Question

For the given cell reaction $\text{BH}_4^- + \text{ClO}_3^- \to \text{Cl}^- + \text{H}_2\text{BO}_3^-$. Cell emf 'E' is given as $\text{E} = \text{E}^0 - \frac{\text{RT}}{\text{nF}} \ln(\text{Q})$. Determine the value of 'n' in above equation.

Hint:

In borohydride oxidations, the hydrogen atoms typically change from $-1$ to $+1$, contributing significantly to the n-factor.

Answer 1

Correct Answer: 24

Determine oxidation states and electron change.
Reduction: $\text{ClO}_3^- \to \text{Cl}^-$.
Cl state: $+5 \to -1$. Change $= 6 e^-$ per Cl.
Oxidation: $\text{BH}_4^- \to \text{H}_2\text{BO}_3^-$ (Assuming borate species like $\text{H}_2\text{BO}_3^-$ or similar).
Check H oxidation. In borohydride ($\text{BH}_4^-$), H is hydridic ($-1$). In the oxidized oxyanion product, H is protonic ($+1$) attached to Oxygen.
For 1 mole of $\text{BH}_4^-$, there are 4 H atoms.
Change per H atom: $-1 \to +1$, loss of $2 e^-$.
Total loss for 4 H atoms: $4 \times 2 = 8 e^-$.
Does Boron change? Usually B is $+3$ in both. $-5 + 3 = -2$ no.
In $\text{BH}_4^-$: B is $+3$, H is $-1$. Net: $+3 + 4(-1) = -1$. Correct.
In Product (e.g. $\text{H}_3\text{BO}_3$ or ion): B is $+3$, H is $+1$.
Main redox is $4\text{H}^- \to 4\text{H}^+ + 8e^-$.
So oxidation n-factor is 8.
Balance electron transfer between Oxidant (6e) and Reductant (8e).
LCM of 6 and 8 is 24.
Reaction stoichiometry: $3 \times (\text{BH}_4^- \to 8e^-)$ and $4 \times (\text{ClO}_3^- \to 6e^-)$.
Total electrons $n = 24$.