Question

The equilibrium constant of the reaction :
`Zn(s)+2Ag^(+)(aq)toZn(aq)+2Ag(s),E^(@)=1.50V` at 298 K is
A. `2.6xx10^(49)`
B. `8.7xx10^(51)`
C. `6.1xx10^(30)`
D. `6.6xx10^(50)`

Answer 1

Correct Answer - D
`nFE^(@)=2.303RT"log" K`
`"log"K=(nFE^(@))/(2.303RT)=(2xx96500)/(2.303xx8.3xx298)`
`K=6.6xx10^(50)`