Question:medium
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).
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).
Show Hint
In the Nernst equation, \(n\) is always equal to the total number of electrons exchanged in the balanced redox reaction.
Updated On: Jun 6, 2026
Show Solution
Correct Answer: 24
Solution and Explanation
Step 1: Understanding the Concept:
The value of \(n\) in the Nernst equation represents the total number of moles of electrons transferred in the balanced redox reaction. This is found by equating the number of electrons lost in the oxidation half-reaction to the number of electrons gained in the reduction half-reaction.
Step 2: Key Formula or Approach:
1. Write separate oxidation and reduction half-reactions.
2. Balance the number of electrons for both.
3. Find the Least Common Multiple (LCM) of the electrons to get the overall value of \(n\).
Step 3: Detailed Explanation:
1. Reduction half-reaction:
\(\text{ClO}_3^- \rightarrow \text{Cl}^-\)
The oxidation state of Chlorine changes from \(+5\) in \(\text{ClO}_3^-\) to \(-1\) in \(\text{Cl}^-\).
Number of electrons gained per Cl atom = \(5 - (-1) = 6e^-\).
2. Oxidation half-reaction:
\(\text{BH}_4^- \rightarrow \text{H}_2\text{BO}_3^-\)
In \(\text{BH}_4^-\), Boron is in \(+3\) state and Hydrogen is in \(-1\) state (Hydride).
In \(\text{H}_2\text{BO}_3^-\), Boron is in \(+3\) state, Hydrogen is in \(+1\) state, and Oxygen is in \(-2\) state.
Boron's oxidation state remains unchanged. However, four Hydrogen atoms change from \(-1\) to \(+1\).
Change per H atom = \(1 - (-1) = 2e^-\).
For four H atoms, total electrons lost = \(4 \times 2 = 8e^-\).
3. Determine overall \(n\):
Reduction requires \(6e^-\) and Oxidation releases \(8e^-\).
To balance the overall reaction, we find the LCM of 6 and 8.
LCM(6, 8) = 24.
This means the balanced equation involves the transfer of 24 electrons.
Step 4: Final Answer:
The value of \(n\) is 24.
The value of \(n\) in the Nernst equation represents the total number of moles of electrons transferred in the balanced redox reaction. This is found by equating the number of electrons lost in the oxidation half-reaction to the number of electrons gained in the reduction half-reaction.
Step 2: Key Formula or Approach:
1. Write separate oxidation and reduction half-reactions.
2. Balance the number of electrons for both.
3. Find the Least Common Multiple (LCM) of the electrons to get the overall value of \(n\).
Step 3: Detailed Explanation:
1. Reduction half-reaction:
\(\text{ClO}_3^- \rightarrow \text{Cl}^-\)
The oxidation state of Chlorine changes from \(+5\) in \(\text{ClO}_3^-\) to \(-1\) in \(\text{Cl}^-\).
Number of electrons gained per Cl atom = \(5 - (-1) = 6e^-\).
2. Oxidation half-reaction:
\(\text{BH}_4^- \rightarrow \text{H}_2\text{BO}_3^-\)
In \(\text{BH}_4^-\), Boron is in \(+3\) state and Hydrogen is in \(-1\) state (Hydride).
In \(\text{H}_2\text{BO}_3^-\), Boron is in \(+3\) state, Hydrogen is in \(+1\) state, and Oxygen is in \(-2\) state.
Boron's oxidation state remains unchanged. However, four Hydrogen atoms change from \(-1\) to \(+1\).
Change per H atom = \(1 - (-1) = 2e^-\).
For four H atoms, total electrons lost = \(4 \times 2 = 8e^-\).
3. Determine overall \(n\):
Reduction requires \(6e^-\) and Oxidation releases \(8e^-\).
To balance the overall reaction, we find the LCM of 6 and 8.
LCM(6, 8) = 24.
This means the balanced equation involves the transfer of 24 electrons.
Step 4: Final Answer:
The value of \(n\) is 24.
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