Calculate the emf for the following reaction. Will the reaction occur spontaneously at 25°C, given that [Fe2+] = 0.600 M and [Cd2+] = 0.00550 M? Cd(s) + Fe2+(aq)→Cd2+(aq) + Fe(s) E o Cd2+/Cd = −0.40 V E o Fe2+/Fe = −0.44 V

The EMF of the cell is -0.100 V and the reaction is non-spontaneous

Explanation:

The half reaction for the equation follows:

Oxidation half reaction:

Reduction half reaction:

Net cell reaction:

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the of the reaction, we use the equation:

Putting values in above equation, we get:

where,

= electrode potential of the cell =  ?

= standard electrode potential of the cell = -0.04 V

n = number of electrons exchanged = 2

Putting values in above equation, we get:

For the reaction to be spontaneous, the Gibbs free energy of the reaction must come out to be negative.

Relationship between standard Gibbs free energy and standard electrode potential follows:

For a reaction to be spontaneous, the standard electrode potential must be positive.

Hence, the EMF of the cell is -0.100 V and the reaction is non-spontaneous

the answer is option a.