1.

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

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


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