Calculate the standard cell potential (in V) of the cell in which following reaction takes place:

Fe^{2+}\left ( aq \right )+Ag^{+}\left ( aq \right )\rightarrow Fe^{3+}\left ( aq \right )+Ag\left ( s \right )

Given that

E_{Ag^{+}/Ag}^{0}=xV

E_{Fe^{2+}/Fe}^{0}=yV

E_{Fe^{3+}/Fe}^{0}=zV

  • Option 1)

    x-z

  • Option 2)

    x-y

  • Option 3)

    x+2y-3z

  • Option 4)

    x+y-z

Answers (1)

             

    Fe^{2+}\left ( aq \right )+Ag^{+}\left ( aq \right )\rightarrow Fe^{3+}\left ( aq \right )+Ag\left ( s \right )      

                

Given E^{0}_{Ag^{+}}/ag=xV       E^{0}_{Fe^{2+}}/Fe=yV     E^{0}_{Fe^{3+}}/Fe=zV

\Rightarrow Cell equation :- Fe^{2+}/Fe^{+3}//Ag^{+}/Ag

                           (Oxidation)                    (Cathode Reduction)

\Rightarrow Std. EMF of given cell reaction:-

          E^{0}=E_{Ag^{+}/Ag}^{0}-E_{Fe^{3+}/Fe^{2+}}^{0}----(i)  

E_{Ag^{+}/Ag}^{0}=x(given)

Fe^{+2}+2e^{-}\rightarrow Fe\;\;\; E^{0}=y\;\;\Delta G^{0}=-2Fy--(ii)

Fe^{+3}+2e^{-}\rightarrow Fe\;\;\; E^{0}=z\;\;\Delta G^{0}=-3Fz--(iii)

(iii-ii)

Fe^{+3}+e^{-}\rightarrow Fe^{+2}\;\;\; \;\Delta G^{0}=-3Fz+2Fy

\Delta G^{0}=-nFE^{0}

-1 \times F \times E^{0}=-3Fz+2Fy

E^{0}=3z-2y   

Put this in eqn (i)

\left ( Fe^{3+}/Fe^{2+} \right )

E^{0}=x-(3z-2y)=x+2y-3z

 


Option 1)

x-z

Option 2)

x-y

Option 3)

x+2y-3z

Option 4)

x+y-z

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