Question

Consider the following redox reaction taking place in acidic medium: \[ \mathrm{BH_4^- (aq) + ClO_3^- (aq) \rightarrow H_2BO_3^- (aq) + Cl^- (aq)} \] If the Nernst equation for the above balanced reaction is \[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{RT}{nF}\ln Q, \] then the value of \(n\) is ________ (Nearest integer).

Hint:

In the Nernst equation, \(n\) is always equal to the total number of electrons exchanged in the balanced redox reaction.

Answer 1

Correct Answer: 24

Concept: In the Nernst equation, \(n\) represents the number of electrons transferred in the {balanced redox reaction}. Hence, we must balance the given reaction in acidic medium using the {ion–electron method} and determine the total electrons exchanged.
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 B: \[ \text{In } \mathrm{BH_4^-}: \quad B = -3 \] \[ \text{In } \mathrm{H_2BO_3^-}: \quad B = +3 \] Change in oxidation state: \[ -3 \rightarrow +3 \quad (\text{loss of } 6\ e^-) \] Balance the oxidation half-reaction in acidic medium: \[ \mathrm{BH_4^- + 3H_2O \rightarrow H_2BO_3^- + 8H^+ + 6e^-} \]
Step 3: Balance the reduction half-reaction Oxidation state of Cl: \[ \text{In } \mathrm{ClO_3^-}: \quad Cl = +5 \] \[ \text{In } \mathrm{Cl^-}: \quad Cl = -1 \] Change in oxidation state: \[ +5 \rightarrow -1 \quad (\text{gain of } 6\ e^-) \] Balanced reduction half-reaction: \[ \mathrm{ClO_3^- + 6H^+ + 6e^- \rightarrow Cl^- + 3H_2O} \]
Step 4: Add the two half-reactions Electrons cancel directly (\(6e^-\) each): \[ \mathrm{BH_4^- + ClO_3^- + 2H_2O \rightarrow H_2BO_3^- + Cl^- + 2H^+} \] This is the fully balanced redox reaction in acidic medium.
Step 5: Determine the value of \(n\) From the balanced equation: \[ \boxed{6\ \text{electrons are transferred}} \] Final Answer: \[ \boxed{n = 6} \]