1.

For the cell Mg_((s))abs(Mg^(2+)""_((aq)))abs(Ag^(+)""_((aq)))Ag_((s)), calculate the equilibrium constant at 25^(@)C and maximum work that can be obtained during operation of cell. Given : E_(Mg^(2+)|Mg)^(@)=-237V " and " E_(Ag^(2+)|Ag)^(@)=0.80V

Answer»

Solution :oxidation at anode
`MG rarr mg^(2+)+2e^(-)"...(1)"`
`""(E_("ox")^(@))=2.37V`
Reduction at cathode
`Ag^(+)+e^(-) rarr Ag "...(2)"`
`""(E_("red")^(2))=0.80V`
`therefore E_("CELL")^(@)=(E_("ox")^(@))_("anode")+(E_("red")^(@))_("cathode")`
`""=2.37+0.80`
`""=3.17V`
Overall REACTION
Equation (1) + `2 times " equation (2)" rArr`
`Mg+2Ag^(2+) rarr Mg^(2+)+2Ag`
`""DELTAG^(@)=-nfE^(@)`
`""=-2 times 96500 times 3.17`
`""=611.810J`
`DeltaG^(@)=-6.12 times 10^(5)J`
`W=6.12 times 10^(5)J`
`DeltaG^(@)=-2.803" RT log " K_(c)`
`rArr logK_(c)=(6.12 times 10^(5))/(2.803 times 8.314 times 298)`
`""K_(c)=` Antilog of (107.2).


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