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 \(\mathrm{E}_{\mathrm{Sn}(\mathrm{OH})_{6}^{2-} / \mathrm{HSnO}_{2}^{-}, \mathrm{OH}}^{0}=-0.90 \mathrm{~V}\)

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Redox Reactions and Electrochemistry

Consider the following electrochemical cell at 298 K

Pt HSnO_{2}^{-} (aq) Sn (OH)_{6}^{2 -} (aq) ∣ Bi_{2} O_{3} (s) ∣ Bi (s)

.
If the reaction quotient at a given time is

1 0^{6}

, then the cell EMF (

E_{cell}

) is ____

\times 1 0^{- 1} V

(Nearest integer).
Given the standard half-cell reduction potential as

E_{Bi_{2} O_{3} / Bi_{, OH^{-}}}^{0} = - 0.44 V

and

E_{Sn (OH)_{6}^{2 -} / HSnO_{2}^{-}, OH}^{0} = - 0.90 V

Answer: 4

At the left electrode tin is oxidised,

HSn O_{2}^{-} \to Sn (OH)_{6}^{2 -}

(tin from

+ 2

+ 4

, two electrons), and at the right electrode bismuth is reduced,

B i_{2} O_{3} \to 2 Bi

(six electrons). Balancing the electrons, the cell reaction transfers

n = 6

. The standard cell potential is

E_{cell}^{\circ} = E_{cathode}^{\circ} - E_{anode}^{\circ} = - 0.44 - (- 0.90) = + 0.46 V

. By the Nernst equation with

Q = 1 0^{6}

E_{cell} = 0.46 - \frac{0.06}{6} lo g 1 0^{6} = 0.46 - 0.01 \times 6 = 0.40 V

, which is

4 \times 1 0^{- 1} V

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Integer answer