Calculate emf and `DeltaG` for the following reaction at 298 K `Mg(s)|Mg^(2+)(0.01 M)||Ag^(+)(0.0001 M)|Ag(s)` Given `E_((Mg^(2+)//Mg))^(@)=-2.37" V ", E_((Ag^(+)//Ag))^(@)=+0.80" V "`

Calculate emf and `DeltaG` for the following reaction at 298 K `Mg(s)|

1.	Calculate emf and `DeltaG` for the following reaction at 298 K `Mg(s)\|Mg^(2+)(0.01 M)\|\|Ag^(+)(0.0001 M)\|Ag(s)` Given `E_((Mg^(2+)//Mg))^(@)=-2.37" V ", E_((Ag^(+)//Ag))^(@)=+0.80" V "`
Answer» Correct Answer - `"emf"=2.9930" V ";DeltaG=-577.649" kJ mol^(-1)` Step I. Calculation of `E_(cell)` `E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=(+0.80" V")-(-2.37" V")=3.17" V"` According to Nernst equation `E_(cell)=E_(cell)^(@)-(0.0591)/(n)"log"(["Anode"])/(["Cathode"])` `=3.17V-(0.0591(V))/(2)"log"((0.01))/((0.0001)^(2))` `=3.17" V"-(0.02955" V") log10^(6)=3.17" V"-(0.02955" V")+6` `=3.17" V"-0.1770" V"=2.9930" V"` Step II. Calculation of `DeltaG` `DeltaG=-nFE_(cell)=(-2)xx(96500" C mol"^(-1))xx(2.9930" V")` `=-(577649"V")mol^(-1)=-577649" J mol"^(-1)=-577.649" kJ mol^(-1)`

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Calculate emf and `DeltaG` for the following reaction at 298 K `Mg(s)|Mg^(2+)(0.01 M)||Ag^(+)(0.0001 M)|Ag(s)` Given `E_((Mg^(2+)//Mg))^(@)=-2.37" V ", E_((Ag^(+)//Ag))^(@)=+0.80" V "`

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