The measured potential of the following cell was 0.1776 volts.
Pb|Pb2+ (?M )||Pb2+ (1.0M)|Pb
Calculate the electrode potential of the unknown half-cell (anode).
got this on google unknown half-cell(anode). ... 2Ag+(aq)+Pb(s)→2Ag(s)+Pb2+(aq) E∘cell.
1 answer
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Top answer:
Ecell = Eocell + (0.0592/#e)log(Pb^2+)/(Pb^2+) Ecell = 0.1776 Eocell= I don't have my text handy but I think this is -0.126. You should verify that. #e ...
if it does not work feel free to tell me here or on
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This is a concentration cell.
Pb ==> Pb^2+ + 2e.......................Eox = ?
Pb^2+ + 2e ==> Pb...................... Ered = - 0.126
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Pb + Pb^2+ ==> Pb^2+ + Pb........ Ecell = 0.1776
Eox + Ered = Ecell
Eox + (-0.126) = 0.1776 which I will round to 0.178
Eox = 0.304 and written as a reduction it is -0.304
-0.304 = Eocell -(0.0592/2)*log (Pb)/(Pb^2+)
-0.304 = - 0.126 - (0.0592/2)*log (1)/(Pb^2+).
Solve for (Pb^2+) in moles/L.
To calculate the electrode potential of the unknown half-cell (anode), we can use the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
Where:
- Ecell is the measured potential of the cell (0.1776 V)
- E°cell is the standard cell potential of the cell
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (usually 298 K, but check the question for the specific temperature)
- n is the number of moles of electrons transferred in the balanced equation for the cell reaction
- F is Faraday's constant (96485 C/mol)
- Q is the reaction quotient, which can be given by the concentrations of the species involved in the cell reaction.
First, we need to write the balanced equation for the cell reaction. From the given cell notation:
Pb|Pb2+ (unknown concentration)||Pb2+(1.0M)|Pb
The half-cell reaction would be:
Pb2+ + 2e- -> Pb
Now, we need to calculate the reaction quotient (Q). The reaction quotient can be calculated using the concentrations of the species involved in the half-cell reaction. Since the concentration of Pb2+ in the unknown half-cell is given as unknown (?M), we will use x to represent the concentration:
Q = [Pb2+]unknown / [Pb2+]standard
= x / 1.0M
= x
Substituting the values into the Nernst equation:
0.1776V = E°cell - (8.314 J/(mol·K)) * 298 K / (2 * 96485 C/mol) * ln(x)
Simplifying the equation:
0.1776V = E°cell - (0.025695 V) * ln(x)
Now, we need the value of E°cell to solve for x. The standard cell potential can be found in a reference table or given in the question. If the standard cell potential (E°cell) is not provided, it cannot be calculated.
Once you have the value of E°cell, you can rearrange the equation to solve for x and find the unknown concentration of Pb2+ in the unknown half-cell.
To calculate the electrode potential of the unknown half-cell (anode), we need to use the Nernst equation. The Nernst equation relates the measured cell potential to the concentrations of the species involved in the half-cell reaction.
The Nernst equation can be written as:
E = E° - (RT / nF) * ln(Q)
Where:
- E is the electrode potential
- E° is the standard electrode potential
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of electrons transferred in the half-reaction
- F is the Faraday constant (96485 C/mol)
- Q is the reaction quotient
In this case, we have the following half-reaction at the unknown anode:
Pb2+ (? M) + 2e- -> Pb
First, we need to determine the value of Q, which is the reaction quotient. For the half-reaction at the anode, Q is the concentration of Pb2+ ions (unknown) divided by the concentration of Pb2+ ions (1.0M) raised to the power of 2 (since 2 electrons are involved in the half-reaction).
Now, let's substitute the known values into the Nernst equation:
E = E° - (RT / nF) * ln(Q)
Since we don't have the standard electrode potential (E°) for the unknown half-cell, we need additional information to calculate it. Please provide the standard electrode potential for Pb2+ -> Pb half-cell or any other relevant information you have, and I'll help you calculate the electrode potential of the unknown half-cell (anode).