A voltaic cell consists of Mn/Mn2+ and Cd/Cd2+ half-cells with the following initial concentrations: [Mn2+] = 0.090 M; [Cd2+] = 0.060 M.
A)What is the intital Ecell?
B)What is Ecell when [Cd2+} reaches .050M.
C) What is [Mn2+] when Ecell reaches 0.055V?
D) What are the equilibirum concentrations of the ions?
I only need to know part C and D. How is it done, the answers are:
C) 0.150M and D)[Mn2+]= 0.150M and [Cd2+]=6.4X10^(-28)M
Relevant data: Mn(S) to its ion is 1.18V and for Cd ion to solid, it is -.40V, the total is .78V.
Thanks in advance.
C)
Set up the equation with known information:
E cell = E naught -(0.0592/n)*log Q
0.055=0.78-(0.0592/2)*log Q
SOLVE for Q so that you will be able to equate Q to the concentration of Mn2+ over the concentration of Cd2+ (according to chemical equilibria rules):
-0.725 = -(0.0592/2)*log Q
log Q = 24.493
Q = 3.11*10^(24)
NOW set equal to [Mn2+]/[Cd2+] with concentrations from question:
3.11*10^(24) = (0.090+x)/(0.060-x)
SOLVE
[1.868*10^(23)]- (3.11*10^(24))x = 0.09+x
the 0.09 has barely any affect (same with the +1x), so:
1.868*10^(23) = [3.11*10^(24)]x
x= 0.06
ADD to original concentration of Mn2+ (=0.090):
0.06 + 0.090 = 0.150M = [Mn2+]
PART D)
At equilibrium, Ecell= 0. Why? Because the term (0.0592V/n)*log Q becomes so big that it equals E naught which means that E cell = 0.
So, we set E cell equal to zero.
0 = E naught - (0.0592/n)*log Q
Now, fill in what we know:
0 = 0.78-(0.0592/2)*log((0.090+x)/(0.060-x))
Then we find through the same methods in part C that since [Mn2+]= 0.150M, that [Cd2+]= 6.68*10^(-28)M
fiyu
C) To find the concentration of Mn2+ when Ecell reaches 0.055V, we can use the Nernst equation:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
Ecell = Cell potential (0.055V)
E°cell = Standard cell potential (0.78V)
n = Number of electrons transferred (in this case, 2)
Q = Reaction quotient
Since we want to find the concentration of Mn2+, Q will be [Mn2+]/[Cd2+]. We can rearrange the equation to solve for [Mn2+]:
0.055V = 0.78V - (0.0592/2) * log([Mn2+]/[Cd2+])
Simplifying, we get:
log([Mn2+]/[Cd2+]) = (0.78V - 0.055V) * 2 / 0.0592
log([Mn2+]/[Cd2+]) = 13.8514
[Mn2+]/[Cd2+] = 10^13.8514
[Mn2+] = [Cd2+] * 10^13.8514 (Since [Mn2+]:[Cd2+] = 1:1)
Substituting the initial concentrations, [Cd2+] = 0.060M and [Mn2+] = 0.090M:
[Mn2+] = 0.060M * 10^13.8514
[Mn2+] = 0.150M
Therefore, [Mn2+] when Ecell reaches 0.055V is 0.150M.
D) To find the equilibrium concentrations of the ions, we know that at equilibrium, the cell potential (Ecell) will be 0V. Therefore, we can use the Nernst equation again:
Ecell = E°cell - (0.0592/n) * log(Q)
Since Ecell is 0V, we have:
0V = 0.78V - (0.0592/2) * log(Q)
Solving for Q:
log(Q) = 0.78V / (0.0592/2)
Q = 10^(0.78V / (0.0592/2))
Now, Q is the reaction quotient, which is equal to ([Mn2+]/[Cd2+])^2.
Therefore:
([Mn2+]/[Cd2+])^2 = 10^(0.78V / (0.0592/2))
Solving for [Mn2+]/[Cd2+]:
[Mn2+]/[Cd2+] = sqrt(10^(0.78V / (0.0592/2)))
Substituting the initial concentration [Cd2+] = 0.060M:
[Mn2+] = [Cd2+] * sqrt(10^(0.78V / (0.0592/2)))
[Mn2+] = 0.060M * sqrt(10^(0.78V / (0.0592/2)))
[Mn2+] ≈ 0.150M
Therefore, at equilibrium, the concentrations are approximately [Mn2+] = 0.150M and [Cd2+] ≈ 6.4 x 10^(-28)M.
To solve part C, we need to use the Nernst equation:
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 redox reaction
Q is the reaction quotient
In this case, since we have the standard cell potential (0.78V) and we want to find the concentration [Mn2+] when Ecell reaches 0.055V, we can rearrange the equation as follows:
E°cell = Ecell + (0.0592 / n) * log(Q)
Substituting the known values:
0.78V = 0.055V + (0.0592 / n) * log(Q)
Rearranging the equation to solve for log(Q):
log(Q) = (0.78V - 0.055V) * n / 0.0592
log(Q) = (0.725V) * n / 0.0592
Taking the antilog of both sides:
Q = 10^((0.725V) * n / 0.0592)
Since [Mn2+] is the reactant in the Mn/Mn2+ half-cell, Q can be written as:
Q = [Mn2+] / [Cd2+]
Substituting this into the equation:
[Mn2+] / [Cd2+] = 10^((0.725V) * n / 0.0592)
Given that the standard cell potential is 0.78V and the number of electrons transferred is 2 (as Mn2+ gains 2 electrons to become Mn), we can calculate the value inside the exponent:
(0.725V) * 2 / 0.0592 = 23.20
Now, we can substitute this value back into the equation:
[Mn2+] / [Cd2+] = 10^(23.20)
To find [Mn2+], we need to know the value of [Cd2+]. Since we don't have that information, we need to find it using the given data.
To solve part D, we can use the Nernst equation again. At equilibrium, the cell potential (Ecell) is 0V. Using the same equation as above, but with Ecell = 0V:
0V = E°cell - (0.0592 / n) * log(Q)
Rearranging the equation to solve for log(Q):
log(Q) = (0.0592 / n) * E°cell
log(Q) = (0.0592 / 2) * 0.78V
log(Q) = 0.0231
Taking the antilog of both sides:
Q = 10^0.0231
Since Q = [Mn2+] / [Cd2+], we have a relationship between their concentrations at equilibrium:
[Mn2+] / [Cd2+] = 10^0.0231
Now, we can substitute known values and solve for [Mn2+]:
0.090M / [Cd2+] = 10^0.0231
Multiplying both sides by [Cd2+]:
0.090M = 10^0.0231 * [Cd2+]
Now, we can calculate [Cd2+]:
[Cd2+] = 0.090M / 10^0.0231
[Cd2+] ≈ 0.090M / 1.146
[Cd2+] ≈ 0.0786M
Therefore, at equilibrium, the concentrations are approximately:
[Mn2+] = 0.090M
[Cd2+] ≈ 0.0786M