JEE Main 2026 — Jan 22, Shift 1 — Chemistry solution

JEE Main2026Jan 22, Shift 1Chemistry

Consider the following electrochemical cell at 298 K

$P t ∣ H S n O_{2}^{-} (a q) ∣ S n (O H)_{6}^{2 -} (a q) ∣∣ B i_{2} O_{3} (s) ∣ B i (s)$ .

If the reaction quotient at a given time is $1 0^{6}$ , then the cell EMF ( $E_{ce l l}$ ) is _______ $\times 1 0^{- 1}$ V (Nearest integer).

Given the standard half-cell reduction potential as $E_{B i_{2} O_{3} / B i, O H^{-}}^{0} = - 0.44$ V and $E_{S n (O H)_{6}^{2 -} / H S n O_{2}^{-}, O H^{-}}^{0} = - 0.90$ V

Your answer

Solution

The cathode is the $B i_{2} O_{3} / B i$ couple and the anode is the $S n (O H)_{6}^{2 -} / H S n O_{2}^{-}$ couple.

$E_{ce l l}^{0} = E_{c a t h o d e}^{0} - E_{an o d e}^{0} = - 0.44 - (- 0.90) = + 0.46$ V.

Number of electrons transferred (balancing the overall cell reaction): $n = 6$ .

Nernst equation at 298 K: $E_{ce l l} = E_{ce l l}^{0} - \frac{0.06}{n} lo g Q = 0.46 - \frac{0.06}{6} lo g 1 0^{6} = 0.46 - 0.01 \times 6 = 0.46 - 0.06 = 0.40$ V.

Expressed in units of $1 0^{- 1}$ V: $E_{ce l l} = 4 \times 1 0^{- 1}$ V, so the answer is 4.

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