JEE Main 2026 — Jan 28, Shift 1 — Chemistry solution

JEE Main2026Jan 28, Shift 1Chemistry

Consider the following redox reaction taking place in acidic medium

$BH_{4}^{-} (a q) + ClO_{3}^{-} (a q) \to H_{2} BO_{3}^{-} (a q) + Cl^{-} (a q)$

If the Nernst equation for the above balanced reaction is

$E_{cell} = E_{cell}^{\circ} - \frac{R T}{n F} ln Q$ ,

Then the value of n is ______. (Nearest integer)

Your answer

Solution

Assign oxidation numbers to find the electron change per species.

In BH $_{4}^{-}$ , B is in oxidation state +3 (treating H as -1 here is the convention for borohydride), but the relevant change is for the overall species: B remains +3 going to H $_{2}$ BO $_{3}^{-}$ , while the four H atoms in BH $_{4}^{-}$ (each at -1) become +1 in the borate product, so each B-bound H loses 2 electrons, giving 8 electrons lost per BH $_{4}^{-}$ . Thus n-factor of BH $_{4}^{-}$ = 8.

In ClO $_{3}^{-}$ , Cl is +5; it is reduced to Cl $^{-}$ ( $- 1$ ), a gain of 6 electrons. n-factor of ClO $_{3}^{-}$ = 6.

Balance electrons by LCM(8, 6) = 24: take 3 BH $_{4}^{-}$ (3 × 8 = 24 e $^{-}$ lost) and 4 ClO $_{3}^{-}$ (4 × 6 = 24 e $^{-}$ gained).

Balanced reaction: $3 BH_{4}^{-} + 4 ClO_{3}^{-} \to 4 Cl^{-} + 3 H_{2} BO_{3}^{-} + 3 H_{2} O$ , with $n = 24$ electrons transferred per equation as written.

$∴ n = 24$ .

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