A voltaic cell consists of an Mn/Mn2+ half-cell and a Pb/Pb2+ half-cell. Calculate [Pb2+] when [Mn2+] is 2.3 M and Ecell = 0.49 V.
To calculate the concentration of Pb2+ in the half-cell, 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 in the balanced equation
Q is the reaction quotient, which is the ratio of products over reactants raised to the power of their stoichiometric coefficients
In this case, the reaction taking place in the Mn/Mn2+ half-cell is:
Mn2+ + 2e- ⇌ Mn
And the reaction taking place in the Pb/Pb2+ half-cell is:
Pb2+ + 2e- ⇌ Pb
Since there are no coefficients in front of the half-cell reactions, n = 2 for both reactions.
First, let's find the standard cell potential (E°cell) by using the given value of Ecell:
Ecell = 0.49 V
Next, we need to calculate the reaction quotient (Q). The reaction quotient can be calculated using the concentrations of the species involved in the reaction. In this case, the concentrations given are [Mn2+] = 2.3 M and [Pb2+] is what we're trying to find.
Q = [Pb2+] / [Mn2+]
Now, let's substitute the values into the Nernst equation:
0.49 V = E°cell - (0.0592/2) * log(Q)
Simplifying the equation:
0.49 V = E°cell - (0.0296) * log(Q)
We can rearrange the equation to solve for log(Q):
log(Q) = (E°cell - 0.49 V) / 0.0296
Now, let's convert log(Q) back to Q by taking the anti-log:
Q = 10^[(E°cell - 0.49 V) / 0.0296]
Finally, substitute the known concentrations and the value of Q into the original equation to solve for [Pb2+]:
0.49 V = E°cell - (0.0296) * log( [Pb2+] / [Mn2+] )
Simplifying the equation:
0.49 V = E°cell - (0.0296) * log( [Pb2+] / 2.3 M )
Now, rearrange the equation to solve for [Pb2+]:
[ Pb2+] = 2.3 M * 10^[(E°cell - 0.49 V) / 0.0296]
Plug in the values of E°cell (which you need to find from a table of standard reduction potentials) and the given concentration of [Mn2+] to calculate [Pb2+].