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The equilibrium constant of the reaction...

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)`

Text Solution

Verified by Experts

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