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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. |
Answer» Solution : 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` |
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