A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 2.1 M and Ecell = 0.13 V.
What values are you using for Ered.
To solve this question, we need to use the Nernst equation, which relates the cell potential (Ecell) to the concentrations of the reactants and products in the half-cells. The Nernst equation is given as:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell = Cell potential
E°cell = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
n = Number of electrons transferred in the balanced cell reaction
F = Faraday's constant (96500 C/mol)
Q = Reaction quotient
In this case, we know the value of Ecell from the problem (0.13 V), but we need to determine the concentration of Pb2+ ([Pb2+]) when [Mn2+] is 2.1 M.
First, we need to write the balanced equation for the reaction occurring at each half-cell:
Mn2+ + 2e- -> Mn (reduction half-reaction)
Pb -> Pb2+ + 2e- (oxidation half-reaction)
Given that the balanced cell reaction is:
Mn2+ + Pb -> Mn + Pb2+
We can see that the number of electrons transferred (n) is 2.
Now, we need to calculate the reaction quotient (Q), which is the ratio of the concentrations of the products to the concentrations of the reactants, each raised to the power of their stoichiometric coefficients. In this case, Q is given by:
Q = ([Mn] * [Pb2+]) / ([Mn2+] * [Pb])
Since we are given [Mn2+] as 2.1 M, we can substitute all the known values into the Nernst equation and solve for [Pb2+].
0.13 V = E°cell - (RT/nF) * ln(([Mn] * [Pb2+]) / ([Mn2+] * [Pb]))
Now we can plug in the values we know:
0.13 V = E°cell - (8.314 J/(mol·K) * T / (2 * 96500 C/mol)) * ln((2.1 M * [Pb2+]) / ([Mn2+] * [Pb]))
Unfortunately, the temperature (T) is not given in the problem, so we cannot calculate the exact concentration of [Pb2+] without knowing the temperature. Please provide the temperature to proceed with the calculation.