JEE Chemistry 2026 Question with Solution

Given: The redox reaction is $\mathrm{BH_4^- (aq) + ClO_3^- (aq) \rightarrow H_2BO_3^- (aq) + Cl^- (aq)}$ in acidic medium.

Find: The value of $n$ in the Nernst equation, where $n$ is the number of electrons transferred in the balanced redox reaction.

Concept: In the Nernst equation, $n$ represents the total number of electrons exchanged in the balanced redox reaction.

Step 1: Identify oxidation and reduction

• $\mathrm{BH_4^- \rightarrow H_2BO_3^-}$: Boron is oxidised.
• $\mathrm{ClO_3^- \rightarrow Cl^-}$: Chlorine is reduced.

Step 2: Balance the oxidation half-reaction Oxidation state of boron changes from $-3$ in $\mathrm{BH_4^-}$ to $+3$ in $\mathrm{H_2BO_3^-}$. So, the loss is $6$ electrons.

Step 3: Balance the reduction half-reaction Oxidation state of chlorine changes from $+5$ in $\mathrm{ClO_3^-}$ to $-1$ in $\mathrm{Cl^-}$. So, the gain is $6$ electrons.

Step 4: Add the two half-reactions Since both half-reactions involve $6e^-$, the electrons cancel directly.

Step 5: Determine $n$ From the balanced equation, the number of electrons transferred is $6$.

Therefore, $n = 6$.

Given: $\mathrm{BH_4^-}$ is converted to $\mathrm{H_2BO_3^-}$ and $\mathrm{ClO_3^-}$ is converted to $\mathrm{Cl^-}$.

Find: The total electrons transferred, that is, $n$.

Boron changes from $-3$ to $+3$, so it loses $6e^-$. Chlorine changes from $+5$ to $-1$, so it gains $6e^-$.

Because the electron loss and gain are both $6$, the balanced redox process involves transfer of $6$ electrons in total.

Therefore, the value of $n$ in the Nernst equation is $6$.

The solution lists Correct Answer: 24, but the balancing shown in the working clearly gives $n = 6$. Hence the working is taken as authoritative.