Question

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 1

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)`