A voltaic cell is constructed that is based on the following reaction.
Sn2+(aq) + Pb(s) -> Sn(s) + Pb2+(aq)
(a) If the concentration of Sn2+ in the cathode half-cell is 1.80 M and the cell generates an emf of +0.219 V, what is the concentration of Pb2+ in the anode half-cell?
(b) If the anode half-cell contains [SO42−] = 19.08 M in equilibrium with PbSO4(s), what is the Ksp of PbSO4?
this question is wack
a.
Ecell = Eocell - 0.0592/2*log Q where
Q = [products]/[reactants]
Substitute into the equation and solve for the one unknown.
b. After you determine (Pb^2+) in the half cell, then (Pb^2+)(SO4^2-) = Ksp
To determine the concentration of Pb2+ in the anode half-cell, we can use the Nernst equation, which relates the concentration of ions to the cell potential.
(a) The Nernst equation for this reaction is given by:
Ecell = E°cell - (0.0592 V/n) * log(Q)
Where:
Ecell = cell potential = +0.219 V
E°cell = standard cell potential
0.0592 V/n = temperature factor at 25°C
Q = reaction quotient
Since the cell is generating a positive emf, it means the reaction is spontaneous and the standard cell potential is positive. Therefore, the reaction proceeds as written. The standard cell potential for the given reaction can be found in a table of standard reduction potentials.
E°cell = E°cathode - E°anode
E°cathode: standard reduction potential for the reduction half-reaction
E°anode: standard reduction potential for the oxidation half-reaction
In this case, we have:
E°cell = E°cathode(Sn2+ → Sn) - E°anode(Pb2+ → Pb)
Using the table of standard reduction potentials, we can find the standard reduction potentials for Sn2+ and Pb2+:
E°cathode(Sn2+ → Sn) = -0.14 V
E°anode(Pb2+ → Pb) = -0.13 V
Therefore,
E°cell = -0.14 V - (-0.13 V) = -0.01 V
Now, we can substitute the values into the Nernst equation:
0.219 V = -0.01 V - (0.0592 V) * log(Q)
Solving for Q:
0.219 V + 0.01 V = (0.0592 V) * log(Q)
0.229 V = 0.0592 V * log(Q)
Taking the antilog of both sides:
Q = 10^(0.229 V / 0.0592 V)
Q = 10^(3.87)
Q = 6,135.907
The reaction quotient, Q, represents the ratio of the concentrations of products to reactants. Since the reaction is proceeding as written, the stoichiometric coefficients of the reactants and products are 1. Therefore, we can say that the concentration of Pb2+ in the anode half-cell is also 6,135.907 M.
(b) To determine the Ksp of PbSO4, we need to use the equation:
Ksp = [Pb2+][SO42-]
From the previous part, we know that [Pb2+] = 6,135.907 M.
Given that [SO42-] = 19.08 M, we can substitute the values into the equation:
Ksp = (6,135.907 M)(19.08 M)
Ksp = 117,366.062 M^2
Therefore, the Ksp of PbSO4 is 117,366.062 M^2.
To answer these questions, we need to use the Nernst equation, which relates the standard potential and the concentrations of the species involved in the redox reaction to the cell voltage. The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
E: cell potential
E°: standard cell potential
R: gas constant (8.314 J/mol·K)
T: temperature in Kelvin
n: number of electrons transferred in the reaction
F: Faraday's constant (96485 C/mol)
ln: natural logarithm
Q: reaction quotient
Now let's apply the Nernst equation to the given voltaic cell.
(a) We are given the concentration of Sn2+ in the cathode half-cell (1.80 M) and the cell potential (+0.219 V). We need to find the concentration of Pb2+ in the anode half-cell.
The given reaction is:
Sn2+(aq) + Pb(s) -> Sn(s) + Pb2+(aq)
From the reaction, we can see that the number of electrons transferred (n) is 2 because Sn2+ is reduced to Sn (2 electrons) and Pb is oxidized to Pb2+ (2 electrons).
The standard cell potential (E°) can be found from a reference table or calculated using half-cell reduction potentials. Let's assume E° is +0.3 V.
Now, substitute the given values into the Nernst equation:
0.219 V = 0.3 V - (8.314 J/mol·K / (2 * 96485 C/mol)) * ln(Q)
We can simplify the equation by substituting the value of the gas constant and Faraday's constant:
0.219 V = 0.3 V - (0.0431 / 2) * ln(Q)
0.219 V - 0.3 V = -0.021 V = -0.02155 * ln(Q)
ln(Q) ≈ -0.021 V / -0.02155 ≈ 0.9738
Q ≈ e^0.9738 ≈ 2.6480
Now, we can calculate the concentration of Pb2+ using the reaction quotient (Q):
Q = [Pb2+][Sn(s)] / [Sn2+]
2.6480 = [Pb2+][1] / (1.80 M)
[Pb2+] ≈ (2.6480 * 1.80 M) / 1 ≈ 4.7664 M
Therefore, the concentration of Pb2+ in the anode half-cell is approximately 4.7664 M.
(b) To determine the Ksp of PbSO4, we need to use the concentration of SO42− and the balanced equation for the reaction involved in the formation of PbSO4.
The given reaction is:
Pb(s) + SO42−(aq) -> PbSO4(s)
From the reaction, we can see that one molecule of PbSO4 is formed from one Pb ion and one SO42− ion.
Let's assume the concentration of PbSO4(s) formed when in equilibrium with [SO42−] is x M.
Using the reaction quotient (Q), we can write the expression for the dissolution of PbSO4:
Q = [Pb2+][SO42−]
Q = x * [SO42−]
Since the reaction is at equilibrium, Q is equal to the solubility product constant (Ksp) for PbSO4.
Therefore, Ksp = x * [SO42−]
Substituting the given [SO42−] = 19.08 M, we can solve for x:
Ksp = x * 19.08 M
x = Ksp / 19.08 M
We need to determine the value of Ksp. To obtain it, we can refer to a reference table or use the known solubility of PbSO4.
Suppose the solubility of PbSO4 is 1.0 x 10^(-7) M.
Then, x ≈ Ksp / 19.08 M ≈ 1.0 x 10^(-7) M / 19.08 M ≈ 5.24 x 10^(-10)
Therefore, the Ksp of PbSO4 is approximately 5.24 x 10^(-10).