Question

The given redox reaction

$B{r}_2+O{H}^{-}\to Br{O}_3^{-}+B{r}^{-}+H_{2}O$
is balanced by ion electron method. Choose the correct statement(s) regarding stoichiometric ratios $\frac{n_a}{n_b}$ :
Here $a,b$ represents different species in the reaction.

• A. $\frac{n_{B{r}_2}}{n_{Br{O}_3^{-}}}=\frac{3}{2}$
• B. $\frac{n_{B{r}_2}}{n_{Br{O}_3^{-}}}=3$
• C. $\frac{n_{O{H}^{-}}}{n_{H_{2}O}}=2$
• D. $\frac{n_{O{H}^{-}}}{n_{H_{2}O}}=3$

Answer 1

Answer: (B), (C)

The correct options are
B $\frac{n_{B{r}_2}}{n_{Br{O}_3^{-}}}=3$
C $\frac{n_{O{H}^{-}}}{n_{H_{2}O}}=2$

$\stackrel{0}{B}{r}_2+O{H}^{-}\to \stackrel{+5}{B}r{O}_3^{-}+B{r}^{-}+H_{2}O$

Oxidation half : $B{r}_2\to Br{O}_3^{-}$
Bromine changes its oxidation state from $0$ to $+5$

Reduction half : $B{r}_2\to 2B{r}^{-}$.
Bromine changes its oxidation state from $0$ to $-1$

Taking the oxidation half reaction :
$B{r}_2\to Br{O}_3^{-}$
Balancing $Br$
$B{r}_2\to 2Br{O}_3^{-}$

Balancing $O$ by adding $H_{2}O$
$B{r}_2+6H_{2}O\to 2Br{O}_3^{-}$

Balancing $H$ by adding $H^{+}$
$B{r}_2+6H_{2}O\to 2Br{O}_3^{-}+12H^{+}$

Adding similar amount of $O{H}^{-}$ to both sides and combining $H^{+}$ and $O{H}^{-}$ to form $H_{2}O$
$B{r}_2+6H_{2}O+12O{H}^{-}\to 2Br{O}_3^{-}+12H^{+}+12O{H}^{-}$
$B{r}_2+6H_{2}O+12O{H}^{-}\to 2Br{O}_3^{-}+12H_{2}O$

Cancelling the extra $H_{2}O$
$B{r}_2+12O{H}^{-}\to 2Br{O}_3^{-}+6H_{2}O$

Balancing the charge on RHS by adding $10e^{-}$
$B{r}_2+12O{H}^{-}\to 2Br{O}_3^{-}+6H_{2}O+10e^{-}$ ......(i)


Taking the reduction half :
$B{r}_2\to B{r}^{-}$

Balancing $Br$
$B{r}_2\to 2B{r}^{-}$

Balancing charge by adding $2e^{-}$ to LHS
$B{r}_2+2e^{-}\to 2B{r}^{-}$......(ii)

Multiplying eq(ii) by 5 and then adding to eq (i)
$5B{r}_2+10e^{-}\to 10B{r}^{-}$
$B{r}_2+12O{H}^{-}\to 2Br{O}_3^{-}+6H_{2}O+10e^{-}$
On adding
__________________________________________
$6B{r}_2+12O{H}^{-}\to 2Br{O}_3^{-}+10B{r}^{-}+6H_{2}O$

$\frac{n_{B{r}_2}}{n_{Br{O}_3^{-}}}=\frac{6}{2}=3$

$\frac{n_{O{H}^{-}}}{n_{H_{2}O}}=\frac{12}{6}=2$