Given that E°_red = -0.40 V for Cd²⁺/Cd at 25°C,find E° and E for the concentration cell expressed using shorthand notation below. Cd(s)∣ Cd²⁺(1.0 × 10^-5 M)∣∣ Cd²⁺(0.100 M)∣ Cd(s)

Question

Quizplus · Accepted Answer

The shorthand notation represents a concentration cell where the anode and cathode compartments have different concentrations of the same species. In this case, the anode compartment has a concentration of 1.0 × 10-5 M CdS1P12+S/0.1 M Cd, while the cathode compartment has a concentration of 0.1 M CdS1P12+S/0.1 M Cd.

To find E° and E, we need to use the Nernst equation:

E = E° - (RT/nF) ln(Q)

where Q is the reaction quotient, defined as the product of the concentrations of the species on the right side of the half-reaction over the product of the concentrations of the species on the left side of the half-reaction.

For the anode half-reaction:

CdS1P12+S + 2e- → Cd + S1P0

we have:

E°anode = -0.40 V
n = 2
Qanode = [Cd2+]/[CdS1P12+S][S1P0]2

Since the anode compartment has an excess of Cd, [Cd2+] = 0.1 M, while [CdS1P12+S] and [S1P0] are as given above. Therefore:

Qanode = (0.1 M)/(1.0 × 10-5 M)(0.1 M)2 = 1000

For the cathode half-reaction:

CdS1P12+S + 2e- → Cd + S1P0

we have:

E°cathode = 0.00 V
n = 2
Qcathode = [Cd2+]/[CdS1P12+S][S1P0]2

Since the cathode compartment has a lower concentration of CdS1P12+S, [CdS1P12+S] = 0.1 M, while [Cd2+] and [S1P0] are as given above. Therefore:

Qcathode = (0.1 M)/(0.1 M)(0.1 M)2 = 1

Substituting into the Nernst equation, we get:

Eanode = -0.40 V - (0.0257 V/K)(298 K)/(2 F) ln(1000) = -0.64 V
Ecathode = 0.00 V - (0.0257 V/K)(298 K)/(2 F) ln(1) = 0.00 V

The overall cell potential is given by:

Ecell = Ecathode - Eanode = 0.64 V

And since the shorthand notation represents a concentration cell, we know that E°cell = 0.00 V, so:

E° = E°cell - (RT/nF) ln(Kcell)

where Kcell = [Cd2+]anode/[Cd2+]cathode = 1000/1 = 1000

Substituting the values and solving for E°, we get:

E° = 0.00 V - (0.0257 V/K)(298 K)/(2 F) ln(1000) = -0.64 V

Therefore, the answer is B: E° = 0.00 V and E = +0.12 V.

Given That E°red = -0

Given That E°_red = -0