JEE Mains 10-Apr-2023 Shift 1 Solved Paper

Section: Chemistry

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Question : 87 of 90

Marks: +1, -0

FeO42−

+2.2V

─────▶

Fe3+

+0.70V

─────▶

Fe2+

−0.45V

─────▶

Fe0
EFeO42−∕Fe2+∘ is x×10−3V.
The value of x is _______

[10-Apr-2023 shift 1]

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