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Chemistry · 2026

JEE Main · 22 January 2026, Shift 1 · Q72

Consider the following electrochemical cell at 298 K Pt | HSnO_2^-( aq )| Sn ( OH )_6^2-( aq )| OH^-( aq )| Bi_2 O_3( s ) ∣ Bi ( s ). If the reaction…

Consider the following electrochemical cell at $\displaystyle 298$ K $\displaystyle \mathrm{Pt}\left|\mathrm{HSnO}_2{ }^{-}(\mathrm{aq})\right| \mathrm{Sn}(\mathrm{OH})_6{ }^{2-}(\mathrm{aq})\left|\mathrm{OH}^{-}(\mathrm{aq})\right| \mathrm{Bi}_2 \mathrm{O}_3(\mathrm{~s}) \mid \mathrm{Bi}(\mathrm{s})$. If the reaction quotient at a given time is $\displaystyle 10^6$, then the cell $\displaystyle \operatorname{EMF}\left(\mathrm{E}_{\text {cell }}\right)$ is $\displaystyle \_\_\_\_$ $\displaystyle \times 10^{-1} \mathrm{~V}$ (Nearest integer).Given the standard half-cell reduction potential as $$\mathrm{E}_{\mathrm{Bi}_2 \mathrm{O}_3 / \mathrm{Bi}, \mathrm{OH}^{-}}^{\circ}=-0.44 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_6^{2-} / \mathrm{HSnO}_2^{-}, \mathrm{OH}^{-}}^{\circ}=-0.90 \mathrm{~V} $$
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JEE Main 2026 Chemistry question, with the answer from NTA’s final answer key. Where our answers come from.