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Electrochemistry question

2023 · 10 Apr · Shift 1 · Q22
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Electrochemistry question

2023 · 10 Apr · Shift 1 · Q22

JEE MainChemistryElectrochemistryNumerical+4 / −1
FeO42−→+2.2VFe3+→+0.70VFe2+→−0.45VFe0\mathrm{FeO_4^{2 - }\xrightarrow{+ 2.2V} F{e^{3 + }}\xrightarrow{+ 0.70V} F{e^{2 + }}\xrightarrow{- 0.45V} F{e^0}}FeO42−​+2.2V​Fe3++0.70V​Fe2+−0.45V​Fe0 EFeO42−/Fe2+θE_{FeO_4^{2 - }/F{e^{2 + }}}^\thetaEFeO42−​/Fe2+θ​ is x×10−3x \times {10^{ - 3}}x×10−3 V. The value of xxx is ‾\underline{\hspace{2cm}}​
Numerical answer
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