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)?
Assistance needed.
yes please
To answer these questions, we need to use the Nernst equation, which relates the potential of an electrochemical cell to the concentrations of the species involved:
E = E° - (RT/nF) * ln(Q)
Where:
- E is the potential of the cell
- E° is the standard reduction potential of the cell
- 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 balanced equation
- F is Faraday's constant (96485 C/mol)
- ln(Q) is the natural logarithm of the reaction quotient
- Q is the reaction quotient, which is the ratio of the concentrations of the species involved raised to their stoichiometric coefficients
Now let's solve each question step by step:
1) What is the potential of this cell at 25�‹C when [CrO42-] = 1.00 M?
First, we need to determine the number of electrons transferred in the balanced reaction:
Ag2CrO4(s) + 2e- -> 2Ag(s) + CrO42-(aq)
The number of electrons transferred is 2.
Now, we substitute the values into the Nernst equation:
E = E° - (RT/nF) * ln(Q)
E° is given as 0.242 V relative to the standard hydrogen electrode (SHE), so E° = 0.242 V.
R = 8.314 J/(mol·K)
T = 25°C = 298 K
n = 2 (from the balanced reaction)
F = 96485 C/mol
ln(Q) = ln([Ag+]^2[CrO42-]/[Ag2CrO4])
At equilibrium, the expression Q should equal 1. So, we can simplify ln(Q) to zero in this case.
Therefore, the potential of the cell at 25�‹C when [CrO42-] = 1.00 M is:
E = E° = 0.242 V.
2) Calculate ƒ¢G for the full-cell reaction at 25�‹C when [CrO42-] = 1.00 M?
The change in Gibbs free energy (ΔG) can be calculated using the equation:
ΔG = -nF*E
Substituting the values:
n = 2 (from the balanced reaction)
F = 96485 C/mol
E = 0.242 V (potential of the cell at 25�‹C calculated in the previous question)
ΔG = -2 * 96485 C/mol * 0.242 V
= -46604 J/mol
= -46.6 kJ/mol (approximately)
Therefore, the value of ΔG for the full-cell reaction at 25�‹C when [CrO42-] = 1.00 M is approximately -46.6 kJ/mol.
3) What is the potential of this cell at 25�‹C when [CrO42-] = 1.10�~10-6 M?
Using the same steps as in question 1, we can calculate the potential.
Substituting the new concentration ([CrO42-] = 1.10x10^-6 M) into the Nernst equation, we can find the potential.
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)?
In this case, we need to rearrange the Nernst equation to solve for the concentration. The equation becomes:
ln(Q) = ((E° - E) * nF) / (RT)
Now, plug in the known values and solve for Q. Then, based on the stoichiometry of the balanced equation, you can determine the concentration of [CrO42-].