At 298 K,
Pb(IO3)2(s) ↔ Pb2+(aq) + 2IO3-(aq) Ksp = 2.6x 10-13
Pb2+(aq) + 2e- → Pb(s) Eo = (-0.126) V.
Find the standard potential of the half-reaction:
Pb(IO3)2(s) + 2e- ↔ Pb(s) + 2IO3-(aq) Eo = ?
I'm really confused. Please help.
To find the standard potential of the half-reaction, we can use the Nernst equation, which relates the standard potential (Eo) of the reaction to the actual potential (E) under non-standard conditions.
The Nernst equation is given by:
E = Eo - (RT/nF) * ln(Q)
where:
E = actual potential of the half-reaction
Eo = standard potential of the half-reaction
R = gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
n = number of electrons transferred
F = Faraday constant (96485 C/mol)
ln = natural logarithm
Q = reaction quotient
Since the given reaction involves the transfer of 2 electrons, n = 2.
Now, let's calculate Q, the reaction quotient. The reaction quotient is the ratio of the concentrations of the products raised to their stoichiometric coefficients (IO3-) to the concentration of the reactant raised to its stoichiometric coefficient (Pb(IO3)2):
Q = [Pb2+][IO3-]^2
Since the reaction is solid Pb(IO3)2, its concentration remains constant, so we can omit it in the reaction quotient. Therefore,
Q = [Pb2+][IO3-]^2
To relate Q to Ksp, we use the following relationship:
Ksp = [Pb2+][IO3-]^2
Now, let's substitute the given values into the Nernst equation:
E = Eo - (RT/nF) * ln(Q)
Since we want to find Eo, we can rearrange the equation:
Eo = E + (RT/nF) * ln(Q)
Substituting the given values:
Eo = -0.126 V + (8.314 J/(mol·K) * 298 K / (2 * 96485 C/mol)) * ln(2.6 x 10^-13)
Eo ≈ -0.126 V + (0.00212) * ln(2.6 x 10^-13)
Eo ≈ -0.126 V + (0.00212) * (-29.83)
Eo ≈ -0.126 V + (-0.0618)
Eo ≈ -0.1878 V
Therefore, the standard potential of the half-reaction:
Pb(IO3)2(s) + 2e- ↔ Pb(s) + 2IO3-(aq) is approximately -0.1878 V.