JEE Main 2023 · 25 Jan · Shift 1 · numerical answer

JEE Main 2023 Chemistry question on Redox and Electrochemistry

Consider the cell

Pt(s)∣H2(g) (1 atm)∣H+(aq, [H+]=1)∥Fe3+(aq), Fe2+(aq)∣Pt(s)\text{Pt}(s) \mid \text{H}_2(g)\ (1\ \text{atm}) \mid \text{H}^+(aq,\ [\text{H}^+]=1) \parallel \text{Fe}^{3+}(aq),\ \text{Fe}^{2+}(aq) \mid \text{Pt}(s)

Given EFe3+/Fe2+∘=0.771 VE^\circ_{\text{Fe}^{3+}/\text{Fe}^{2+}} = 0.771\ \text{V} and EH+/12H2∘=0 VE^\circ_{\text{H}^+/\frac{1}{2}\text{H}_2} = 0\ \text{V}, T=298 KT = 298\ \text{K}

If the potential of the cell is 0.712 V0.712\ \text{V}, the ratio of concentration of Fe2+\text{Fe}^{2+} to Fe3+\text{Fe}^{3+} is ________ (Nearest integer)
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Answer: 10

[JEE Main 2023 · 25 Jan · Shift 1]

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