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.

Calculate the potential of the following cell at 298 K Zn//Zn^(2+)(a=

1.	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.
Answer» Solution :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."`