Calculate the potential of the following cell at 298 K
`Zn//Zn^(2+)(a=0.1)//Cu^(2+)(a=0.01)//Cu`
`E_(Zn^(2+)//Zn)^(@)=-0.762V`
`E_(Cu^(2+)//Cu)^(@)=+0.337V`
Compare the free energy change for this cell with the free enegy of the cell in the standard state.

Explanation

The overall cell reaction is
`Zn+Cu^(2+)(a=0.01)rarrZn^(2+)(a=0.1)+Cu`
The cell potential given by nernst equation
`E_("cell")=E_("cell")^(@)-(RT)/(2F)ln""(a_(Zn)^(2+)a_(Cu))/(a_(Zn)a_(Cu)^(2+))`
`=E_("cell")^(@)-(RT)/(2F)ln""(a_(Zn)^(2+))/(a_(Cu)^(2+))`
(Since activity of a pure metal is unity)
`E_("cell")^(@)=0.337-(-0.762)=1.099V`
`E_("cell")=1.099-(0.0591)/(2)log""(0.2)/(0.01)`
`=1.099-(0.0591)/(2)log10`
`=1.099-0.02956`
`=1.0694V`
The free energy change `DeltaG` is given by`DeltaG=-nFE`
`DeltaG=-("2 equi/mol")("1.0694 V")("96495 coulomb equiv"^(-1))`
`=-206.6" kJ mol"^(-1)`
The standard free energy change :
`DeltaG^(@)=-("2 equi/mol")(1.099V)("96495 coulomb/equiv")`
`=-212.1" kJ/mol."`