[xtheunknown0]

Hello,

http://bouman.chem.georgetown.edu/S02/lect27/lect27.htm says:
"2H₂O (l) + 2e^- H₂ (g) + 2OH^-(aq)  E⁰ = -0.83 V
and
2H₂O (l) 4H⁺(aq) + O₂(g) + 4e^-  E⁰ = -1.23 V

These reactions can easily become the main reaction in an electrolysis setup when aqueous solutions are employed.  Thus, any metal that falls under the following category,

Mⁿ⁺ + ne^- M(s) E⁰ < -0.83 V

is not extractable from an aqueous solution (Examples are Al, Na and Li)."

Why is the last statement true?

TIA,
xtheunknown0

[Professor 0110]

"Extractable" means the metal will form at the cathode (by being reduced). The higher the E (reduction potential value) of an element, the more like that it will be reduced. The element with the more POSITIVE standard reduction value will be more readily reduced at the cathode. So, if you had an aqueous solution with Chromium, the standard reduction potential value is -0.73 volts.

Cr3+ + 3e-  --> Chromium metal (-0.73 volts). So at the cathode, the possibilities for reduction are:

2H2O (l) + 2e- -->  H2 (g) + 2OH-(aq)  E0 = -0.83 V

Cr3+ + 3e-  --> Chromium metal (-0.73 volts)

Which is the more positive standard reduction value? Chromium does! Therefore, chromium will be reduced at the cathode.

On the other hand, the reaction 2H2O (l) + 2e- --> H2 (g) + 2OH-(aq)  E0 = -0.83 V has a much more positive value than the process for the reduction of aluminum:

Al3+ + 3e- --> Al (E value of -1.66 volts).

Therefore, what do you think will be produced at the cathode? Hydrogen gas! From the equation:  2H2O (l) + 2e- --> H2 (g) + 2OH-(aq)

In all aqueous solutions, you have to take into account the water ions present.

Hope that helps.

[Borek]

Note that comparing just standard electrode potentials is not enough, you need to check formal potentials, taking into account concentrations of ions present and using Nernst equation.

[DevaDevil]

in what way neutral?