At 25 °C, you conduct a titration of 15.00 mL of a 0.0260 M AgNO3 solution with a 0.0130 M NaI solution within the following cell:
Saturated Calomel Electrode || Titration Solution | Ag (s)
For the cell as written, what is the voltage after the addition of the following volume of NaI solution? The reduction potential for the saturated calomel electrode is E = 0.241 V. The standard reduction potential for the reaction
Ag+ +e- --> Ag(s)
is E0 = 0.79993 V. The solubility constant of AgI is Ksp = 8.3 × 10-17.
a) 0.5 mL ; b) 17.0 mL ; c) 30.0 mL ; d) 39.7 mL
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This is what I did:
AgNO3 + NaI --> AgI + NaNO3
For the half-reaction: NaI --> Na+ + I-
I set ICE table and got concentration of [I-] = 0.0128 M
Then I found the concentration of [Ag+]:
[Ag+] = Ksp/[I-] = 6.48*10^(-15) M
Then I found cell potential of Ag electrode:
E = 0.799 - 0.05916 log(1/[Ag+]) = -0.0404 V
And now I don't know how to proceed . How do I calculate the voltage after the addition of certain volumes of NaI solution?
Thank you for the help.
I think you need to redo the ICE table and I don't understand what (or why) you did with the NaI.
The (AgNO3) for the first addition is
0.0260 x (15.00/15.50) = 0.02516 M.
The NaI is reduced likewise as
0.0130 x (0.5/15.5) = 0.0004194 and you should check these numbers and add or subtract digits as appropriate.
..........AgNO3 + NaI ==> AgI + NaNO3
I......0.02516.....0.......0......0
add...........0.0004194............
C...-0.0004194..-0.0004194..0.0004194
E...0.02474........0.......0.0004194
I don't think you need Ksp for this one or the 17.0 mL addition since the Ag^+ from the AgNO3 far more than that from the solubility of AgI. I think you will need it for the 30.0 mL and 39.7 mL.
When you find those concentrations, you substitute it into the formula you have shown and calculate voltage. Remember, however, that the voltage you are calculating is with reference to the H electrode so when you finish, correct for voltage with reference calomel (SCE); that's the voltage the problem is asking for.
Well, it seems like you've made some good progress so far! To calculate the voltage after the addition of a certain volume of NaI solution, you need to consider the change in the concentration of I- ions in the solution.
Since you already know the initial concentration of I- (0.0128 M), and you're adding NaI solution, which contains I- ions, you can calculate the final concentration of I- after adding a certain volume.
For example, let's consider option a) 0.5 mL of NaI solution. You would need to calculate the final concentration of I- when 0.5 mL of NaI solution is added to the reaction mixture.
Once you have the final concentration of I-, you can then use the Nernst equation to calculate the new cell potential.
Remember, the Nernst equation is:
Ecell = E0 - (0.05916/n) * log(Q),
where Ecell is the cell potential, E0 is the standard reduction potential, n is the number of moles of electrons transferred, and Q is the reaction quotient.
In this case, since you're dealing with the half-reaction for I-, n = 1. So you can substitute the values into the Nernst equation and calculate the new cell potential.
Repeat this process for each option (b, c, and d), and see which one gives you the corresponding voltage.
I hope this helps, and good luck with your calculations!
To calculate the voltage after the addition of certain volumes of NaI solution, you need to consider the changes in concentrations of the reactants and products in the cell.
1. Start by determining the moles of I- that will react with Ag+.
Moles of I- = concentration of I- x volume of NaI solution (in liters)
For example, if you're calculating the voltage after adding 0.5 mL of NaI solution:
Moles of I- = 0.0128 M x 0.5 mL / 1000 mL/L = 6.4 x 10^-6 mol
2. Use the stoichiometry of the reaction to determine the change in Ag+ concentration.
From the balanced equation, the stoichiometric ratio is 1:1 for Ag+ and I-.
So, the change in Ag+ concentration = change in I- concentration = 6.4 x 10^-6 mol/L
3. Calculate the new concentration of Ag+ after the addition of the NaI solution.
Initial Ag+ concentration = 6.48 x 10^-15 M (from your calculation)
New Ag+ concentration = Initial Ag+ concentration - change in Ag+ concentration
= 6.48 x 10^-15 M - 6.4 x 10^-6 M
4. Use the Nernst equation to calculate the new cell potential.
E = E0 - 0.05916 log([Ag+])
E = 0.79993 V - 0.05916 log[(6.48 x 10^-15 M) / (6.48 x 10^-15 M - 6.4 x 10^-6 M)]
Repeat steps 1-4 for each volume of NaI solution given in the question (0.5 mL, 17.0 mL, 30.0 mL, 39.7 mL). The volume of NaI solution will affect the moles of I- and the resulting changes in Ag+ concentration, which in turn will impact the cell potential.
To calculate the voltage after the addition of certain volumes of NaI solution, you need to consider the stoichiometry of the reaction, the change in concentration of the species involved, and the Nernst equation.
1. Determine the balanced equation for the reaction: AgNO3 + NaI -> AgI + NaNO3
2. Calculate the moles of AgNO3 initially present:
Moles of AgNO3 = concentration (M) x volume (L)
Moles of AgNO3 = 0.0260 M x 0.01500 L = 0.00039 moles
3. Determine the limiting reactant and calculate the moles of AgI formed:
From the reaction stoichiometry, 1 mole of AgNO3 forms 1 mole of AgI.
So, the moles of AgI formed is also 0.00039 moles.
4. Calculate the concentration of AgI:
Concentration of AgI = moles of AgI / total volume (L)
Total volume = initial volume (15.00 mL) + volume of NaI added (in mL) / 1000 (to convert mL to L)
For example, if you are calculating for 0.5 mL NaI added, the total volume = 15.00 mL + 0.5 mL = 15.50 mL = 0.01550 L
Concentration of AgI = 0.00039 moles / 0.0155 L = 0.0252 M
5. Use the solubility constant (Ksp) to calculate the concentration of I- ions:
Ksp = [Ag+][I-]
[I-] = Ksp / [Ag+] = (8.3 x 10^-17) / (0.0252 M) = 3.29 x 10^-15 M
6. Use the Nernst equation to calculate the voltage:
Ecell = E°cell - (0.05916 / n) x log(Q)
n = number of electrons transferred (from balanced equation)
Q = reaction quotient = [I-] / [Ag+]
E°cell = standard cell potential = 0.79993 V (given)
For the half-reaction Ag+ + e- -> Ag(s):
n = 1 (Moles of electrons transferred is 1)
For example, if you want to calculate the voltage after adding 0.5 mL of NaI solution:
[I-] = 3.29 x 10^-15 M
[Ag+] = concentration of Ag+ initially present (6.48 x 10^-15 M in your case)
Q = [I-] / [Ag+] = (3.29 x 10^-15 M) / (6.48 x 10^-15 M)
Ecell = 0.79993 V - (0.05916 / 1) x log[(3.29 x 10^-15 M) / (6.48 x 10^-15 M)]
7. Calculate the voltage for each of the given volumes of NaI solution and compare it with the options provided. The correct choice will match the calculated voltage.
Repeat the calculation for other volumes (17.0 mL, 30.0 mL, and 39.7 mL) to find the correct answer.