Given that E°red = - 1.66 V for Al3+ / Al at 25°C, find E° and E for the concentration cell expressed using shorthand notation below.

Al (s) Al3 + (1.0 * 10-5 M) Al3 + (0.100 M) Al (s)

A) E° = 0.00 V and E = + 0.24 V

B) E° = 0.00 V and E = + 0.12 V

C) E° = - 1.66 V and E = - 1.42 V

D) E° = - 1.66 V and E = - 1.54 V

E° = 0.00 V

E = 0.079 V

Explanation:

We can identify both half-reactions occurring in a concentration cell.

Anode (oxidation) : Al (s) → Al³⁺ (1.0 * 10⁻⁵ M) + 3 e⁻ E°red = - 1.66 V

Cathode (reduction) : Al³⁺ (0.100 M) + 3 e⁻ → Al (s) E°red = - 1.66 V

The global reaction is:

Al (s) + Al³⁺ (0.100 M) → Al³⁺ (1.0 * 10⁻⁵ M) + Al (s)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = - 1.66 V - (-1.66 V) = 0.00 V

To calculate the cell potential (E) we have to use the Nernst equation.

E = E° - (0.05916/n). log Q

where,

n: moles of electrons transferred

Q: reaction quotient

E = 0.00 V - (0.05916/3). log (1.0 * 10⁻⁵/0.100)

E = 0.079 V