Question

Not balanced reaction $B{r}^{-}+Br{O}_3^{-}\to B{r}_2$ in acidic medium shows comproportionation to form Bromine. After balancing, x= no. of moles of $B{r}_2$ produced, y= no. of moles of $Br{O}_3^{-}$ used. The value of $\frac{x}{y}=$

Answer 1

Comproportionation reaction involves a simultaneous reduction and oxidation of two reactants to give one product.
Reaction occurs: $B{r}^{-}+Br{O}_3^{-}\to B{r}_2$
$\text{Oxidation half}:2B{r}^{-}\to {\stackrel{0}{Br}}_2+2e^{-}...\left(i\right)$
$\text{Reduction half}:2\stackrel{+5}{Br}{O}_3^{-}+10e^{-}\to {\stackrel{0}{Br}}_2...\left(ii\right)$
Multiplying eq. (i) by 5 and adding to eq. (ii), we get
$10B{r}^{-}+2Br{O}_3^{-}\to 6B{r}_2$
In acidic medium overall reaction can be represented as:
$10B{r}^{-}+2Br{O}_3^{-}+12H^{+}\to 6B{r}_2+6H_{2}O$
$5B{r}^{-}+Br{O}_3^{-}+6H^{+}\to 3B{r}_2+3H_{2}O$
Here $x=3$ and $y=1$ so the ratio becomes$\frac{x}{y}=3$