Question

When the following redox reaction
$C{l}_2+O{H}^{-}\to Cl{O}_3^{-}+C{l}^{-}+H_{2}O$
is balanced by ion-electron method , what will be the coefficient of $C{l}_2$ and $O{H}^{-}$ respectively?

• A. $3,6$
• B. $2,4$
• C. $1.2$
• D. $5,5$

Answer 1

The correct option is A $3,6$

$\stackrel{0}{C}{l}_2+O{H}^{-}\to \stackrel{+5}{C}l{O}_3^{-}+C{l}^{-}+H_{2}O$

Oxidation half : $C{l}_2\to Cl{O}_3^{-}$
Chlorine changes its oxidation state from $0$ to $+5$

Reduction half : $C{l}_2\to 2C{l}^{-}$.
Chlorine changes its oxidation state from $0$ to $-1$

Taking the oxidation half reaction :
$C{l}_2\to Cl{O}_3^{-}$
Balancing $Cl$
$C{l}_2\to 2Cl{O}_3^{-}$

Balancing $O$ by adding $H_{2}O$
$C{l}_2+3H_{2}O\to 2Cl{O}_3^{-}$

Balancing $H$ by adding $H^{+}$
$C{l}_2+3H_{2}O\to 2Cl{O}_3^{-}+6H^{+}$

Adding similar amount of $O{H}^{-}$ to both sides and combining $H^{+}$ and $O{H}^{-}$ to form $H_{2}O$
$C{l}_2+3H_{2}O+6O{H}^{-}\to 2Cl{O}_3^{-}+6H^{+}+6O{H}^{-}$
$C{l}_2+3H_{2}O+6O{H}^{-}\to 2Cl{O}_3^{-}+6H_{2}O$

Cancelling the extra $H_{2}O$
$C{l}_2+6O{H}^{-}\to 2Cl{O}_3^{-}+3H_{2}O$

Balancing the charge on RHS by adding $4e^{-}$
$C{l}_2+6O{H}^{-}\to 2Cl{O}_3^{-}+3H_{2}O+4e^{-}$ ......(i)


Taking the reduction half :
$C{l}_2\to C{l}^{-}$

Balancing $Cl$
$C{l}_2\to 2C{l}^{-}$

Balancing charge by adding $2e^{-}$ to LHS
$C{l}_2+2e^{-}\to 2C{l}^{-}$......(ii)

Multiplying eq(ii) by 2 and then adding to eq (i)
$2C{l}_2+4e^{-}\to 4C{l}^{-}$
$C{l}_2+6O{H}^{-}\to 2Cl{O}_3^{-}+3H_{2}O+4e^{-}$
On adding
__________________________________________
$3C{l}_2+6O{H}^{-}\to 2Cl{O}_3^{-}+4C{l}^{-}+3H_{2}O$

So the correct answer is 3,6