Question

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

Answer 1

Answer:

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