A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 C. The initial concentrations of Pb+2 and Cu+2 are 0.0500 M and 1.50 M, respectively.
A. What is the initial cell potential?
B. What is the cell potential when the concentration of Cu+2 has fallen to 0.200 M?
C. What are the concentrations of Pb+2 and Cu+2 when the cell potential falls to 0.35 V?
I answered A and B. For A I got: 0.43 V and for B I got 0.46 V. I cannot figure out part C though. I tried 0.35V=O.47V-((0.0592/2)*log(0.0500+X/1.50-X))but the answer doesn't seem right. I found X to equal 1. When I plug it all back in, it doesn't work though. Is my equation wrong? So confused...
To solve part C of the problem, you need to use the Nernst equation, which relates the cell potential to the concentrations of the reactants and products involved in the cell reaction.
The Nernst equation is given by:
E = E° - (RT/nF) * ln(Q)
Where:
E = Cell potential at given conditions
E° = Standard cell potential
R = Gas constant (8.314 J/(mol·K))
T = Temperature in Kelvin
n = Number of moles of electrons transferred in the balanced cell reaction
F = Faraday's constant (96,485 C/mol)
Q = Reaction quotient
In this case, the cell reaction is:
Pb(s) + Cu2+(aq) -> Pb2+(aq) + Cu(s)
The balanced cell reaction has a stoichiometry of 2:1, meaning 2 moles of electrons are transferred.
A. For the initial cell potential (part A):
To calculate the initial cell potential, you need to use the standard cell potentials (E°) for the half-reactions. The standard cell potential for the Pb/Pb2+ half-cell is -0.13 V, and for the Cu/Cu2+ half-cell is +0.34 V.
E° cell = E° cathode - E° anode
= 0.34 V - (-0.13 V)
= 0.47 V
So, the initial cell potential is 0.47 V.
B. For the cell potential when the concentration of Cu+2 has fallen to 0.200 M (part B):
To calculate the new cell potential, you need to use the Nernst equation. Plugging in the given information:
E = E° - (RT/nF) * ln(Q)
E = 0.47 V - (0.0592/2) * ln([Pb2+][Cu]/[Cu2+])
E = 0.47 V - (0.0296) * ln([0.0500]/[0.200])
E ≈ 0.46 V
So, the cell potential when the concentration of Cu+2 has fallen to 0.200 M is approximately 0.46 V, which matches your calculation.
C. For the concentrations of Pb+2 and Cu+2 when the cell potential falls to 0.35 V (part C):
To find the concentrations of the ions when the cell potential falls to 0.35 V, you need to rearrange the Nernst equation and solve for the concentrations.
E = E° - (RT/nF) * ln(Q)
0.35 V = 0.47 V - (0.0296) * ln([Pb2+][Cu]/[Cu2+])
Solving this equation for [Pb2+] and [Cu], you'll end up with a non-linear equation, which cannot be solved analytically. You'll need to use numerical methods or iterative techniques to approximate the concentrations of the ions.
To solve part C, you need to calculate the concentrations of Pb+2 and Cu+2 when the cell potential falls to 0.35 V.
First, let's determine the cell potential expression for the redox reaction occurring in the voltaic cell. The balanced half-reactions for this cell are:
Pb → Pb2+ + 2e- (oxidation half-reaction)
Cu2+ + 2e- → Cu (reduction half-reaction)
The overall reaction is the sum of these half-reactions:
Pb + Cu2+ → Pb2+ + Cu
The cell potential at any time can be calculated using the Nernst Equation:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
Ecell = cell potential at a given time
E°cell = standard cell potential
R = ideal gas constant (8.314 J/(mol·K))
T = temperature in Kelvin (25 °C = 298 K)
n = number of electrons transferred in the balanced half-reactions (2 in this case)
F = Faraday's constant (96,485 C/mol)
Q = reaction quotient, which is the ratio of product concentrations to reactant concentrations
Since it is given that Ecell = 0.35 V, you can rearrange the equation as follows to solve for Q:
Q = exp((E°cell - Ecell) / (RT/nF))
Now, plug in the values:
E°cell = 0.47 V
Ecell = 0.35 V
R = 8.314 J/(mol·K)
T = 298 K
n = 2
F = 96,485 C/mol
Q = exp((0.47 - 0.35) / ((8.314 * 298) / (2 * 96,485)))
Calculate Q using this formula to find its value. Then, solve it for x in the equation:
Q = (Pb2+ concentration + x) / (Cu+2 concentration - x)
Substitute the value of Q and solve for x algebraically. Once you find x, use it to calculate the concentrations of Pb+2 and Cu+2.