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

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:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Cd^{2+}]}{[Fe^{2+}]}$

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:

$E_{cell}=-0.04-\frac{0.059}{2}\times \log(\frac{0.00550}{0.600})$

$E_{cell}=-0.100V$

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