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 quotient at a given time is 10^ 6 , then the cell EMF ( E_ cell ) is

JEE Main 202622 JanuaryMorning Shift

JEE Main 2026 · 22 January · Morning Shift · Chemistry · Electrochemistry

Consider the following electrochemical cell at 298 K Pt|HSnO_ 2 ^ -...

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}

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}

0.401 = 4.01 \times 10^{-1}

V, so

x = 4

(nearest integer).