Question

For the equation in acidic medium: $S_2{O}_3^{2-}\left(aq\right)+C{r}_2{O}_7^{2-}\left(aq\right)\to S_4{O}_6^{2-}\left(aq\right)+C{r}^{3+}\left(aq\right)$
The coefficient of $H^{+}$ in the balanced reaction by oxidation number method will be

Answer 1

${\stackrel{+2}{S}}_2{O}_3^{2-}\left(aq\right)+\stackrel{+6}{C}{r}_2{O}_7^{2-}\left(aq\right)\to {\stackrel{+2.5}{S}}_4{O}_6^{2-}\left(aq\right)+\stackrel{+3}{C}{r}^{3+}\left(aq\right)$
$C{r}_2{O}_7^{2-}$ is oxidising agent.
$S_2{O}_3^{2-}$ is reducing agent.


${\stackrel{+2}{S}}_2{O}_3^{2-}\left(s\right)\to {\stackrel{+2.5}{S}}_4{O}_6^{2-}\left(aq\right)$ oxidation
$n_f=(|+2.5-2|\times 2=1$
$\stackrel{+6}{C}{r}_2{O}_7^{2-}\left(aq\right)\to \stackrel{+3}{C}{r}^{3+}\left(aq\right)$ reduction
$n_f=(|3-6|\times 2=6$

Cross multiplying:
$6S_2{O}_3^{2-}\left(aq\right)+1C{r}_2{O}_7^{2-}\left(aq\right)\to S_4{O}_6^{2-}\left(aq\right)+C{r}^{3+}\left(aq\right)$

Balancing elements except $O$ and $H$
$6S_2{O}_3^{2-}\left(aq\right)+C{r}_2{O}_7^{2-}\left(aq\right)\to 3S_4{O}_6^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)$


Balancing oxygen atoms by adding $H_{2}O$
$6S_2{O}_3^{2-}\left(aq\right)+C{r}_2{O}_7^{2-}\left(aq\right)\to 3S_4{O}_6^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)+7H_{2}O$

Balancing hydrogen by adding $H^{+}$
$6S_2{O}_3^{2-}\left(aq\right)+C{r}_2{O}_7^{2-}\left(aq\right)+14H^{+}\to 3S_4{O}_6^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)+7H_{2}O$

Balancing charge:
charge in reactant side = -12-2+14=0
charge in product side = -6+6=0
so the balanced equation is $6S_2{O}_3^{2-}\left(aq\right)+C{r}_2{O}_7^{2-}\left(aq\right)+14H^{+}\to 3S_4{O}_6^{2-}\left(aq\right)+2C{r}^{3+}\left(aq\right)+7H_{2}O$

The coefficient of $H^{+}$ in balanced equation is is 14