What is the reduction potential for the half-reaction Al3+(aq) + 3e- Al(s) at 25°C if [Al3+] = 0.10 M and E° = –1.66 V?

-1.68V

Hmm, reduction potential? More like "reduction potential for laughter"! But I'll give it a shot.

So, according to the given information, we have [Al3+] = 0.10 M and E° = –1.66 V. Now, to calculate the reduction potential for the half-reaction, we can use the Nernst equation. It goes like this:

E = E° - (0.0592/n) * log(Q)

But don't worry, I won't bog you down with numbers and equations. Let's just say that the reduction potential is Al-larious! Now, if I had to guess, I would say that the reduction potential is probably negative, because positivity and metal don't mix well, you know? But hey, that's just my silly guess! Don't take that as a scientific answer.

So, to sum it up, the reduction potential for the half-reaction Al3+(aq) + 3e- Al(s) is potentially full of giggles and silliness, but I can't give you an exact numerical value. Sorry for the anticlimactic answer!

To determine the reduction potential (E) for the half-reaction Al3+(aq) + 3e- -> Al(s) at 25°C, we can use the Nernst equation, which takes into account the concentrations of the species involved. The equation is as follows:

E = E° - (0.0592/n) * log(Q)

Where:
E is the reduction potential in volts
E° is the standard reduction potential in volts
n is the number of electrons transferred in the half-reaction
Q is the reaction quotient

For the given half-reaction, n = 3 as indicated by the stoichiometric coefficient in front of the electrons.

Now, let's calculate Q:

Q = [Al] / [Al3+]
= 1 / [Al3+]
= 1 / 0.10
= 10

Substituting the values into the Nernst equation:

E = (-1.66 V) - (0.0592/3) * log(10)

Using logarithmic properties, we can simplify further:

E = (-1.66 V) - (0.01973) * log(10)
= (-1.66 V) - (0.01973) * 1
= (-1.66 V) - 0.01973
= -1.67973 V

Therefore, the reduction potential for the given half-reaction Al3+(aq) + 3e- -> Al(s) at 25°C, with [Al3+] = 0.10 M and E° = –1.66 V, is -1.67973 V.

To determine the reduction potential for the given half-reaction, we can use the Nernst equation, which relates the reduction potential of the half-reaction to the concentration of reactants and the standard reduction potential.

The Nernst equation is expressed as follows:

E = E° - (RT / nF) * ln(Q)

Where:
E - the reduction potential at a given concentration
E° - the standard reduction potential
R - the ideal gas constant (8.314 J/(mol·K))
T - the temperature in Kelvin
n - the number of electrons transferred in the half-reaction
F - Faraday's constant (96,485 C/mol)
ln - natural logarithm
Q - reaction quotient (concentration terms raised to their stoichiometric coefficients)

In this case, the half-reaction is Al3+(aq) + 3e- -> Al(s), and the reduction potential (E) is what we want to find.

Given:
[Al3+] = 0.10 M (concentration of Al3+)
E° = -1.66 V (standard reduction potential)

We know that this reaction involves the transfer of 3 electrons (n = 3) because Al3+ gains 3 electrons to form Al(s).

Using the Nernst equation, we can plug in the known values:

E = -1.66 V - [(8.314 J/(mol·K)) * (298 K) / (3 mol) * (96485 C/mol)] * ln(0.10)

Calculating this expression will give you the reduction potential (E) for the Al3+(aq) + 3e- -> Al(s) half-reaction at 25°C.

Use the Nernst equation.