Question

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

Answer 1

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

Step-by-step explanation:

The half reaction for the equation follows:

Oxidation half reaction:
`Cd(s)\rightarrow Cd^(2+)(aq,0.00550M)+2e^-;E^o_(Cd^(2+)/Cd)=-0.40V`

Reduction half reaction:
`Fe^(2+)(aq,0.600M)+2e^-\rightarrow Fe(s);E^o_(Fe^(2+)/Fe)=-0.44V`

Net cell reaction:
`Cd(s)+Fe^(2+)(aq,0.600M)\rightarrow Cd^(2+)(aq,0.00550M)+Fe(s)`

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

To calculate the
`E^o_(cell)` of the reaction, we use the equation:

`E^o_(cell)=E^o_(cathode)-E^o_(anode)`

Putting values in above equation, we get:

`E^o_(cell)=-0.44-(-0.40)=-0.04V`

• To calculate the EMF of the cell, we use the Nernst equation, which is:

`E_(cell)=E^o_(cell)-(0.059)/(n)\log ([Cd^(2+)])/([Fe^(2+)])`

where,

`E_(cell)` = electrode potential of the cell = ?

`E^o_(cell)` = standard electrode potential of the cell = -0.04 V

n = number of electrons exchanged = 2

`[Cd^(2+)]=0.00550M`

`[Fe^(2+)]=0.600M`

Putting values in above equation, we get:

`E_(cell)=-0.04-(0.059)/(2)* \log((0.00550)/(0.600))`

`E_(cell)=-0.100V`

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:

`\Delta G^o=-nFE^o_(cell)`

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