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.
(a) What is the initial E
cell
?
(b) What is E
cell
when [Cd2+ ] reaches 0.035 M?
(c) What is [Mn2+ ] when E
cell
reaches 0.055 V?
(d) What are the equilibrium concentrations of the ions? (Enter your answer in scientific notation.)
(a) The standard reduction potentials for Mn2+/Mn and Cd2+/Cd are as follows:
Mn2+ + 2e- -> Mn E° = -1.18 V
Cd2+ + 2e- -> Cd E° = -0.40 V
Since the cell reaction is:
2Mn2+ + Cd -> 2Mn + Cd2+
The standard cell potential (E°cell) is calculated by adding the reduction potentials of the half-reactions:
E°cell = E°cathode - E°anode
E°cell = (-0.40 V) - (-1.18 V) = +0.78 V
(b) The Nernst equation is used to calculate the non-standard cell potential when [Cd2+] reaches 0.035 M:
Ecell = E°cell - (0.0592/n) * log(Q)
Q = [Mn]^2 / [Cd2+]
Ecell = 0.78 V - (0.0592/2) * log(0.090^2 / 0.035)
Ecell ≈ 0.88 V
(c) To find [Mn2+] when Ecell reaches 0.055 V, we can use the Nernst equation again, but this time solve for [Mn2+]:
Ecell = E°cell - (0.0592/n) * log(Q)
Q = [Mn]^2 / [Cd2+]
0.055 = 0.78 - (0.0592/2) * log([Mn]^2 / 0.060)
Solving for [Mn2+] yields:
[Mn2+] ≈ 0.069 M
(d) To find the equilibrium concentrations of the ions, we can use the equilibrium constant expression:
Kc = [Mn]^2 / [Cd2+]
Kc = 10^(0.78 / 0.0592) = 10^13.1757 ≈ 1.43 x 10^13
Since [Cd2+] = 0.035 M at equilibrium, we can calculate [Mn2+] using the equilibrium constant expression:
[Mn2+] = sqrt(Kc * [Cd2+]) = sqrt(1.43 x 10^13 * 0.035) ≈ 5.14 x 10^6 M
Therefore, at equilibrium:
[Mn2+] ≈ 5.14 x 10^6 M
[Cd2+] = 0.035 M