Core Logic
To find 'n', the number of moles of electrons transferred, we must balance the overall redox equation and calculate the net electron exchange.
Step 1: Assign Oxidation States
In BH₄^-$BH_4^-$, Hydrogen is -1$-1$ (hydride), so Boron is +3$+3$. In H₂BO₃^-$H_2BO_3^-$, Boron is +3$+3$. The oxidation state of B doesn't change, but H$H$ oxidizes from -1$-1$ to +1$+1$ (in water/acid).
Alternatively, treat the whole moiety:
Oxidation half-reaction: BH₄^- + 3H₂O arrow H₂BO₃^- + 8H^+ + 8e^-$BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 8e^-$
In ClO₃^-$ClO_3^-$, Chlorine is +5$+5$. In Cl^-$Cl^-$, Chlorine is -1$-1$.
Reduction half-reaction: ClO₃^- + 6H^+ + 6e^- arrow Cl^- + 3H₂O$ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O$
Step 2: Balance Electrons
Multiply the oxidation half-reaction by 3 and the reduction half-reaction by 4 to equalize electrons transferred:
3 × (BH₄^- + 3H₂O arrow H₂BO₃^- + 8H^+ + 8e^-) 24e^-$3 \times (BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 8e^-) \implies 24e^-$
4 × (ClO₃^- + 6H^+ + 6e^- arrow Cl^- + 3H₂O) 24e^-$4 \times (ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O) \implies 24e^-$
The lowest common multiple of electrons exchanged is 24.
Final Conclusion
The balanced equation transfers 24 electrons, so n = 24$n = 24$ in the Nernst equation.
Pattern Recognition
The total electrons 'n$n$' in the Nernst equation is always the lowest common multiple of electrons from the balanced oxidation and reduction half-reactions.
Chapter Mix
Class 12 Chemistry: Electrochemistry
Class 11 Chemistry: Redox Reactions