Question

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

Answer 1

E° = 0.00 V

E = 0.079 V

Step-by-step 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