Question

$Cl{O}_2+Sb{O}_2^{-}\to Cl{O}_2^{-}+Sb(OH{)}_6^{-}$
The balanced form of the above reaction in basic medium using oxidation number method is:

• A. $2Cl{O}_2+Sb{O}_2^{-}+2O{H}^{-}\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}+2H_{2}O$
• B. $Cl{O}_2+Sb{O}_2^{-}+2H_{2}O+O{H}^{-}\to Cl{O}_2^{-}+Sb(OH{)}_6^{-}$
• C. $2Cl{O}_2+Sb{O}_2^{-}+2H_{2}O+2O{H}^{-}\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}$
• D. $2Cl{O}_2+4Sb{O}_2^{-}+2O{H}^{-}\to Cl{O}_2^{-}+4Sb(OH{)}_6^{-}+3H_{2}O$

Answer 1

The correct option is C $2Cl{O}_2+Sb{O}_2^{-}+2H_{2}O+2O{H}^{-}\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}$

$\stackrel{+4}{C}l{O}_2+\stackrel{+3}{S}b{O}_2^{-}\to \stackrel{+3}{C}l{O}_2^{-}+\stackrel{+5}{S}b(OH{)}_6^{-}$

$Cl{O}_2$ is an oxidising agent.
$Sb{O}_2^{-}$ is a reducing agent.

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

$\stackrel{+3}{S}b{O}_2^{-}\to \stackrel{+5}{S}b(OH{)}_6^{-}$ oxidation
$n_f=(|+5-3|)\times 1=2$

$\stackrel{+4}{C}l{O}_2\to \stackrel{+3}{C}l{O}_2^{-}$ reduction
$n_f=(|+3-4|)\times 1=1$

Ratio of $n_f$ for oxidation to reduction is $2:1$.

Cross mutiplying the oxidising and reducing agents with ratio of n-factors.
$2Cl{O}_2+Sb{O}_2^{-}\to Cl{O}_2^{-}+Sb(OH{)}_6^{-}$

Balancing elements except $O$ and $H$
$2Cl{O}_2+Sb{O}_2^{-}\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}$

Balancing oxygen by adding $H_{2}O$
$2Cl{O}_2+Sb{O}_2^{-}+4H_{2}O\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}$

Balancing hydrogen by adding $H^{+}$
$2Cl{O}_2+Sb{O}_2^{-}+4H_{2}O\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}+2H^{+}$

Adding $2O{H}^{-}$ on both sides to balance in basic medium
$2Cl{O}_2+Sb{O}_2^{-}+4H_{2}O+2O{H}^{-}\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}+2H^{+}+2O{H}^{-}$

$2Cl{O}_2+Sb{O}_2^{-}+4H_{2}O+2O{H}^{-}\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}+2H_{2}O$

Canceling $H_{2}O$
$2Cl{O}_2+Sb{O}_2^{-}+2H_{2}O+2O{H}^{-}\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}$

Balancing charge:

charge in reactant side $=-1-2=-3$
charge in product side $=-2-1=-3$

So the balanced equation is
$2Cl{O}_2+Sb{O}_2^{-}+2H_{2}O+2O{H}^{-}\to 2Cl{O}_2^{-}+Sb(OH{)}_6^{-}$