NVAEasyJEE Main 2023 · 30 January, Shift 1Nernst Equation

Chemistry Question from JEE Main 2023 · 30 January, Shift 1

Consider the cell Pt(s)∣H2(g,1 atm)∣H+(aq,1 M)∣∣Fe3+(aq),Fe2+(aq)∣∣Pt(s).\text{Pt}(s) \mid \text{H}_2 (g, 1 \, \text{atm}) \mid \text{H}^+ (aq, 1\,\text{M}) \mid\mid \text{Fe}^{3+} (aq) , \text{Fe}^{2+} (aq) \mid\mid \text{Pt}(s). When the potential of the cell is 0.712 V0.712 \, \text{V} at 298 K298 \, \text{K}, the ratio [Fe2+][Fe3+]\frac{[Fe^{2+}]}{[Fe^{3+}]} is _____. (Nearest integer)

Given: Fe3++e−→Fe2+,EFe3+∣Fe2+∘=0.771 V.\text{Fe}^{3+} + e^- \rightarrow \text{Fe}^{2+}, \quad E^\circ_{Fe^{3+}|Fe^{2+}} = 0.771 \, V. 2.303RTF=0.06 V.\frac{2.303RT}{F} = 0.06 \, V.

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