Electro Chemistry Part 3

Concept:In a balanced redox reaction, the number of electrons transferred in the Nernst equation corresponds to the total electrons exchanged after balancing oxidation and reduction half-reactions.Explanation:Balance the oxidation half: $\mathrm{BH_4^- \rightarrow H_2BO_3^-}$ in acidic medium.Add $3\ \mathrm{H_2O}$ to the left and $8\ \mathrm{H^+}$ to the right to balance O and H.Charge: left $= -1$, right $= +7$, so add $8\ e^-$ to the right.Oxidation half: $\mathrm{BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 8e^-}$ (electrons lost $=8$).Balance the reduction half: $\mathrm{ClO_3^- \rightarrow Cl^-}$ in acidic medium.Add $3\ \mathrm{H_2O}$ to the right and $6\ \mathrm{H^+}$ to the left.Charge: left $= +5$, right $= -1$, so add $6\ e^-$ to the left.Reduction half: $\mathrm{6e^- + 6H^+ + ClO_3^- \rightarrow Cl^- + 3H_2O}$ (electrons gained $=6$).Make electrons equal: LCM of $8$ and $6$ is $24$.Multiply oxidation half by $3$ and reduction half by $4$.Add the two half-reactions and cancel common species to get:$$\mathrm{3BH_4^- + 4ClO_3^- \rightarrow 3H_2BO_3^- + 4Cl^- + 3H_2O}$$Here, $24$ electrons are transferred in the balanced reaction.Thus, $n = 24$.Answer:$n = \boldsymbol{24}$ (Nearest integer). Option A.