a cell is constructed by immersing a strip of silver in .10 M AgNO3 solution and a strip of lead in 1.0 M Pb(NO3)2 solution. A wire and salt bridge complete the cell. What is the potential of the silver electrode in the cell?
To determine the potential of the silver electrode in the cell, we need to use the standard reduction potentials of silver and lead reactions and apply them to the Nernst equation.
1. First, let's write down the balanced half-reactions for the silver and lead electrodes:
- Ag+ (aq) + e- -> Ag (s) (Silver reduction)
- Pb2+ (aq) + 2e- -> Pb (s) (Lead reduction)
2. Next, let's find the standard reduction potentials (E°) for these half-reactions. You can find these values in a standard potentials table. The values are:
- Ag+ (aq) + e- -> Ag (s) : E° = +0.80 V
- Pb2+ (aq) + 2e- -> Pb (s) : E° = -0.13 V
3. Now, we can apply the Nernst equation to calculate the cell potential (Ecell) using the given concentrations and the standard potentials:
Ecell = E°cell - (RT / nF) * ln(Q)
Where:
- Ecell is the cell potential
- E°cell is the standard cell potential (E° of reduction potential of silver minus E° of reduction potential of lead)
- 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 involved in the half-reactions (1 for both silver and lead)
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient, which can be calculated as:
Q = [Pb2+] / [Ag+]
4. Given the concentrations of the silver and lead ions, we have:
[Ag+] = 0.10 M
[Pb2+] = 1.0 M
5. Plugging in the values into the Nernst equation:
Ecell = (0.80 V - (-0.13 V)) - (8.314 J/(mol·K) * T) / (1 * 96,485 C/mol) * ln(1.0 M / 0.10 M)
Note: The value of ln(Q) simplifies to ln(1.0), which is 0.
6. Therefore, the equation becomes:
Ecell = 0.93 V - (8.314 J/(mol·K) * T) / (96,485 C/mol) * 0
7. Since there is no temperature given, we can't calculate the exact potential. However, based on the other information provided, the potential of the silver electrode in the cell would be approximately 0.93 volts.
To determine the potential of the silver electrode in the cell, you need to use the Nernst equation. The Nernst equation is given by:
E = E° - (RT / nF) * ln(Q)
Where:
- E is the cell potential
- E° is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin
- n is the number of electrons transferred in the balanced cell reaction
- F is the Faraday constant (96,485 C/mol)
- Q is the reaction quotient
First, let's determine the balanced cell reaction. Since both silver and lead are in the same ionic state in their respective salts, they will not undergo any oxidation or reduction. Therefore, the balanced cell reaction can be written as:
Ag+(aq) + Pb2+(aq) → Ag(s) + Pb2+(aq)
The number of electrons transferred (n) in this reaction is 1.
Next, we need to calculate the reaction quotient (Q). For a given concentration, Q is the ratio of the products' concentrations to the reactants' concentrations, each raised to the power of their stoichiometric coefficients.
In this case, Q can be simplified to:
Q = [Ag+] / [Pb2+]
We are given the concentrations of the silver and lead ions in the solutions. [Ag+] = 0.10 M and [Pb2+] = 1.0 M.
Substituting the values into the Nernst equation:
E = E° - (RT / nF) * ln(Q)
E = E° - (8.314 J/(mol·K) * T / (1 * 96,485 C/mol) * ln(0.10 / 1.0)
Now, we need to know the standard cell potential (E°) for the balanced cell reaction. This information is typically provided in a reference table or can be found experimentally. Let's assume the standard cell potential for this reaction is 0.50 V.
Substituting all the values into the equation:
E = 0.50 V - (8.314 J/(mol·K) * T / (1 * 96,485 C/mol) * ln(0.10 / 1.0)
Now, you can plug in the value for the temperature (in Kelvin) to calculate the potential of the silver electrode in the cell.
Please note that the actual value of the potential will depend on the temperature.
Use the Nernst equation.
EAg = EoAg + (0.05916/n)*log(1/Ag^+)