A solution is prepared in which a trace or small amount of Fe2+ is added to a much larger amount of solution in which the [OH-] is 1.0 x 10^-2 M. Some Fe(OH)2 precipitates. The value of Ksp for Fe(OH)2 = 8.0 x10^-10.
A. Assuming that the hydroxide is 1.0 x 10^-2 M, calculate the concentration of Fe2+ ions in the solution.
B. A battery is prepared using the above solution with an iron wire dipping into it as one half-cell. The other half-cell is the standard nickel electrode. Write the balanced net ionic equation for the cell reaction.
c. use the nerst equation to calculate the potential of the above cell.
I don't know how to do B & C
For (A) i got 8.0*10^-6M. if not please help
Thank You!
A is right.
What values are you using for Eo Ni and Eo Fe? My best table for those is over 50 years old.
EFe = EoFe -(0.0592/2)log(Fe/Fe^2+)
Substitute 1 for Fe and 8E-6 for Fe^2+, solve for EFe. Then reverse the sign and the half equation and add to the Ni reduction equation.
Fe + Ni^2+ ==> Fe^2+(8E-6M) + Ni
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A. Assuming that the hydroxide concentration is 1.0 x 10^-2 M, we can calculate the concentration of Fe2+ ions using the Ksp expression for Fe(OH)2:
Ksp = [Fe2+][OH-]^2
Since the concentration of OH- ions is 1.0 x 10^-2 M, we can substitute these values into the equation:
8.0 x 10^-10 = [Fe2+] * (1.0 x 10^-2)^2
Simplifying the equation, we get:
8.0 x 10^-10 = [Fe2+] * 1.0 x 10^-4
To solve for [Fe2+], we divide both sides of the equation by 1.0 x 10^-4:
[Fe2+] = 8.0 x 10^-6 M
Therefore, the concentration of Fe2+ ions in the solution is 8.0 x 10^-6 M.
B. The balanced net ionic equation for the cell reaction involving Fe2+ and the standard nickel electrode can be written as:
Fe(s) + 2Ni2+(aq) -> Fe2+(aq) + 2Ni(s)
C. To use the Nernst equation to calculate the potential of the cell, we need to know the concentrations of Fe2+ and Ni2+ ions in the respective half-cells. Since we only have the concentration of Fe2+ ions in the solution (8.0 x 10^-6 M in this case), we would need additional information to calculate the potential of the cell.
For part A: To calculate the concentration of Fe2+ ions in the solution, we can use the equilibrium expression for the dissolution of Fe(OH)2:
Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH-(aq)
The solubility product constant (Ksp) expression for Fe(OH)2 is:
Ksp = [Fe2+][OH-]^2
Given that the [OH-] is 1.0 x 10^-2 M and Ksp = 8.0 x 10^-10, we can rearrange the equation to solve for [Fe2+]:
[Fe2+] = Ksp / [OH-]^2
[Fe2+] = (8.0 x 10^-10) / (1.0 x 10^-2)^2
[Fe2+] = 8.0 x 10^-10 / 1.0 x 10^-4
[Fe2+] = 8.0 x 10^-6 M
So the concentration of Fe2+ ions in the solution is 8.0 x 10^-6 M. Your calculation is correct.
Now, let's move on to parts B and C:
Part B: To write the balanced net ionic equation for the cell reaction, we need to identify the oxidation and reduction half-reactions. The half-reactions can be determined by looking at the change in oxidation states for each element.
The iron (Fe) in the Fe2+ ions is oxidized to Fe3+ ions in the solution, and the nickel (Ni) in the nickel electrode is reduced to Ni2+ ions. The half-reactions can be written as follows:
Oxidation half-reaction:
Fe2+(aq) → Fe3+(aq) + e-
Reduction half-reaction:
Ni(s) + 2e- → Ni2+(aq)
To obtain the balanced net ionic equation, we multiply the oxidation half-reaction by 2 and add it to the reduction half-reaction:
2 Fe2+(aq) + Ni(s) → 2 Fe3+(aq) + Ni2+(aq)
So, the balanced net ionic equation for the cell reaction is 2 Fe2+(aq) + Ni(s) → 2 Fe3+(aq) + Ni2+(aq).
Part C: The Nernst equation relates the cell potential (Ecell) to the standard cell potential (E°cell), the gas constant (R), temperature (T), and the reaction quotient (Q). The Nernst equation is:
Ecell = E°cell - (RT/nF) ln(Q)
In this case, the standard cell potential for the reaction can be determined from standard reduction potentials. However, to calculate the potential of the cell, you need the concentrations of the species involved, which we already know for this problem.
The reaction quotient (Q) for the given cell reaction is calculated using the concentrations of the species involved:
Q = [Fe3+]^2/[Fe2+]^2
Q = (Fe3+)^2 / (Fe2+)^2
Substituting the concentrations (assuming they are the same as the initial concentrations determined in part A):
Q = (8.0 x 10^-6)^2 / (8.0 x 10^-6)^2
Q = 1
Now, you can use the Nernst equation to calculate the potential of the cell:
Ecell = E°cell - (RT/nF) ln(Q)
Since the standard cell potential (E°cell) is not given, I cannot provide you with a specific value for the potential of the cell. However, you can insert the known values, such as the gas constant (R), temperature (T), and the number of electrons transferred in the half-reactions (n) to calculate it.
For Ni its -0.28V
& for Fe its -.440V
Is it right?