Be sure to answer all parts.
A voltaic cell with Mn/Mn2+ and Cd/Cd2+ half-cells has the following initial concentrations:
[Mn2+]= 0.090 M; [Cd2+ ] = 0.060 M.
What are the equilibrium concentrations of the ions?
To find the equilibrium concentrations of the ions, we need to determine the cell potential (Ecell) of the voltaic cell using the Nernst equation and then calculate the equilibrium concentrations of Mn2+ and Cd2+ using the cell potential.
1. Calculate the cell potential (Ecell) of the voltaic cell:
The cell reaction is: Mn2+ + 2e- -> Mn and Cd2+ + 2e- -> Cd
The standard reduction potentials are:
Mn2+ + 2e- -> Mn: E° = -1.18 V
Cd2+ + 2e- -> Cd: E° = -0.40 V
The cell potential can be calculated as:
Ecell = Ecathode - Eanode
Ecell = E°(Cd) - E°(Mn)
Ecell = -0.40 V - (-1.18 V)
Ecell = 0.78 V
2. Calculate the equilibrium concentrations of the ions using the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
where n is the number of electrons transferred (2 in this case) and Q is the reaction quotient.
At equilibrium, Ecell = 0.
0 = 0.78 V - (0.0592/2) * log(Q)
0 = 0.78 - 0.0296 * log(Q)
log(Q) = 0.78 / 0.0296
log(Q) = 26.35
Q = 10^26.35
Q = 2.03 * 10^26
Since Q = [Mn2+]/[Cd2+]^2
0.090 / (0.060)^2 = 2.03 * 10^26
Now, we can solve for the equilibrium concentrations of Mn2+ and Cd2+:
Let x be the change in concentration of Cd2+ and 2x be the change in concentration of Mn2+.
0.090 - 2x = equilibrium concentration of Mn2+
0.060 - x = equilibrium concentration of Cd2+
Using the value of Q we found earlier, we can now solve for x:
0.090 - 2x = 0.060
2x = 0.030
x = 0.015
Equilibrium concentrations:
[Mn2+] = 0.090 - 2(0.015) = 0.060 M
[Cd2+] = 0.060 - 0.015 = 0.045 M
Therefore, the equilibrium concentrations of Mn2+ and Cd2+ ions are 0.060 M and 0.045 M, respectively.