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Electrochemistry question

2025 · 23 Jan · Shift 1 · Q5
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  5. /2025 · 23 Jan · Shift 1 · Q5

Electrochemistry question

2025 · 23 Jan · Shift 1 · Q5

JEE MainChemistryElectrochemistryMCQ+4 / −1
FeO42−→+2.0 VFe3+→0.8 VFe2+→−0.5 VFe0\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{~V}} \mathrm{Fe}^0FeO42−​+2.0 V​Fe3+0.8 V​Fe2+−0.5 V​Fe0 In the above diagram, the standard electrode potentials are given in volts (over the arrow). The value of EFeO42−/Fe2+\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}EFeO42−​/Fe2+​ is :
  1. A
    1.2 V
  2. B
    2.1 V
  3. C
    1.4 V
  4. D
    1.7 V
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