For the reaction \(2 \mathrm{Fe}^{3+}(\mathrm{aq})+2 \mathrm{I}^{-}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{I}_{2}(\mathrm{~s})\) The magnitude of the standard molar free energy change, \(\Delta_{\mathrm{r}} \mathrm{G}_{\mathrm{m}}^{\mathrm{o}}=-\) \_\_\_\_ kJ (Round off to the nearest integer). \(\left[\begin{array}{ll}\mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}(\mathrm{s})}^{\mathrm{o}}=-0.440 \mathrm{~V} ; & \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}(\mathrm{s})}^{\mathrm{o}}=-0.036 \mathrm{~V} \\ \mathrm{E}_{\mathrm{I}_{2} / 2 \mathrm{I}^{-}}^{\mathrm{o}}=0.539 \mathrm{~V} ; & \mathrm{F}=96500 \mathrm{C}\end{array}\right]\)

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Redox Reactions and Electrochemistry

For the reaction

2 Fe^{3 +} (aq) + 2 I^{-} (aq) \to 2 Fe^{2 +} (aq) + I_{2} (s)

The magnitude of the standard molar free energy change,

Δ_{r} G_{m}^{o} = -

____ kJ (Round off to the nearest integer).

[E_{Fe^{2 +} / Fe (s)}^{o} = - 0.440 V; E_{I_{2} /2 I^{-}}^{o} = 0.539 V; E_{Fe^{3 +} / Fe (s)}^{o} = - 0.036 V F = 96500 C]

Answer: 45

Free energies, not potentials, are additive, so the

F e^{3 +} /F e^{2 +}

couple follows from the two iron couples given:

Δ G_{F e^{3 +} /F e^{2 +}}^{\circ} = Δ G_{F e^{3 +} /Fe}^{\circ} - Δ G_{F e^{2 +} /Fe}^{\circ}

- 1 \cdot F \cdot E^{\circ} = - 3 F (- 0.036) + 2 F (- 0.440) = - 0.772 F

E_{F e^{3 +} /F e^{2 +}}^{\circ} = 0.772 V

. Iron(III) is reduced and iodide oxidised, giving

E_{cell}^{\circ} = 0.772 - 0.539 = 0.233 V

Two electrons are transferred as written, so

Δ_{r} G_{m}^{\circ} = - n F E_{cell}^{\circ} = - 2 (96500) (0.233) = - 4.5 \times 1 0^{4} J

The magnitude of the standard molar free energy change is

45 kJ

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer