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

the dependant variable is spilled water

the independent is the heat

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