Calculate the standard cell potentials of galvanic cells in which the following reaction take place:
(i) `2Cr_((S))+3Cd_((aq))^(2+) to 2Cr_((aq))^(3+)+3Cd`
(ii) `Fe_((aq))^(2+)+Ag_((aq))^(+) to Fe_((aq))^(3+)+Ag_((S))`
Calculate the `Delta_(r)G^(@)` and equilibrium constant of the reactions.

Explanation

Cell representation : `Cr_((S))|Cr_((aq))^(3+)||Cd_((aq))^(2+)|Cd_((S))`
(i) So oxidation half reaction :
`Cr_((S)) to Cr_((aq))^(3+)+3e^(-)`
So, this half cell is on left side of galvanic cell.
`E_(L)^(Theta)=E_(Cr^(3+)|Cr)^(Theta)=-0.74V`
* Reduction half reaction : `Cd^(2+)+2e^(-) to Cd_((S))`
In this, reduction is occurred so this half-cell is present on the right side of the galvanic cell.
`E_(R)^(Theta)=E_(Cd^(2+)|Cd)^(Theta)=-0.40V`
* So The standard potential,
`Delta_(cell)^(Theta)=(E_(R)^(Theta)-E_(L)^(Theta))`
`=[-0.40-(-0.74)]V`
`=[-0.40+0.74]V`
`=0.34V`
* Calculation of Gibbs free energy:
Where, `Delta_(r)G^(Theta)=`Cell of Gibbs free energy, n=6 mole
F=96500 Coulomb `"mole"^(-1)`, `E_(cell)^(Theta)=0.34V`
`Delta_(r)G^(Theta)=-nFE_(cell)^(Theta)`
`Delta_(r)G^(Theta)=-(6mol)xx(96500" C "mol^(-1))xx(0.34V)`
`=-196860CV`
`=-196860J`
`=-196.860kJ`
* Calculation for equilibrium constant K :
`Delta_(r)G^(Theta)=-2.303RT" log "k`
`therefore log " "k=(-196860)/(2.303xx8.314xx298)=34.5014`
`therefore k="Antilog "34.5014=3.174xx10^(34)`
(ii) Calculation for cell potential of :
`Fe_((aq))^(2+)+Ag_((aq))^(+) to Fe_((aq))^(3+)+Ag_((S))`

Cell representation : `Fe_((aq))^(2+)|Fe_((aq))^(3+)||Ag_((aq))^(+)|Ag_((S))`
* So oxidation half reaction : `Fe_((aq))^(2+) to Fe_((aq))^(3+)+e^(-)`
`E_(L)^(Theta)=E_(Fe^(3+)|Fe^(2+))^(Theta)=0.77V`
* Reduction half reaction: `Ag_((aq))^(+)+e^(-) to Ag_((S))`
In this, reduction is occurred so this half-cell is present on the right side of the galvanic cell.
`E_(R)^(Theta)=E_(Ag^(+)|Ag)^(Theta)=0.80V`
So the standard potential,
`DeltaE_(cell)^(Theta)=(E_(R)^(Theta)-E_(L)^(Theta))`
`=0.80-0.77V`
`=0.03V`
* Calculation of Gibbs free energy :
Where, `Delta_(r)G^(Theta)=`Cell of Gibbs free energy, n=1 mole
`F=96500` Coulomb `"mole"^(-1)`, `E_(cell)^(Theta)=Delta_(r)E_(cell)^(Theta)=0.03V`
`Delta_(r)G^(Theta)=-nFE_(cell)^(Theta)`
`therefore Delta_(r)G^(Theta)=-(1mol)xx(96500" C "mol^(-1))xx(0.03V)`
`=-2895CV`
`=-2895J`
`=-2.895kJ`
* Calculation for equilibrium constant K :
`Delta_(r)G^(Theta)=-2.303RT " log "k `
`therefore -2895J=-(2.303)xx(8.314J)xx(298k)log" "k_(C)`
`therefore log" "k_(C)=(-2895)/(2.303xx8.314xx298)=0.5074`
`therefore k_(C)="Antilog "0.5074=3.216~~3.22`