Question

An electrochemical cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminium metal electrode immersed in a solution with [Al3+] = 1.0 M. Sodium hydroxide is added to the aluminium compartment causing Al(OH)3(s) to precipitate. After precipitation of Al(OH)3 has ceased, the concentration of OH- is 1.0 . 10-4 M and the measured cell potential is 1.82 V. Calculate Ksp for Al(OH)3.

Please help me! thank you so much!

Answer 1

1. Find the the standard cell potential of the initial Al-Ni electrochemical cell (before the addition of NaOH). Consult your General Chemistry textbook on how to do this.
2. The solubility equilibrium is:
Al(OH)3(s) <=>Al^3+(aq) + 3OH-(aq)
Ksp = [Al+3][OH]^3
[OH-](eq) = 1.0x10^-4 (given)
[Al+3](eq) = ____? (see below)
3. The electrochemical reaction is:
2Al(s) + 3Ni^+2(aq) ---> 2Al^+3(aq) + 3Ni(s)
According to the Nernst Equation,
E(cell) = Eo - [(0.0592)/n)]log{[Al+3]/[Ni+2]}
E(cell) is given, Eo you calculate, n = 3 electrons for Al, and [Ni+2] is given. Solve for [Al+3].
Substitute into the expression for the Ksp.
4. Look up the Ksp for Al(OH)3 and see if your answer has the correct power of 10. If it does, your calculations should be OK.

Answer 2

1.62 x 10^-33