Ag2CrO4(s) + 2e- 2Ag(s) + CrO42-(aq)
A chemist wishes to determine the concentration of CrO42- electrochemically. A cell is constructed consisting of a saturated calomel electrode (SCE; which has a reduction potential of 0.242 V relative to the SHE) and a silver wire coated with Ag2CrO4 and suspended in a CrO42- solution.
1) What is the potential of this cell at 25�‹C when [CrO42-] = 1.00 M?
2) Calculate ƒ¢G for the full-cell reaction at 25�‹C when [CrO42-] = 1.00 M?
3) What is the potential of this cell at 25�‹C when [CrO42-] = 1.10�~10-6 M?
4) For a solution of unknown [CrO42-], the measured potential for the cell at 25�‹C is 0.363 V. What is [CrO42-] (in mol/L)?
To determine the concentration of CrO42- electrochemically using the given cell, we can follow these steps:
1) Calculate the cell potential at 25°C when [CrO42-] = 1.00 M:
The cell reaction is: Ag2CrO4(s) + 2e- -> 2Ag(s) + CrO42-(aq)
The half-cell reduction potential for Ag2CrO4/Ag is 0.242 V (given). The half-cell reduction potential for CrO42-/Cr3+ is not provided, so we will assume it to be zero.
Using the Nernst equation: Ecell = E°cell - (0.0592/n) * log(Q)
Where:
E°cell is the standard cell potential (0.242 V)
n is the number of electrons transferred (2)
Q is the reaction quotient [CrO42-]/[Cr3+]
Since CrO42- is the only species in the CrO42- solution, [Cr3+] can be assumed to be negligible.
So, Q = [CrO42-]/(1) = [CrO42-]
Ecell at 25°C = 0.242 V - (0.0592/2) * log(1.00 M)
Ecell = 0.242 V - (0.0296) * log(1.00)
Ecell ≈ 0.242 V - (0.0296) * 0
Ecell ≈ 0.242 V
2) Calculate ΔG for the full-cell reaction at 25°C:
ΔG = -n * F * Ecell
Where:
n is the number of electrons transferred (2)
F is Faraday's constant (96485 C/mol)
Ecell is the cell potential (0.242 V)
ΔG = -2 * 96485 C/mol * 0.242 V
ΔG ≈ -46,707 J/mol
3) Calculate the cell potential at 25°C when [CrO42-] = 1.10 × 10^(-6) M:
Using the same process as in step 1:
Ecell = E°cell - (0.0592/2) * log(1.10 × 10^(-6) M)
Ecell ≈ 0.242 V - (0.0296) * log(1.10 × 10^(-6))
Ecell ≈ 0.242 V - (0.0296) * (-5.9588)
Ecell ≈ 0.242 V - (-0.17643)
Ecell ≈ 0.242 V + 0.17643
Ecell ≈ 0.418 V
4) Calculate [CrO42-] for a cell potential of 0.363 V at 25°C:
Using the Nernst equation, we need to rearrange it to solve for [CrO42-]:
Ecell = E°cell - (0.0592/n) * log([CrO42-]/[Cr3+])
0.363 V = 0.242 V - (0.0592/2) * log([CrO42-]/[Cr3+])
0.363 V - 0.242 V = (0.0296/2) * log([CrO42-]/[Cr3+])
0.121 V = 0.0148 * log([CrO42-]/[Cr3+])
8.175 = log([CrO42-]/[Cr3+])
[CrO42-]/[Cr3+] = 10^(8.175)
Since [Cr3+] is negligible relative to [CrO42-], we can assume it to be zero. Therefore, [CrO42-] = 10^(8.175) mol/L.
So, the concentration of CrO42- in the solution is approximately 10^(8.175) mol/L.
To determine the concentration of CrO42- electrochemically using the given cell, we need to apply the Nernst equation. The Nernst equation relates the potential of a cell to the concentrations of the species involved. The equation is:
E = E° - (RT/nF) ln([oxidized]/[reduced])
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 equation
F is the Faraday constant (96,485 C/mol)
[oxidized] is the concentration of the oxidized species
[reduced] is the concentration of the reduced species
Let's proceed step by step to answer each question:
1) What is the potential of this cell at 25�‹C when [CrO42-] = 1.00 M?
First, we need to find the balanced equation for the cell reaction. The given equation is:
Ag2CrO4(s) + 2e- ⟶ 2Ag(s) + CrO42-(aq)
From the equation, we can see that 2 electrons are transferred. So, n = 2.
The standard cell potential, E°, is the potential of the cell under standard conditions. It can be found by looking up the reduction potentials of the half-reactions involved. In this case, we need to find the reduction potential for Ag2CrO4(s) and CrO42-(aq).
The reduction potential for Ag2CrO4(s) is not given, but we are given that the saturated calomel electrode (SCE) has a reduction potential of 0.242 V relative to the standard hydrogen electrode (SHE). The SHE is our reference electrode with an E° = 0 V.
Therefore, the E° for the half-reaction Ag2CrO4(s) + 2e- ⟶ 2Ag(s) is 0.242 V.
Now we can use the Nernst equation:
E = E° - (RT/nF) ln([oxidized]/[reduced])
Here, [oxidized] represents CrO42- concentration, and [reduced] represents Ag concentration, which is related to Ag2CrO4 concentration.
At 25�‹C, we have:
T = 25 + 273 = 298 K
Plugging in the values:
E = 0.242 V - [(8.314 J/(mol·K))/(2 * 96,485 C/mol)] * ln(1.00 M/([Ag]/1 M)^2)
We don't have the exact concentration of Ag, but let's assume it is negligibly small compared to 1 M since it is a solid coating on the silver wire. In that case, ([Ag]/1 M)^2 will be close to zero and can be ignored in the calculation.
Therefore, our final equation becomes:
E ≈ 0.242 V - [(8.314 J/(mol·K))/(2 * 96,485 C/mol)] * ln(1.00 M)
Calculating this equation will give you the cell potential.
2) Calculate ƒ¢G for the full-cell reaction at 25�‹C when [CrO42-] = 1.00 M?
The standard Gibbs free energy change, ƒ¢G°, can be found using the equation:
ƒ¢G° = -nF E°
Here, n = 2, F is the Faraday constant (96,485 C/mol), and E° is the standard cell potential we found earlier (0.242 V).
Calculate the value of ƒ¢G° using the given equation.
3) What is the potential of this cell at 25�‹C when [CrO42-] = 1.10�~10-6 M?
For this question, we follow the same process as in question 1, but this time, we use the given concentration of [CrO42-] = 1.10�~10-6 M.
Plugging in the new [CrO42-] value into the Nernst equation, we can calculate the potential of the cell.
4) For a solution of unknown [CrO42-], the measured potential for the cell at 25�‹C is 0.363 V. What is [CrO42-] (in mol/L)?
To find the concentration of CrO42-, we can rearrange the Nernst equation and solve for [CrO42-]:
ln([CrO42-]/1 M) = (E° - E) * (nF/RT)
Given the measured potential (E) as 0.363 V and all the other values from the previous steps, we can calculate the concentration of CrO42-.
WEBASSIGN....
too bad no one answered this question...