Q21 / 75 · JEE Main 2026 (22 January Shift 1) · Chemistry
Consider the following electrochemical cell at 298 K
$\mathrm{Pt}\left|\mathrm{HSnO}_{2}{ }^{-}(\mathrm{aq})\right| \mathrm{Sn}(\mathrm{OH})_{6}{ }^{2-}(\mathrm{aq})\left|\mathrm{OH}^{-}(\mathrm{aq})\right| \mathrm{Bi}_{2} \mathrm{O}_{3}(\mathrm{~s}) \mid \mathrm{Bi}(\mathrm{s})$.
If the reaction quotient at a given time is $10^{6}$, then the cell EMF ($\mathrm{E}_{\text {cell }}$) is $\_\_\_\_$ $\times 10^{-1} \mathrm{~V}$ (Nearest integer).
Given the standard half-cell reduction potential as
$\mathrm{E}_{\mathrm{Bi}_{2} \mathrm{O}_{3} / \mathrm{Bi}, \mathrm{OH}^{-}}^{\circ}=-0.44 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_{6}^{2-} / \mathrm{HSnO}_{2}^{-}, \mathrm{OH}^{-}}^{\circ}=-0.90 \mathrm{~V}$
$\mathrm{Pt}\left|\mathrm{HSnO}_{2}{ }^{-}(\mathrm{aq})\right| \mathrm{Sn}(\mathrm{OH})_{6}{ }^{2-}(\mathrm{aq})\left|\mathrm{OH}^{-}(\mathrm{aq})\right| \mathrm{Bi}_{2} \mathrm{O}_{3}(\mathrm{~s}) \mid \mathrm{Bi}(\mathrm{s})$.
If the reaction quotient at a given time is $10^{6}$, then the cell EMF ($\mathrm{E}_{\text {cell }}$) is $\_\_\_\_$ $\times 10^{-1} \mathrm{~V}$ (Nearest integer).
Given the standard half-cell reduction potential as
$\mathrm{E}_{\mathrm{Bi}_{2} \mathrm{O}_{3} / \mathrm{Bi}, \mathrm{OH}^{-}}^{\circ}=-0.44 \mathrm{~V} \text { and } \mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_{6}^{2-} / \mathrm{HSnO}_{2}^{-}, \mathrm{OH}^{-}}^{\circ}=-0.90 \mathrm{~V}$
Correct answer: 4
Solution
The electrochemical cell has cathode:
$Bi_2O_3 + 3H_2O + 6e^- \rightarrow 2Bi + 6OH^-$ with $E° = -0.44$ V
and anode:
$Sn(OH)_6^{2-} - 2e^- \rightarrow HSnO_2^- + H_2O + OH^-$ with $E° = -0.90$ V.
The standard cell EMF is $E°_{cell} = -0.44 - (-0.90) = 0.46$ V.
Using the Nernst equation with $Q = 10^6$ and $n = 6$ electrons transferred:
$E_{cell} = 0.46 - \frac{0.059}{6} \log(10^6) = 0.46 - \frac{0.059 \times 6}{6} = 0.46 - 0.059 = 0.401$ V.
Expressing as $x \times 10^{-1}$ V:
$0.401 = 4.01 \times 10^{-1}$ V, so $x = 4$ (nearest integer).
$Bi_2O_3 + 3H_2O + 6e^- \rightarrow 2Bi + 6OH^-$ with $E° = -0.44$ V
and anode:
$Sn(OH)_6^{2-} - 2e^- \rightarrow HSnO_2^- + H_2O + OH^-$ with $E° = -0.90$ V.
The standard cell EMF is $E°_{cell} = -0.44 - (-0.90) = 0.46$ V.
Using the Nernst equation with $Q = 10^6$ and $n = 6$ electrons transferred:
$E_{cell} = 0.46 - \frac{0.059}{6} \log(10^6) = 0.46 - \frac{0.059 \times 6}{6} = 0.46 - 0.059 = 0.401$ V.
Expressing as $x \times 10^{-1}$ V:
$0.401 = 4.01 \times 10^{-1}$ V, so $x = 4$ (nearest integer).
Hence, the correct answer is 4.