JEE Main 2021 · ChemistryNumerical answerNumerical valueHardMulti-step

JEE Main 18 March 2021, Shift 1, Chemistry Q56

Question 56 of 90 in this shift, Chemistry question 26 of 30, Section B.

For the reaction

\mathrm{2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s)}

the magnitude of the standard molar free energy change,

\Delta_rG^\circ_m = -

__________ kJ (Round off to the Nearest Integer).

\left[E^\circ_{Fe^{2+}/Fe(s)} = -0.440\ \mathrm{V};\ E^\circ_{Fe^{3+}/Fe(s)} = -0.036\ \mathrm{V};\ E^\circ_{I_2/2I^-} = 0.539\ \mathrm{V};\ F = 96500\ \mathrm{C}\right]

Official answer

NTA final key.

Same idea in other shifts