JEE MAIN - Chemistry (2021 - 18th March Morning Shift - No. 23)
For the reaction
2Fe3+(aq) + 2I$$-$$(aq) $$ \to $$ 2Fe2+(aq) + I2(s)
the magnitude of the standard molar Gibbs free energy change, $$\Delta$$rG$$_m^o$$ = $$-$$ ___________ kJ (Round off to the Nearest Integer).
$$\left[ {\matrix{ {E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \cr {E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \cr } } \right]$$
2Fe3+(aq) + 2I$$-$$(aq) $$ \to $$ 2Fe2+(aq) + I2(s)
the magnitude of the standard molar Gibbs free energy change, $$\Delta$$rG$$_m^o$$ = $$-$$ ___________ kJ (Round off to the Nearest Integer).
$$\left[ {\matrix{ {E_{F{e^{2 + }}/Fe(s)}^o = - 0.440V;} & {E_{F{e^{3 + }}/Fe(s)}^o = - 0.036V} \cr {E_{{I_2}/2{I^ - }}^o = 0.539V;} & {F = 96500C} \cr } } \right]$$
Answer
46
Explanation
$$E_{F{e^{3 + }}/Fe}^o = 0.036V$$
$$F{e^{3 + }} + 3{e^ - } \to Fe$$
$$ \therefore $$ $$\Delta G_1^o = - nF{E^o}$$
$$ = - 3F( - 0.036)$$
$$E_{F{e^{2 + }}/Fe}^o = 0.440V$$
$$F{e^{2 + }} + 2{e^ - } \to Fe;{E^o} = - 0.440V$$
$$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$$
$$ \therefore $$ $$\Delta G_2^o = - 2F(0.440)$$
$$E_{{I_2}/2{I^ - }}^o = 0.539V$$
$${I_2} + 2{e^ - } \to 2{I^ - };{E^o} = 0.539V$$
$$2{I^ - } \to {I_2} + 2{e^ - };{E^o} = - 0.539V$$
$$\Delta G_3^o = - 2F( - 0.539)$$
$$ \therefore $$ $$\Delta {G^o} = 2\left[ {\Delta G_1^o + \Delta G_2^o} \right] + \Delta G_3^o$$
$$ = 2\left[ {3F(0.036) - 2F(0.440)} \right] + 2F(0.539)$$
$$ = - 45934 = - 45.9KJ \simeq - 46KJ$$
$$F{e^{3 + }} + 3{e^ - } \to Fe$$
$$ \therefore $$ $$\Delta G_1^o = - nF{E^o}$$
$$ = - 3F( - 0.036)$$
$$E_{F{e^{2 + }}/Fe}^o = 0.440V$$
$$F{e^{2 + }} + 2{e^ - } \to Fe;{E^o} = - 0.440V$$
$$Fe \to F{e^{2 + }} + 2{e^ - };{E^o} = 0.440V$$
$$ \therefore $$ $$\Delta G_2^o = - 2F(0.440)$$
$$E_{{I_2}/2{I^ - }}^o = 0.539V$$
$${I_2} + 2{e^ - } \to 2{I^ - };{E^o} = 0.539V$$
$$2{I^ - } \to {I_2} + 2{e^ - };{E^o} = - 0.539V$$
$$\Delta G_3^o = - 2F( - 0.539)$$
$$ \therefore $$ $$\Delta {G^o} = 2\left[ {\Delta G_1^o + \Delta G_2^o} \right] + \Delta G_3^o$$
$$ = 2\left[ {3F(0.036) - 2F(0.440)} \right] + 2F(0.539)$$
$$ = - 45934 = - 45.9KJ \simeq - 46KJ$$
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