Question

Given redox reaction :

$Cl{O}^{-}+Cr{O}_2^{-}\to C{l}^{-}+Cr{O}_4^{2-}$

When balanced in a basic medium by oxidation number method, the ratio of stoichiometric coefficient of $O{H}^{-}$ and $H_{2}O\left(\frac{n_{O{H}^{-}}}{n_{H_{2}O}}\right)$ will be:

• A. $1:2$
• B. $1:1$
• C. $2:1$
• D. $1:4$

Answer 1

The correct option is C $2:1$

$Cl{O}^{-}+Cr{O}_2^{-}\to C{l}^{-}+Cr{O}_4^{2-}$

Oxidation state of $Cl$ in $Cl{O}^{-}=+1$
Oxidation state of $Cl$ in $C{l}^{-}=-1$
Oxidation state of $Cr$ in $Cr{O}_2^{-}=+3$
Oxidation state of $Cr$ in $Cr{O}_4^{2-}=+6$

$Cr{O}_2^{-}$ is undergoing oxidation and is the reducing agent.
$Cl{O}^{-}$ is undergoing reduction and is the oxidising agent.

$n_f$ of $Cl{O}^{-}=2$
$n_f$ of $Cr{O}_2^{-}=3$

Cross multiplying these with $n_f$ of each other.
we get,
$3Cl{O}^{-}+2Cr{O}_2^{-}\to C{l}^{-}+Cr{O}_4^{2-}$


Balancing the elements other than oxygen and hydrogen on both sides,
$3Cl{O}^{-}+2Cr{O}_2^{-}\to 3C{l}^{-}+2Cr{O}_4^{2-}$

Adding the $H_{2}O$ to balance the oxygen,


$3Cl{O}^{-}+2Cr{O}_2^{-}+H_{2}O\to 3C{l}^{-}+2Cr{O}_4^{2-}$
adding $H^{+}$ to balance hydrogen,


$3Cl{O}^{-}+2Cr{O}_2^{-}+H_{2}O\to 3C{l}^{-}+2Cr{O}_4^{2-}+2H^{+}$

Now adding $O{H}^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2}O$,

$3Cl{O}^{-}+2Cr{O}_2^{-}+H_{2}O+2O{H}^{-}\to 3C{l}^{-}+2Cr{O}_4^{2-}+2H^{+}+2O{H}^{-}$

$3Cl{O}^{-}+2Cr{O}_2^{-}+H_{2}O+2O{H}^{-}\to 3C{l}^{-}+2Cr{O}_4^{2-}+2H_{2}O$

Eliminating the unrequired common $H_{2}O$ which is present on the both sides,
$3Cl{O}^{-}+2Cr{O}_2^{-}+2O{H}^{-}\to 3C{l}^{-}+2Cr{O}_4^{2-}+H_{2}O$

Charge on both sides is -7
So, this is the final balanced equation.
Ratio $\frac{n_{O{H}^{-}}}{n_{H_{2}O}}=\frac{2}{1}$