Question

The given unbalanced redox reaction
$S_2{O}_3^{2-}+S{b}_2{O}_5+H^{+}\to SbO+H_{2}S{O}_3$
is balanced by oxidation number method. The stoichiometric coefficient of $S{b}_2{O}_5$ and $SbO$ respectively, will be:

• A. $2,4$
• B. $4,8$
• C. $3,6$
• D. $1,2$

Answer 1

The correct option is A $2,4$

${\stackrel{+2}{S}}_2{O}_3^{2-}+\stackrel{+5}{S}{b}_2{O}_5\to \stackrel{+2}{S}bO+H_2\stackrel{+4}{S}{O}_3$

$S{b}_2{O}_5$ is an oxidising agent.
$S_2{O}_3^{2-}$ is a reducing agent.

$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|)\times \text{number of atom}$

${\stackrel{+2}{S}}_2{O}_3^{2-}\to H_2\stackrel{+4}{S}{O}_3$ oxidation
$n_f=(|+4-2|)\times 2=4$

$\stackrel{+5}{S}{b}_2{O}_5\to \stackrel{+2}{S}bO$ reduction
$n_f=(|+2-5|)\times 2=6$

Ratio of $n_f$ for oxidation to reduction is $2:3$.

Cross mutiplying the oxidising and reducing agents with ratio of n-factors.
$3S_2{O}_3^{2-}+2S{b}_2{O}_5\to SbO+H_{2}S{O}_3$

Balancing elements except $O$ and $H$
$3S_2{O}_3^{2-}+2S{b}_2{O}_5\to 4SbO+6H_{2}S{O}_3$

Balancing oxygen by adding $H_{2}O$
$3S_2{O}_3^{2-}+2S{b}_2{O}_5+3H_{2}O\to 4SbO+6H_{2}S{O}_3$

Balancing hydrogen by adding $H^{+}$
$3S_2{O}_3^{2-}+2S{b}_2{O}_5+3H_{2}O+6H^{+}\to 4SbO+6H_{2}S{O}_3$

Balancing charge:

charge in reactant side $=-6+6=0$
charge in product side $=0$

So the balanced equation is
$3S_2{O}_3^{2-}+2S{b}_2{O}_5+3H_{2}O+6H^{+}\to 4SbO+6H_{2}S{O}_3$