QJEEM21Ch97ef: JEE Main 2021 Chemistry - Class 12 JEEM21Ch97ef Solution: For the reaction 2Fe 3+ (aq) + 2I (aq) 2Fe 2+ (aq) + I 2 (s)…

Question

Verified Solution • 2025–26 Session moderate

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).

[\begin{matrix} E_{F e^{2 +} / F e (s)}^{o} = - 0.440 V; & E_{F e^{3 +} / F e (s)}^{o} = - 0.036 V \\ E_{I_{2} / 2 I^{-}}^{o} = 0.539 V; & F = 96500 C \end{matrix}]

Accepted Answer

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

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