JEE Main 2026 — 22 January, Morning Shift — Question 67

Consider the following electrochemical cell at 298 K $\mathrm{Pt}\left|\mathrm{HSnO}_{2}^{-}(\mathrm{aq})\right| \mathrm{Sn}(\mathrm{OH})_{6}{ }^{2-}(\mathrm{aq})\left|\mathrm{Bi}_{2} \mathrm{O}_{3}(\mathrm{~s})\right| \mathrm{Bi}(\mathrm{s})$. If the reaction quotient at a given time is $10^{6}$, then the cell EMF ( $\mathrm{E}_{\text {cell }}$ ) is $\_\_\_\_$ $\times 10^{-1} \mathrm{~V}$ (Nearest integer). Given the standard half-cell reduction potential as $\mathrm{E}_{\mathrm{Bi}_{2} \mathrm{O}_{3} / \mathrm{Bi}_{, \mathrm{OH}^{-}}}^{0}=-0.44 \mathrm{~V}$ and $\mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_{6}^{2-} / \mathrm{HSnO}_{2}^{-}, \mathrm{OH}}^{0}=-0.90 \mathrm{~V}$