A voltaic cell is constructed as follows:
Ag(s)|Ag^+,(satd Ag2CrO4)||Ag^+(0.110 M)|Ag(s).
What is the value of Ecell? For Ag2CrO4, Ksp=1.1*10^-12.
Calculate Ag^+ in satd Ag2CrO4 from Ksp.
Use Eo for Ag/Ag^+ and the Nernst equation to calculate E for the left half of the cell.
Use the Nernst equation to calculate E for the right half of the cell.
Use the values to determine which half is oxidized and which half is reduced, add the oxidation half to the reduction half and calculate the cell potential.
To find the value of Ecell, we can use the Nernst equation:
Ecell = E°cell - (RT / nF) * ln(Q)
Where:
Ecell is the cell potential
E°cell 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 moles of electrons transferred in the balanced equation
F is Faraday's constant (96485 C/mol)
ln is the natural logarithm
Q is the reaction quotient
In this case, the balanced equation for the cell reaction is:
2Ag^+(aq) + (saturated Ag2CrO4) + 2Ag(s)
Since there are no gases involved in this reaction, the reaction quotient Q is equal to the equilibrium constant K.
The reaction can be written as:
Ag2CrO4(s) ⇌ 2Ag^+(aq) + CrO4^2-(aq)
Using the solubility product constant (Ksp) given, we can calculate the concentration of Ag^+ ions from the solubility of Ag2CrO4.
1.1 * 10^(-12) = [Ag^+]^2
Taking the square root of both sides:
[Ag^+] = 1.05 * 10^(-6) M
Now we have all the values we need to calculate Ecell:
E°cell is not given, so we need to look it up in a table of standard reduction potentials.
The standard reduction potential for the Ag^+ + e^- --> Ag half-reaction is 0.799 V.
Plugging in the values into the Nernst equation:
Ecell = 0.799 V - (8.314 J/mol·K * temperature / (2 * 96485 C/mol)) * ln(1.05 * 10^(-6))
Note: Since the temperature is not given, you need to substitute the actual temperature in Kelvin in the equation to get the final value of Ecell.
To calculate the value of Ecell for the given voltaic cell, we need to use the Nernst equation. The Nernst equation relates the cell potential (Ecell) to the concentrations of the species involved in the cell reaction.
The voltaic cell is represented as:
Ag(s)|Ag^+,(satd Ag2CrO4)||Ag^+(0.110 M)|Ag(s)
To use the Nernst equation, we need to identify the half-cell reactions that occur at the anode and the cathode.
Anode: Ag(s) → Ag^+(satd Ag2CrO4) + e^-
Cathode: Ag^+(0.110 M) + e^- → Ag(s)
The standard reduction potential (E°) for the half-cell reactions can be obtained from a standard reduction potential table.
The Nernst equation is:
Ecell = E°cell - (0.0592/n)logQ
where Ecell is the cell potential, E°cell is the standard cell potential, n is the number of moles of electrons transferred, and Q is the reaction quotient.
In this case, the number of moles of electrons transferred (n) is 1 for both half-cell reactions.
Now, let's calculate the reaction quotient (Q) using the concentrations of the species involved:
Q = [Ag^+(satd Ag2CrO4)] / [Ag^+(0.110 M)]
Given that Ksp = [Ag^+(satd Ag2CrO4)]^2, we can solve for [Ag^+(satd Ag2CrO4)]:
Ksp = (1.1 × 10^-12) = [Ag^+(satd Ag2CrO4)]^2
[Ag^+(satd Ag2CrO4)] = sqrt(1.1 × 10^-12)
Now we can substitute the values in the Nernst equation to calculate Ecell:
Ecell = E°cell - (0.0592/n)log([Ag^+(satd Ag2CrO4)] / [Ag^+(0.110 M)])
= E°cell - (0.0592/1)log(sqrt(1.1 × 10^-12) / (0.110))
Remember to also consider the sign conventions for the half-cell reactions when plugging in the values.
Using the reference table and the concentrations provided, you can calculate Ecell.
omg. there is thing called chemistry solutions manual ... should have came with your book..
Plz consult that and subsitute the values you see on your MC page.
Cheers!!!