485 C/mol)
RT / nF = 0.025693 V
Step 2: Rearrange the Nernst equation to solve for ln(Q).
ln(Q) = (E° - E) / (RT / nF)
ln(Q) = (0.00 V - 0.021 V) / 0.025693 V
ln(Q) = -0.818
Step 3: Determine the value of Q.
Q = e^(ln(Q))
Q = e^(-0.818)
Q = 0.442
Step 4: Calculate the concentration of Zn2+ ions at the cathode.
Since the concentration of Zn2+ ions at the anode is 0.100 M, the concentration of Zn2+ ions at the cathode can be represented as x.
Q = [Zn2+(cathode)] / [Zn2+(anode)]
0.442 = x / 0.100
Solving for x:
x = 0.442 * 0.100
x = 0.0442 M
Therefore, the concentration of Zn2+ ions at the cathode is 0.0442 M.