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← JEE Main 2022 · 25 Jun · Shift 1
ChemistryInteger+4 / -1Question 87 of 90

JEE Main 2022 · 25 Jun · Shift 1, Chemistry question 87

In a cell, the following reactions take place
Fe2+→Fe3++e−EFe3+/Fe2+∘=0.77 V\mathrm{Fe^{2+}\rightarrow Fe^{3+}+e^-}\qquad E^\circ_{\mathrm{Fe^{3+}/Fe^{2+}}}=0.77\ \mathrm V
2I−→I2+2e−EI2/I−∘=0.54 V\mathrm{2I^-\rightarrow I_2+2e^-}\qquad E^\circ_{\mathrm{I_2/I^-}}=0.54\ \mathrm V
The standard electrode potential for the spontaneous reaction in the cell is x×10−2x\times10^{-2} V 298 K. The value of xx is ______. (Nearest Integer)

Answer and worked solution

Official answer: 23

Spontaneous direction. Fe3+\mathrm{Fe^{3+}} (0.77 V) is reduced and I−\mathrm{I^-} (0.54 V) oxidised: 2Fe3++2I−→2Fe2++I2\mathrm{2Fe^{3+}+2I^-\rightarrow2Fe^{2+}+I_2}.

Cell potential. 0.77−0.54=0.230.77-0.54=0.23 V =23×10−2=23\times10^{-2} V.

Answer: 23