A voltaic cell consists of an Al/Al3+ half-cell and a Cd/Cd2+ half-cell. Calculate {Al3+} when {Cd2+} = 0.401 M and Ecell = 1.299 V.Use reduction potential values of Al3+ = -1.66 V and for Cd2+ = -0.40 V.
2Al(s) ==> 2Al^3+ + 3e Eo ox = +1.66 v
3Cd^2+ + 2e ==> 3Cd(s) Eo red = -0.40
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2Al(s) + 3Cd^2+ ==> 2Al^3+ + 3Cd(s)
So Eo cell = Eo ox + Eo red = +1.66-0.40 = ?
You know Ecell from the problem of 1.299. Plug those values into
Ecell = Eocell + (0.05916/n)log(K)
n = 6 for number of electrons transferred and
K = (Al^3+)^2((Cd)^3/((Al)^2(Cd^2|+)^3
Remember that (Al) and (Cd) in the solid state are 1 by definition. You know (Cd^2+) from the problem and you can calculate (Al^3+).
Post your work if you get stuck.
Thanks for your reply. I was able to get the answer.
Thank you for letting me know.
To calculate the concentration of Al3+ in the voltaic cell, we need to use the Nernst equation. The Nernst equation is given by:
Ecell = E˚cell - (0.0592 / n) * log(Q),
where Ecell is the cell potential, E˚cell is the standard cell potential, n is the number of electrons transferred in the balanced cell reaction, and Q is the reaction quotient.
First, let's find the standard cell potential, E˚cell. The standard cell potential is the difference between the reduction potentials of the two half-cells.
E˚cell = E˚(Al3+/Al) - E˚(Cd2+/Cd).
Given E˚(Al3+/Al) = -1.66 V and E˚(Cd2+/Cd) = -0.40 V:
E˚cell = (-1.66 V) - (-0.40 V)
E˚cell = -1.26 V.
Next, we need to determine the number of electrons transferred in the cell reaction. By examining the half-cell reactions, we can see that Al loses three electrons, while Cd gains two electrons. Therefore, n = 3.
Now, let's calculate the reaction quotient, Q, using the concentrations of Cd2+ and Al3+:
Q = [Cd2+] / [Al3+].
Given [Cd2+] = 0.401 M (provided in the question).
To calculate [Al3+], we will rearrange the Nernst equation and solve for [Al3+]:
Ecell = E˚cell - (0.0592 / n) * log(Q).
Rearranging, we get:
log(Q) = (E˚cell - Ecell) / (0.0592 / n).
Substituting the values we have:
log(Q) = (-1.26 V - 1.299 V) / (0.0592 / 3)
log(Q) = -0.1976.
Now, we take the antilog of both sides to find Q:
Q = 10^(-0.1976)
Q ≈ 0.596.
Finally, substitute the value of Q into the equation for Q:
Q = [Cd2+] / [Al3+]
0.596 = 0.401 M / [Al3+].
Rearranging the equation to solve for [Al3+]:
[Al3+] = 0.401 M / 0.596
[Al3+] ≈ 0.673 M.
Therefore, the concentration of Al3+ in the voltaic cell is approximately 0.673 M when [Cd2+] = 0.401 M and Ecell = 1.299 V.