JEE MAIN - Chemistry (2019 - 8th April Evening Slot - No. 24)
Calculate the standard cell potential in (V) of the
cell in which following reaction takes place :
Fe2+(aq) + Ag+(aq) $$ \to $$ Fe3+(aq) + Ag (s)
Given that
$$E_{A{g^ + }/Ag}^o = xV$$
$$E_{Fe^{2+ }/Fe}^o = yV$$
$$E_{Fe^{3+ }/Fe}^o = zV$$
Fe2+(aq) + Ag+(aq) $$ \to $$ Fe3+(aq) + Ag (s)
Given that
$$E_{A{g^ + }/Ag}^o = xV$$
$$E_{Fe^{2+ }/Fe}^o = yV$$
$$E_{Fe^{3+ }/Fe}^o = zV$$
x + 2y - 3z
x - z
x - y
x + y - z
Explanation
Standard emf,
$${E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0$$ ..............(1)
Given $$E_{Fe^{2+ }/Fe}^o = yV$$
$$ \therefore $$ Fe2+ + 2e- $$ \to $$ Fe ..........(2)
$${E^0}$$ = y and $$\Delta {G^0}$$ = -2Fy
Also given $$E_{Fe^{3+ }/Fe}^o = zV$$
$$ \therefore $$ Fe3+ + 3e- $$ \to $$ Fe .........(3)
$${E^0}$$ = z and $$\Delta {G^0}$$ = -3Fz
Performing (2) - (1), we get
Fe3+ + e- $$ \to $$ Fe2+
$$ \therefore $$ $$\Delta {G^0}$$ = -3Fz + 2Fy
$$ \Rightarrow $$ -(1)F$${E^0}$$ = -3Fz + 2Fy
$$ \therefore $$ $$E_{F{e^{3 + }}|F{e^{2 + }}}^0$$ = 3z - 2y
From Equation (1),
$${E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0$$
= x - (3z - 2y) = x + 2y - 3z
$${E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0$$ ..............(1)
Given $$E_{Fe^{2+ }/Fe}^o = yV$$
$$ \therefore $$ Fe2+ + 2e- $$ \to $$ Fe ..........(2)
$${E^0}$$ = y and $$\Delta {G^0}$$ = -2Fy
Also given $$E_{Fe^{3+ }/Fe}^o = zV$$
$$ \therefore $$ Fe3+ + 3e- $$ \to $$ Fe .........(3)
$${E^0}$$ = z and $$\Delta {G^0}$$ = -3Fz
Performing (2) - (1), we get
Fe3+ + e- $$ \to $$ Fe2+
$$ \therefore $$ $$\Delta {G^0}$$ = -3Fz + 2Fy
$$ \Rightarrow $$ -(1)F$${E^0}$$ = -3Fz + 2Fy
$$ \therefore $$ $$E_{F{e^{3 + }}|F{e^{2 + }}}^0$$ = 3z - 2y
From Equation (1),
$${E^0} = E_{A{g^ + }|Ag}^0 - E_{F{e^{3 + }}|F{e^{2 + }}}^0$$
= x - (3z - 2y) = x + 2y - 3z
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