Determine the value of equilibrium constant and for the following reactions:
`Ni_((s)) + 2Ag_((aq))^(+) to Ni_((Aq))^(2+) + 2Ag_(s), E^(@) = 1.05V` `(if = 96500 cm o1^(-1))`

Explanation

The cell reaction is
`Ni(s)+2Ag^(+)(aq)toNi^(2+)(aq)+2Ag(s)" "triangleG^(@)=-nFE^(@)`
`E^(@)=1.05V,n=2,f=96500C//mol`
`triangle=-(2mol)xx(96500C//mol)xx(1.05V)`
`=-202650CV=-202650J`
calculation of K in the form `triangleG^(@)`
`triangleG^(@)=-202650J,R=8.314J//mol//K`
`T=298K`
`logK=(triangleG^(@))/(2.303RT)=(-202650)/((2.303)xx(8.314J//mol//K)xx(298K))`
`K=(-202650)/((2.303)xx(8.314J//mol//K)xx(298K)),K=(-202650)/(5705.8432)`
or `K=0.35xx10^(7)`