A solution is prepared in which a trace or small amount of Fe2+ is added to a mu
ID: 1035421 • Letter: A
Question
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 concentration of OH- is 1.0 x 10^-2 M. Some Fe(OH)2 precipitates. The value of Ksp for Fe(OH)2 is 8.0 x 10^-10. A. Assuming that the hydroxide concentration 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 ion 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.
Explanation / Answer
A) The dissociation of Fe(OH)2 is given as
Fe(OH)2 (s) <======> Fe2+ (aq) + 2 OH- (aq)
The solubility product constant for the dissociation of Fe(OH)2 is given as
Ksp = [Fe2+][OH-]2 = 8.0*10-10
Given [OH-] = 1.0*10-2 M, we have,
8.0*10-10 = [Fe2+]*(1.0*10-2)2
=====> 8.0*10-10 = [Fe2+]*(1.0*10-4)
=====> [Fe2+] = (8.0*10-10)/(1.0*10-4) = 8.0*10-6 M
The concentration of Fe2+ in the solution is 8.0*10-6 M.
B) Write the two half cell reactions.
Ni2+ (aq) + 2 e- --------> Ni (s); E0 = -0.25 V ……(1)
Fe2+ (aq) + 2 e- --------> Fe (s); E0 = -0.44 V …….(2)
The more positive (less negative) the standard reduction potential of a couple, the more is the tendency of the couple to be reduced. Since Ni2+/Ni has a less negative reduction potential, Ni2+ will be reduced while Fe will be oxidized. The cell reaction is given as
Ni2+ (aq) + Fe (s) ---------> Ni (s) + Fe2+ (aq)
E0cell = E0(1) + [-E0(2)] = (-0.25 V) + [-(-0.44 V)] = 0.19 V.
C) The cell can be represented as
Fe (s)?Fe2+ (8.0*10-6 M, aq)??Ni2+ (1 M, aq)?Ni (s)
The Nernst equation is written as
E = E0cell – (0.0592 V/n)*log [Fe2+]/[Ni2+] (n = 2 is the number of electrons involved in the redox)
======> E = (0.19 V) – (0.0592 V/2)*log (8.0*10-6 M)/(1 M)
======> E = (0.19 V) – (0.0296 V)*log (8.0*10-6)
======> E = (0.19 V) – (0.0296 V)*(-5.0969) = (0.19 V) – (-0.1509 V) = (0.19 V) + 0.1509 V = 0.3409 V ? 0.34 V (ans).
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.