NVAMediumJEE Main 2023 · 10 April, Shift 1Electrochemical Cells

Chemistry Question from JEE Main 2023 · 10 April, Shift 1

FeO42−O_4^{2-} + 2.2 V2.2 \, \text{V} → Fe3+^{3+}, Fe3+^{3+} + 0.7 V0.7 \, \text{V} → Fe2+^{2+}, Fe2+^{2+} −0.45 V-0.45 \, \text{V} → Fe°. EFeO42−/Fe2+∘E^\circ_{\mathrm{FeO}_4^{2-}/\mathrm{Fe}^{2+}} is x×10−3 Vx \times 10^{-3} \, \text{V}. The value of xx is:

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