Calculate the equilibrium constant of the reaction at 298 K .Mg(s) + 2

1.	Calculate the equilibrium constant of the reaction at 298 K .Mg(s) + 2Ag+(aq) --> Mg2+(aq) +2Ag(s); E0 cell=+3.16 V
Answer» Dear Student, ΔG° = −nFE°cell and, Δ G° = −RTlnKeq Therefore, nFE°cell = RTlnKeq or, lnKeq =nFE °cell RT Here, N = 2 (as 2 ELECTRONS are involved) F = 96500 E°cell = 3. 16 V R = 8. 314 JK−1 mol−1 T = 298 K or, lnKeq = = 246. 16 2×96500×3.16 8.314×298 or, Keq = e = 8. 05 So we get the answer as 8.05

Calculate the equilibrium constant of the reaction at 298 K .Mg(s) + 2Ag+(aq) --> Mg2+(aq) +2Ag(s); E0 cell=+3.16 V

Answer»

Dear Student,

ΔG° = −nFE°cell

and, Δ G° = −RTlnKeq

Therefore,

nFE°cell = RTlnKeq

or, lnKeq =nFE °cell RT

Here, N = 2 (as 2 ELECTRONS are involved)

F = 96500

E°cell = 3. 16 V

R = 8. 314 JK−1 mol−1

T = 298 K

or, lnKeq = = 246. 16

2×96500×3.16

8.314×298

or, Keq = e = 8. 05 So we get the answer as 8.05