Chemistry · 2023
JEE Main · 10 April 2023, Shift 1 · Q86
FeO_4^2- → +2.2 V Fe^3+ → +0.70 V Fe^2+ → -0.45 V Fe^0 E_FeO_4^2- / Fe^2+^0 is x × 10^-3 V. The value of x is ____
$$\mathrm{FeO}_4^{2-} \xrightarrow{+2.2 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{+0.70 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.45 \mathrm{~V}} \mathrm{Fe}^0
$$
$\displaystyle \mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}^0$ is $\displaystyle x \times 10^{-3} \mathrm{~V}$. The value of $\displaystyle x$ is $\displaystyle \_\_\_\_$
Official answer
From NTA’s final answer key for this paper.
1825
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JEE Main 2023 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.