Calculate E^$\circ$_cell for the following reaction: 2Fe²⁺(aq) + Cd²⁺(aq) $\rarr$ 2Fe³⁺(aq) + Cd(s)

Question

Quizplus · Accepted Answer

First, we need to write the two half-reactions and their standard reduction potentials:

FeS1U1P12 + 2e- → FeS1U1P13 E° = -0.37 V

Cd2+ + 2e- → Cd(s) E° = -0.40 V

Next, we need to determine which reaction is being oxidized and which is being reduced. Looking at the reactants and products, it appears that FeS1U1P12 is being oxidized to FeS1U1P13 (since it is gaining electrons), while Cd2+ is being reduced to Cd(s) (since it is losing electrons).

Now, we can use the Nernst equation to calculate the cell potential under non-standard conditions:

Ecell = E°cell - (RT/nF)ln(Q)

where Q is the reaction quotient, which we can write as:

Q = [FeS1U1P13]^2[Cd]/[FeS1U1P12]^2[Cd2+]

(Note that we don't include the concentration of S1S1P0(aq) because it is a spectator ion and doesn't participate in the reaction.)

Plugging in the values:

Ecell = -0.37 V - (0.0257 V/K)(298 K)/(2)(96485 C/mol)ln([(FeS1U1P13]^2[Cd])/[(FeS1U1P12]^2[Cd2+])

Ecell = -0.37 V - (0.0257 V/K)(298 K)/(2)(96485 C/mol)ln(1/1)

Ecell = -0.37 V

Finally, we take the negative of Ecell to get the potential of the reaction as written:

EF1F1F1S1F1F1F10S1U1B1cellS1U1B0 = -(-0.37 V) = 0.37 V

Therefore, the answer is choice C.

Calculate E ∘\circ∘ cell for the Following Reaction: 2Fe2+(aq) + Cd2+(aq) →\rarr→

Calculate E^$\circ$_cell for the Following Reaction: 2Fe²⁺(aq) + Cd²⁺(aq) $\rarr$