A voltaic cell, with Ecell = 0.180 V at 298K, is constructed as follows:
Ag(s)|Ag+(satdAg3PO4)||Ag+(0.135 M )|Ag(s)
What is the Ksp of Ag3PO4?
Could you type in the example problem of PbI2 and the answer they gave for that example. If so perhaps I can help. Don't make a new post out of it; use this post for that.
*Update* I tried following a similar example in my textbook using the Nernst equation but still got the wrong answer
log(xM/0.135M) = 2(0.180V)/-0.0592 = -6.081081081
xM/0.135M = 10^-6.081 = 8.29E-7(0.135M) = 1.12E-7M = x = [Ag+]anode
1.12E-7(2) = 2.24E-7M
Ksp= 1.12E-7(2.24E-7)^2 = 5.62E-21
My book gave this explanation for the similar example: "Because the electrodes in this cell are identical, the standard electrode potentials are numerically equal and subtracting one from the other leads to the value Eo = 0.000V. However, because the ion concentrations differ, there is a potential difference between the two half cells (non-zero nonstandard voltage for the cell). [Pb2+] = 0.100 M in the cathode compartment, while the anode compartment contains a saturated solution of PbI2 .
We use the Nernst equation (with n = 2 ) to determine Pb2+ in the saturated solution." Not entirely sure how this Nernst equation works because my prof has not covered this yet; nonetheless my assignment is due tonight.
To find the solubility product constant (Ksp) of Ag3PO4, we need to use the Nernst equation and the given cell diagram.
The Nernst equation is given by:
Ecell = E°cell - (0.0592/n) log(Q)
where:
- Ecell is the cell potential
- E°cell is the standard cell potential
- n is the number of electrons transferred in the balanced redox equation
- Q is the reaction quotient
In the given cell diagram, we have:
Anode: Ag(s) | Ag+(satdAg3PO4)
Cathode: Ag+(0.135 M) | Ag(s)
The balanced redox equation for the cell reaction can be written as:
3Ag+(satdAg3PO4) + Ag(s) → 3Ag(s) + Ag3PO4(s)
Since we have 3 Ag+ ions from Ag3PO4 producing 3 Ag(s) in the electrode, the value of n in the Nernst equation is 3.
Since the voltage is given as the standard cell potential, E°cell, we can rewrite the Nernst equation as:
0.180 V = E°cell - (0.0592/3) log(Q)
We are given E°cell = 0.180 V, and we need to find Q to solve for Ksp.
In the anode half-cell, Ag is the pure solid electrode. We can assume its activity is 1, so [Ag+] in Q is equal to the saturation concentration of Ag3PO4.
In the cathode half-cell, Ag+ has a concentration of 0.135 M.
Using the Nernst equation, the equilibrium condition, and the balanced redox equation, we can write:
i) At equilibrium, Ecell = 0.180 V
ii) Q at equilibrium is equal to the ratio of the concentrations of products over reactants.
0.180 V = E°cell - (0.0592/3) log( [Ag+]^3 / Ag+ )
Substituting the given concentration of Ag+ in the cathode half-cell:
0.180 V = E°cell - (0.0592/3) log( [Ag+]^3 / (0.135 M) )
Simplifying the equation:
0.180 V = E°cell - (0.0592/3) log( [Ag+]^3 / 0.135 )
Rearranging the equation to isolate Ksp:
log( [Ag+]^3 / 0.135 ) = (0.180 V - E°cell) * 3 / 0.0592
Taking the antilog of both sides:
[Ag+]^3 / 0.135 = 10^[(0.180 V - E°cell) * 3 / 0.0592]
Simplifying further:
[Ag+]^3 = 0.135 * 10^[(0.180 V - E°cell) * 3 / 0.0592]
Finally, taking the cube root of both sides:
[Ag+] = (0.135 * 10^[(0.180 V - E°cell) * 3 / 0.0592])^ (1/3)
Now, substitute the given value of E°cell = 0.180 V into the equation and calculate [Ag+]. Then, use this concentration to calculate the Ksp of Ag3PO4.
Please note that there might be a small rounding error due to the use of approximation.
To find the Ksp (solubility product constant) of Ag3PO4, we need to use the Nernst equation and the cell potential (Ecell) of the voltaic cell.
First, let's identify the half-reactions occurring at the anode and cathode:
Anode: Ag(s) → Ag+(satdAg3PO4) + e-
Cathode: Ag+(0.135 M) + e- → Ag(s)
From the given information, we know that the cell potential (Ecell) is 0.180 V. We can assign the anode half-reaction as reduction (since Ag+(satdAg3PO4) is being reduced to Ag), and the cathode half-reaction as oxidation.
Next, we need to balance the half-reactions by ensuring that the number of electrons gained in the reduction matches the number of electrons lost in the oxidation.
Balanced half-reactions:
Anode: Ag(s) → Ag+(satdAg3PO4) + 3e-
Cathode: 3Ag+(0.135 M) + 3e- → 3Ag(s)
The stoichiometry of the balanced half-reactions tells us that three moles of Ag+ ions are involved in the formation of one mole of Ag3PO4. Therefore, the balanced net ionic equation for the cell is:
Ag+(satdAg3PO4) + 3Ag+(0.135 M) → Ag3PO4(s) + 3Ag(s)
Now, let's use the Nernst equation:
Ecell = E°cell - (RT/nF) * ln(Q)
where:
- Ecell is the cell potential (0.180 V)
- E°cell is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (298 K)
- n is the number of electrons involved in the reaction (3 in this case)
- F is the Faraday constant (96485 C/mol)
- Q is the reaction quotient, which is the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants, each raised to the power of their stoichiometric coefficient.
At equilibrium, Q = Ksp, and Ecell = 0. Therefore, we can rearrange the Nernst equation to solve for Ksp:
Ksp = exp((E°cell * n * F) / (RT))
Now, substitute the given values into the equation:
Ksp = exp((0.180 V * 3 * 96485 C/mol) / (8.314 J/(mol·K) * 298 K))
Calculating this expression will give us the solubility product constant (Ksp) of Ag3PO4.