Pt | Cr3+ (2.7010-4 M), Cr2+ (0.500 M) || Pb2+ (0.0710 M) | Pb
Consider the following electochemical cell.
Calculate the voltage of the cell, Ecell, including the sign.
To calculate the voltage of the cell (Ecell), we need to use the Nernst equation. The Nernst equation relates the concentration of the species involved in the cell reaction to the standard cell potential (E°cell) and the actual cell potential (Ecell). The equation is as follows:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell: The actual cell potential of the reaction.
- E°cell: The standard cell potential, which can be found in a table of standard reduction potentials.
- n: The number of moles of electrons transferred in the balanced half-reactions.
- Q: The reaction quotient, which is the ratio of the concentrations of the products raised to their stoichiometric coefficients to the concentrations of the reactants raised to their stoichiometric coefficients.
In the given cell, the balanced half-reactions are as follows:
Oxidation (anode): Pt reacts with Cr3+ ions to form Cr2+
Reduction (cathode): Pb2+ ions are reduced to form Pb on Pb electrode
The concentrations of the reactants and products are also given. We have:
[Pt | Cr3+ (2.7010-4 M), Cr2+ (0.500 M) || Pb2+ (0.0710 M) | Pb]
Now we need to calculate the reaction quotient (Q) using the concentrations of the species involved. In this case, Q can be expressed as:
Q = [Cr2+] / [Pb2+]
Plugging in the given values:
Q = (0.500 M) / (0.0710 M)
Now we need to find the standard cell potential (E°cell). Using a table of standard reduction potentials, we can look up the reduction potentials for the given half-reactions:
Cr3+ + 3e- -> Cr2+ (Reduction potential: E°red = +0.407 V)
Pb2+ + 2e- -> Pb (Reduction potential: E°red = -0.126 V)
To determine the overall cell potential (E°cell), we need to subtract the oxidation potential from the reduction potential:
E°cell = E°red(cathode) - E°red(anode)
E°cell = (-0.126) - (+0.407) = -0.533 V
Now we can substitute the values into the Nernst equation and solve for Ecell:
Ecell = -0.533 - (0.0592/n) * log(Q)
Since Q = (0.500 M) / (0.0710 M), we plug in this value into the equation and solve for Ecell. Note that n = 2, as two electrons are transferred in the reduction half-reaction of lead (Pb2+ + 2e- -> Pb):
Ecell = -0.533 - (0.0592/2) * log((0.500 M) / (0.0710 M))
Calculating this expression will give you the voltage of the cell, Ecell, including its sign.