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For the cell reaction Mg|Mg^(2+)(aq)||Ag...

For the cell reaction `Mg|Mg^(2+)(aq)||Ag^(aq)|Ag`
Calcualte the equilibrium constant at `25^(@)C` and maximum work that can be obtained by operating the cell
`[Given E_(Mg^(2+)//Mg)^(@)=-2.37 V , E_(Ag^(+)//Ag)^(@)=+0.80 B]`

Text Solution

Verified by Experts

At equilibrium the given cell reaction is
`Ag_((s))+2Ag^(+) Leftrightarrow Mg^(++) +2Ag_((s))`
`E_("cell")=0`
Here, equilibrium constant (K) `=([Mg^(++)] [Ag (s)]^(2))/([Mg(s)] [Ag^(+)]^(2) or, K-([Mg^(+2)])/([Ag^(+)]^(2))`
For Nerst equation
`E_("cell")=E_("cell")^(0)-(0.092)/(2) log K`
`log K=(2.E_("cell")^(0))/(0.592)`
`E_("cell")^(0)=E_(Mg//Mg^(+2))^(0) +E_(Ag^(+)//Ag)^(0)=2.37+0.80`
`log K=(2 xx 3.17)/(0.0592)`
`log K=107.457`
Now maximum work that can be obtained by cell is given by `W_("maximum")=-triangleG^(@)`
`=-nFE_("cell")^(0)`
`=-2 xx 96,500 xx 3.17J`
`W_("maximum")=6.118 xx 10^(2) kJ`
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