The voltage generated by the zinc concentration cell described by the line notation
Zn(s) || Zn2+(aq,0.100 M) ‖‖ Zn2+(aq,? M) || Zn(s)
is 15.0 mV at 25 °C. Calculate the concentration of the Zn2+(aq) ion at the cathode.
In a concentration cell Ecell = Eo -0.0592/2*log(dil/concd)
0.015 = 0 - 0.0592/2*log(dil/0.1)
solve for dil
After thinking about this problem for an hour I believe that the line notation means that the 0.100 M cell is the smaller concentration and the one on the right is the larger concentration; however, if that is true then the voltage measured should be -0.015 v so perhaps I've misread the problem.
OK. Here's the scoop. In a concentration cell with the line notation as shown, the anode is on the left and is the weaker of the two solutions. The right hand cell is the cathode and is the more concentrated of the two. I just didn't substitute correctly in that equation I gave you. The correct substitution is as follows:
Ecell = Eo -0.0592/2*log(dil/concd)
0.015 = 0 - 0.0592/2*log(0.1/x)
0.015/-0.0296 = -0.507 = log 0.1/x
0.1/x = 0.311
0.1 = 0.311x
x = 0.1/0.311 = 0.322 M which is the concentration of the solution at the cathode.
DrBob222 posted that last post by anonymous. Sorry about that. Forgot to put my name to it.
To calculate the concentration of Zn2+(aq) ion at the cathode, we need to use the Nernst equation, which relates the voltage generated by an electrochemical cell to the concentrations of the ions involved.
The Nernst equation is given as:
E = E° - (RT / nF) * ln(Q)
Where:
E = Cell potential
E° = Standard cell potential (given as 15.0 mV)
R = Gas constant (8.314 J/(mol*K))
T = Temperature in Kelvin (25°C = 298 K)
n = Number of electrons transferred per mole of reaction (for this cell, it is 2)
F = Faraday constant (96485 C/mol)
ln = Natural logarithm
Q = Reaction quotient
In this case, we have a concentration cell with zinc electrodes and zinc ion (Zn2+(aq)) in different concentrations. The key step is determining the reaction quotient (Q) for this cell.
From the given line notation:
Zn(s) || Zn2+(aq,0.100 M) ‖‖ Zn2+(aq,? M) || Zn(s)
We can infer that the half-reaction at the cathode is:
Zn2+(aq) + 2e- → Zn(s)
And the half-reaction at the anode is:
Zn(s) → Zn2+(aq,? M) + 2e-
Since the concentration of Zn2+(aq) at the anode is unknown, we can represent it as "? M" in the notation.
Now, let's substitute the values into the Nernst equation:
E = 15.0 mV
E° = 15.0 mV
R = 8.314 J/(mol*K)
T = 298 K
n = 2
F = 96485 C/mol
ln = Natural logarithm
Q = [Zn2+(aq,0.100 M)] / [Zn2+(aq,? M)]
By rearranging the equation, we get:
ln(Q) = (E° - E) * (nF / RT)
Substituting the given values, we have:
ln(Q) = (15.0 mV - 0) * ((2 * 96485 C/mol) / (8.314 J/(mol*K) * 298 K))
Now we can solve for ln(Q):
ln(Q) = (0.015 V) * 58.164
ln(Q) = 0.87246
Now, taking the natural exponent of both sides:
Q = e^(0.87246)
Q ≈ 2.391
Next, we need to consider the relationship between Q and the concentrations of the ions involved.
Since the ions are in the same phase (aqueous), the relationship is:
Q = [Zn2+(aq,0.100 M)] / [Zn2+(aq,? M)]
Now we can rearrange for the unknown concentration:
[Zn2+(aq,? M)] = [Zn2+(aq,0.100 M)] / Q
Substituting the known values:
[Zn2+(aq,? M)] = 0.100 M / 2.391
[Zn2+(aq,? M)] ≈ 0.0418 M
Therefore, the concentration of Zn2+(aq) ion at the cathode is approximately 0.0418 M.