Jee Main
Electrochemistry
JEE Main 2025 (Online) 23rd January Morning Shift
MCQ+4 / -12025
\(\mathrm{FeO}_4^{2-} \xrightarrow{+2.0 \mathrm{~V}} \mathrm{Fe}^{3+} \xrightarrow{0.8 \mathrm{~V}} \mathrm{Fe}^{2+} \xrightarrow{-0.5 \mathrm{~V}} \mathrm{Fe}^0\)
In the above diagram, the standard electrode potentials are given in volts (over the arrow).
The value of $\mathrm{E}_{\mathrm{FeO}_4^{2-} / \mathrm{Fe}^{2+}}$ is :