JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Redox Reactions and Electrochemistry

Balancing in acidic medium, the oxidation half-reaction is $\mathrm{BH}{}_4{}^{-}+3\mathrm{H}{}_2\mathrm{O}\to \mathrm{H}{}_2\mathrm{BO}{}_3{}^{-}+8\mathrm{H}{}^{+}+8\mathrm{e}{}^{-}$, in which each borohydride centre loses 8 electrons, and the reduction half-reaction is $\mathrm{ClO}{}_3{}^{-}+6\mathrm{H}{}^{+}+6\mathrm{e}{}^{-}\to \mathrm{Cl}{}^{-}+3\mathrm{H}{}_2\mathrm{O}$, in which chlorine gains 6 electrons going from $+5$ to $-1$. Equalising the electrons needs the least common multiple of 8 and 6, namely 24, so three borohydride ions and four chlorate ions take part: $3\mathrm{BH}{}_4{}^{-}+4\mathrm{ClO}{}_3{}^{-}\to 3\mathrm{H}{}_2\mathrm{BO}{}_3{}^{-}+4\mathrm{Cl}{}^{-}+3\mathrm{H}{}_2\mathrm{O}$. The number of electrons transferred in the balanced reaction is $n=24$.