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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)

best_answer

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

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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