SolveIt is under development
SolveItJEE Main
Question 135 / 161 · latest first
Chemistry · 2021

JEE Main · 25 July 2021, Shift 1 · Q58

Consider the cell at 25°C Zn | Zn^2+( aq ),(1 M ) ∥ Fe^3+( aq ), Fe^2+( aq )| Pt ( s ) The fraction of total iron present as Fe^3+ ion at the cell…

Consider the cell at $\displaystyle 25$°C $\displaystyle \mathrm{Zn}\left|\mathrm{Zn}^{2+}(\mathrm{aq}),(1 \mathrm{M}) \| \mathrm{Fe}^{3+}(\mathrm{aq}), \mathrm{Fe}^{2+}(\mathrm{aq})\right| \mathrm{Pt}(\mathrm{s})$ The fraction of total iron present as $\displaystyle \mathrm{Fe}^{3+}$ ion at the cell potential of $\displaystyle 1.500$ V is $\displaystyle x \times 10^{-2}$. The value of $\displaystyle x$ is \_\_\_\_. (Nearest integer) (Given : $\displaystyle \mathrm{E}_{\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}}^0=0.77 \mathrm{~V}, \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^0=-0.76 \mathrm{~V}$ )
ShareWhatsAppTelegram

More from Electrochemistry

JEE Main 2021 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.