Find the emf for the cell
Cr(s) |Cr3+(aq, 0.37 M) ||
Pb2+(aq, 0.0095 M) | Pb(s).
Answer in units of V
I don't know what to do
I posted this answer below.
To find the electromotive force (emf) for the given cell, you can use the Nernst equation. The Nernst equation relates the emf of an electrochemical cell to the concentrations of the species involved in the cell reaction.
The Nernst equation is:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell: The emf of the cell
E°cell: The standard emf of the cell (obtained from reference tables)
R: The ideal gas constant (8.314 J/(mol·K))
T: The temperature in Kelvin (K)
n: The number of moles of electrons transferred in the balanced cell reaction
F: Faraday's constant (96,485 C/mol)
ln: Natural logarithm
Q: The reaction quotient, which is the ratio of the product concentrations to the reactant concentrations, each raised to their respective stoichiometric coefficients.
Let's apply the Nernst equation to the given cell:
Cr(s) |Cr3+(aq, 0.37 M) ||
Pb2+(aq, 0.0095 M) | Pb(s)
The half-cell reactions for this cell are as follows:
Cr(s) -> Cr3+(aq) + 3e- (Anode)
Pb2+(aq) + 2e- -> Pb(s) (Cathode)
The overall balanced equation is:
Cr(s) + Pb2+(aq) -> Cr3+(aq) + Pb(s)
According to reference tables, the standard reduction potential for the Cr3+(aq) + 3e- -> Cr(s) reaction is -0.74 V, and for the Pb2+(aq) + 2e- -> Pb(s) reaction is -0.13 V.
Using the Nernst equation, we can calculate the emf of the cell:
Ecell = E°cell - (RT/nF) * ln(Q)
E°cell = (-0.74 V) - (-0.13 V) (since we have reduction potentials)
E°cell = -0.61 V
Now, let's calculate Q, the reaction quotient:
Q = [Cr3+(aq)] / [Pb2+(aq)]
Q = (0.37 M) / (0.0095 M)
Q = 38.95
Now we can calculate the emf:
Ecell = (-0.61 V) - (8.314 J/(mol·K) * (298 K) / (3 * 96485 C/mol) * ln(38.95)
Calculating this, we find that the emf of the cell is approximately -0.822 V.