Question

What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Ag(s) and Ni2+(aq) to give Ni(s) and Ag+(aq) Use the reduction potential values for Ag+(aq) of +0.80 V and for Ni2+(aq) of -0.25 V Give your answer using E-notation with NO decimal places (e.g., 2 x 10-2 would be 2E-2; and 2.12 x 10-2 would also be 2E-2.). Do NOT include spaces, units, punctuation or anything else silly!

Answer 1

Answer:

The $K_{eq}$ of the reaction is $3.002\times 10^{-36}$.

Explanation:

$2Ag (s)+Ni^{2+}(aq)\rightarrow  2Ag^+(aq)+Ni(s)$

Reduction at cathode :

$Ni^{2+}(aq)+2e^-\rightarrow Ni(s)$

Reduction potential of  $Ni^{2+}$ to Ni=$E^o_{1}=-0.25 V$

Oxidation at anode:

$2Ag(s)\rightarrow Ag^{+}(aq)+e^-$

Reduction potential of  $Ag^{+}$ to Ag=$E^o_{2}=0.80 V$

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{red,cathode}-E^o_{red,anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.25 V-0.80 V =-1.05 V$

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 = -1.05 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:

$2\times 96500\times (-1.05 V)=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=3.002\times 10^{-36}$

Hence, the $K_{eq}$ of the reaction is $3.002\times 10^{-36}$