Given that E°_red = -1.66 V for Al³⁺(aq)/Al(s)at 25 °C,find E° and E for the concentration cell expressed using shorthand notation below: Al(s)∣ Al³⁺(aq,1.0 × 10^-5 mol L^-1)∣∣ Al³⁺(aq,0.100 mol L^-1)∣ Al(s) A)E° = 0.00 V and E = +0.24 V B)E° = 0.00 V and E = +0.0789 V C)E° = -1.66 V and E = -1.42 V D)E° = -1.66 V and E = -1.54 V

Question

Quizplus · Accepted Answer

The shorthand notation can be expanded into two half-cells:
Al(s)∣ AlS1U1P13+S1S1P0(aq,1.0 × 10S1U1P1-5S1S1P0 mol LS1U1P1-1S1S1P0) and Al(s)∣ AlS1U1P13+S1S1P0(aq,0.100 mol LS1U1P1-1S1S1P0)

Using Nernst equation:
Ecell = E°cell - (0.0592 V/n) log(Q)
where Q is the reaction quotient and n is the number of electrons transferred in the half-reaction.

For the anode half-reaction:
E°anode = -1.66 V
n = 3 (based on the number of electrons transferred in the balanced half-reaction)
[Al3+] = 1.0 × 10-5 M
[electrons, e-] = 1.0 × 10-5 M (since the concentration of Al3+ and e- are equal)
Qanode = [products]/[reactants] = [AlS1U1P13+]/([Al3+] [e-]) = 1.0 × 10-5

For the cathode half-reaction:
E°cathode = -1.66 V (since it is the same reaction as the anode half-reaction, but reversed)
n = 3
[Al3+] = 0.100 M
[e-] = 0.100 M (since the concentration of Al3+ and e- are equal)
Qcathode = [products]/[reactants] = [AlS1U1P13+]/([Al3+] [e-]) = 1.0

Plugging in the values:
Eanode = -1.66 - (0.0592/3) log(1.0 × 10-5) = -1.66 V
Ecathode = -(-1.66) - (0.0592/3) log(1.0) = -1.578 V
Ecell = Ecathode - Eanode = (-1.578) - (-1.66) = 0.0789 V

Since E°cell = 0 V (at standard conditions), the only valid option is B) E° = 0.00 V and E = +0.0789 V.

Given That E°red = -1

Given That E°_red = -1