Calculate the standard cell potentials of the galvanic cells in which the following reactions take place.
(a) `" " 2Cr(s)+3Cd^(2+) to 2Cd^(3+)(aq)+3Cd(s)`
Given `E_(Cr^(3+)//Cr)^(@)=-0.74" V" , E_(Cd^(2+)//Cd)^(@)=-0.40" V"`
(b) `" " Fe^(2+)(aq)+Ag^(+)(aq) to Fe^(3+)(aq)+Ag(s)`
Gievn `E_(Ag^(+)//Ag)^(@)=0.80" V" ,E_(Fe^(3+)//Fe^(2+))^(@)=0.77 " V"`
Also calculate `DeltaG^(@)` and equilibrium constant for the reaction.

Explanation

(a) Calculate of `E_(cell)^(@)`
`E_(cell)^(@)=E_(cathode)^(@)-E_(anode)^(@)=-0.40-(-0.74)=+0.34" V"`
Calculation of `DeltaG^(@)`,
`DeltaG^(@)=-nFE_(cell)^(@)=-(6" mol")xx(96500" C mol"^(-1))xx(0.34" V")`
=-196860" CV"`=-196860 "J"=-196.86" kJ"`
Calculation of Equilibrium Constant `(K_(c))`
`DeltaG^(@)=-2.303" RT log "K_(c)`
`logK_(c)=((-)DeltaG^(@))/(2.303" RT")=(-)((-)196860)/(2.303xx8.314xx298)=34.501`
`K_(c)="Antilog"(34.501)=3.17xx10^(34)`
(b) Calculation of `E_(cell)^(@)`
`E_(cell)^(@)=E_(canode)^(@)-E_(anode)^(@)=(0.80-0.77)=0.03" V"`
Calculation of`DeltaG^(@)`,
`DeltaG^(@)=-nFE_(cell)^(@)=-(1" mol")xx(96500" C mol"^(-1))xx(0.03" V")`
`=-2895" CV"=-2895" J"=-2.895" kJ"`
Calculation Equilibrium Constant `(K_(c))`
`DeltaG^(@)=-2.303" RT log K"_(c)`
`logK_(c)=(-)((-DeltaG^(@)))/(2.303" RT")=(-)((-)2895)/(2.303xx8.314xx298)=0.5074`
`K_(c)="Antilog"(0.5074)=3.22`