A concentration cell consists of two Al/Al+3 electrodes. The electrolyte in compartment A is 0.05M Al(NO3)3 and in compartment B is 1.25 M Al(NO3)3. What is the voltage of the cell at 25 degrees Celcius?
ECell = EoCell-(0.059/n)log(dilute soln/concd soln)
EoCell = 0. n = 3. Substitute and solve.
I know that this is an old thread, but if dilute soln/concd soln, then the answer is negative.
To calculate the voltage of the concentration cell, we can use the Nernst equation, which relates the concentrations of the ions in the electrodes and the temperature to the cell potential. The Nernst equation is given by:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell = Cell potential
E°cell = Standard cell potential (measured under standard conditions)
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 (96,485 C/mol)
ln(Q) = Natural logarithm of the reaction quotient
In this case, we have an Al/Al3+ half-cell reaction occurring in each electrode. The balanced equation for the half-cell reaction is:
Al → Al3+ + 3e-
The standard cell potential for this half-cell reaction is 1.66 V.
Now, let's calculate the reaction quotient, Q, using the concentrations given.
In compartment A (0.05 M Al(NO3)3), Al3+ is the oxidized (losing electrons) form, so its concentration will be 0.05 M.
In compartment B (1.25 M Al(NO3)3), Al3+ is the reduced (gaining electrons) form, so its concentration will be 1.25 M.
Q = [Al3+]B / [Al3+]A = 1.25 / 0.05 = 25
Now, we substitute the values into the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
Since the temperature is given as 25 degrees Celsius, we need to convert it to Kelvin:
T = 25 + 273.15 = 298.15 K
Considering n = 3 (from the balanced equation) and plugging in the relevant values, we get:
Ecell = 1.66 V - (8.314 J/(mol·K) * 298.15 K / (3 * 96,485 C/mol)) * ln(25)
Now, we can calculate this expression to determine the voltage of the cell at 25 degrees Celsius.