Given:
PbSO4 + 2e− Pb + SO42− E° = -0.356 V
PbO2 + 4H+ + SO42− + 2e− PbSO4 + 2H2O E° = 1.685 V
Determine the cell voltage of the lead storage battery when the concentration of sulfuric acid is 4.00 M.
Where do I start, thank you
The easiest way to do this is to reverse equation 1 and add in equation 2 for
PbO2 + 4H^+ + SO4^2- + Pb + SO4^2- --> PbSO4 + PbSO4 or
PbO2 + 2H2SO4 --> 2PbSO4 + 2H2O Ecell = 2.041
Then Ecell = Eocell - (0.05916/n)*log(1/(H2SO4). Plug in the numbers and solve for Ecell.
I'm confused. Why do you use log and why do you do 1/H2SO4? I found this equation in our textbook:
Ecell=Eocell-(RT/nFl)nQ
When I use that equation I get 2.00, when I use the equation you've given I get 2.05. I know it's a small difference but why?
Thank you
To determine the cell voltage of the lead storage battery, you can use the Nernst equation. The Nernst equation relates the standard cell potential (E°) to the cell potential under non-standard conditions.
The Nernst equation is:
E = E° - (RT/nF) * ln(Q)
Where:
E is the cell potential
E° is the standard cell potential
R is the ideal gas constant (8.314 J/(mol·K))
T is the temperature in Kelvin
n is the number of moles of electrons transferred in the balanced equation
F is the Faraday constant (96,485 C/mol)
ln is the natural logarithm
Q is the reaction quotient
In this case, we have two half-reactions:
1. PbSO4 + 2e− → Pb + SO42− E° = -0.356 V
2. PbO2 + 4H+ + SO42− + 2e− → PbSO4 + 2H2O E° = 1.685 V
To calculate the cell potential, we first need to determine the reaction quotient (Q) for the overall cell reaction, which is the ratio of product concentrations to reactant concentrations.
Since we are given the concentration of sulfuric acid (H2SO4) in the battery, we only need to consider the concentrations of the other species involved in the two half-reactions.
Next, we balance the two half-reactions, and determine the number of moles of electrons transferred (n) in each half-reaction.
Finally, we plug the values of E°, R, T, n, F, and Q into the Nernst equation to calculate the cell potential (E).
To determine the cell voltage of the lead storage battery when the concentration of sulfuric acid is 4.00 M, you can use the Nernst equation.
The Nernst equation is:
E = E° - (RT / nF) * ln(Q)
Where:
E = cell voltage
E° = standard cell potential
R = gas constant (8.314 J/(mol*K))
T = temperature in Kelvin
n = number of electrons transferred in the balanced equation
F = Faraday constant (96,485 C/mol)
ln = natural logarithm
Q = reaction quotient
Here's the step-by-step approach to solve the problem:
1. Determine the balanced equation for the redox reaction in the lead storage battery:
PbO2 + 4H+ + SO42− + 2e− -> PbSO4 + 2H2O
2. Identify the number of electrons transferred in the balanced equation. In this case, it's 2.
3. Convert the given concentration of sulfuric acid to the reaction quotient (Q). Since the concentration of sulfuric acid is 4.00 M, Q = [H+]^4.
4. Convert the temperature to Kelvin, if necessary.
5. Substitute the values into the Nernst equation and calculate the cell voltage (E).
Let's go through an example calculation:
Given:
E° = 1.685 V
Concentration of sulfuric acid = 4.00 M
Assuming the temperature is 298 K:
T = 298 K
n = 2 (from the balanced equation)
First, calculate the natural logarithm of Q:
Q = [H+]^4 = (4.00 M)^4 = 256 M^4
ln(Q) = ln(256)
Next, substitute the values into the Nernst equation:
E = E° - (RT / nF) * ln(Q)
E = 1.685 V - (8.314 J/(mol*K) * 298 K / (2 * 96,485 C/mol) * ln(256)
Calculate the right side of the equation to find the cell voltage (E).
Remember to convert the units as necessary.
Once you have the value of E, it will represent the cell voltage of the lead storage battery under the given conditions.