Question

What will be the ratio $\frac{n_{C{r}_2{O}_7^{2-}}}{n_{S{O}_3^{2-}}}$ when the following redox reaction
$C{r}_2{O}_7^{2-}\left(aq\right)+S{O}_3^{2-}\left(aq\right)\to C{r}^{3+}\left(aq\right)+S{O}_4^{2-}\left(aq\right)$
is balanced in an acidic medium. Where $n$ represents the stoichiometric coefficient of the respective species.

• A. $2:1$
• B. $1:3$
• C. $1:2$
• D. $2:3$

Answer 1

The correct option is B $1:3$

The redox reaction is given below :
${\stackrel{+6}{Cr}}_2{O}_7^{2-}+3\stackrel{+4}{S}{O}_3^{2-}\to \stackrel{+3}{C{r}^{3+}}+\stackrel{+6}{S}{O}_4^{2-}$

So , oxidizing agent : $C{r}_2{O}_7^{2-}$
Reducing agent : $S{O}_3^{2-}$

n-factor :
for $C{r}_2{O}_7^{2-}:6$ ,
for $S{O}_3^{2-}:2$
$n_f$ ratio is $1:3$
Cross multiply the oxidising and reducing agent with $n_f$ ratio,

$C{r}_2{O}_7^{2-}+3S{O}_3^{2-}\to C{r}^{3+}+S{O}_4^{2-}$

Balancing all atoms except $O$ and $H$ :

$C{r}_2{O}_7^{2-}+3S{O}_3^{2-}\to 2C{r}^{3+}+3S{O}_4^{2-}$

Balance oxygen atoms by adding $H_{2}O$

$C{r}_2{O}_7^{2-}+3S{O}_3^{2-}\to 2C{r}^{3+}+3S{O}_4^{2-}+4H_{2}O$
Balance hydrogen atoms by adding $H^{+}$

$C{r}_2{O}_7^{2-}+3S{O}_3^{2-}+8H^{+}\to 2C{r}^{3+}+3S{O}_4^{2-}+4H_{2}O$

Balance charge
charge in reactant side = 0
charge in product side = 0

So the balanced equation is :

$C{r}_2{O}_7^{2-}+3S{O}_3^{2-}+8H^{+}\to 2C{r}^{3+}+3S{O}_4^{2-}+4H_{2}O$
So the ratio $\frac{n_{C{r}_2{O}_7^{2-}}}{n_{S{O}_3^{2-}}}=\frac{1}{3}$