Consider the following redox reaction taking | vedclass

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(C) First, balance the redox reaction in acidic medium: Oxidation half-reaction: $BH_{4}^{-} + 3H_{2}O \rightarrow H_{2}BO_{3}^{-} + 8H^{+} + 8e^{-}$

Vedclass · Accepted Answer

$24$

Vedclass · Answer

(C) First, balance the redox reaction in acidic medium: Oxidation half-reaction: $BH_{4}^{-} + 3H_{2}O \rightarrow H_{2}BO_{3}^{-} + 8H^{+} + 8e^{-}$ Reduction half-reaction: $ClO_{3}^{-} + 6H^{+} + 6e^{-} \rightarrow Cl^{-} + 3H_{2}O$ To balance the electrons, multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $4$: $3BH_{4}^{-} + 9H_{2}O \rightarrow 3H_{2}BO_{3}^{-} + 24H^{+} + 24e^{-}$ $4ClO_{3}^{-} + 24H^{+} + 24e^{-} \rightarrow 4Cl^{-} + 12H_{2}O$ Adding these, we get the balanced equation: $3BH_{4}^{-} + 4ClO_{3}^{-} \rightarrow 3H_{2}BO_{3}^{-} + 4Cl^{-} + 3H_{2}O$ In the Nernst equation, $n$ represents the number of moles of electrons transferred in the balanced redox reaction. From the balanced equation, $n = 24$.

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