Be sure to answer all parts. Consider the following reaction: Cd (s) + Fe2 + (aq) → Cd2 + (aq) + Fe (s) E o (Fe2 + / Fe) = - 0.4400 V, E o (Cd2 + / Cd) = - 0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+] = 0.80 M and [Cd2+] = 0.010 M. E = V Will the reaction occur spontaneously at these conditions? yes no cannot predict

EMF = - 0.17 V

No

Explanation:

For the reaction:

Cd (s) + Fe2 + (aq) → Cd2 + (aq) + Fe (s)

we will be using the Nersnt equation to calculate Ecell:

Ecell = Eº - 0.0592 V/2 ln Q

where Q = (Fe⁺²) / (Cd²⁺)

and (Fe²⁺) and (Cd²) are the molar concentration of Fe²⁺ and Cd²⁺

Reduction

Fe²⁺ + 2e⁻ ⇒ Fe (s) Eº red = - 0.4400 V

Oxidation

Cd (s) ⇒ Cd²⁺ + 2e⁻ Eº ox = + 0.4000 V

Eºcell = Eox + Ered = 0.4000 V + (-0.4400 V) = - 0. 0400 V

Ecell = Eº - 0.0692 V/2 ln Q = - 0.0400 V - 0.0592/2 ln (0.80/0.010) = - 0.17V

No because Eºcell is negative and ΔºG will be positive since

ΔºG = - nFEºcell