Question

In the activity, click on the E∘cell and Keq quantities to observe how they are related. Use this relation to calculate Keq for the following redox reaction that occurs in an electrochemical cell having two electrodes: a cathode and an anode. The two half-reactions that occur in the cell areCu2+(aq)+2e−→Cu(s) and Fe(s)→Fe2+(aq)+2e−

Answer 1

Answer: The $E^o_{cell}\text{ and }K_{eq}$ of the reaction is 0.78 V and $2.44\times 10^{26}$ respectively.

Explanation:

For the given half reactions:

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.34V$

Net reaction: $Fe(s)+Cu^{2+}\rightarrow Fe^{2+}+Cu(s)$

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:

$E^o_{cell}=0.34-(-0.44)=0.78V$

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

$\Delta G^o=-nfE^o_{cell}$

and,

$\Delta G^o=-RT\ln K_{eq}$

Equating these two equations, we get:

$nfE^o_{cell}=RT\ln K_{eq}$

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

$E^o_{cell}$ = standard electrode potential of the cell = 0.78 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = $25^oC=[273+25]=298K$

$K_{eq}$ = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

$2\times 96500\times 0.78=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=2.44\times 10^{26}$

Hence, the $E^o_{cell}\text{ and }K_{eq}$ of the reaction is 0.78 V and $2.44\times 10^{26}$ respectively.