Question

How many moles of $O{H}^{-}$ and $H_{2}O$ respectively required for the balanced equation for $H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to Cl{O}_2^{-}\left(aq\right)+O_2\left(g\right)$

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

Answer 1

The correct option is A 2, 5

Steps for Balancing redox reactions:

• Identify the oxidation and reduction halves.
• Find the oxidising and reducing agent.
• Find the n-factor of oxidising and reducing agents.
• Cross multiply the oxidising and reducing agent with the n-factor of each other.
• Balance the atoms other than oxygen and hydrogen.
• Balance oxygen atoms.
• Balance hydrogen atoms.

For basic medium:

If x oxygens are less on one side than the other side add x $H_{2}O$ units to that side. As soon as we add x $H_{2}O$ units for balancing oxygen, we add generally 2x (or whatever suitable) $H^{+}$ ions on the opposite side to balance hydrogen.Then after that we add equal (equal to number of $H^{+}$ added) number of $O{H}^{-}$ ions to the both sides and combine $H^{+}$ and $O{H}^{-}$ ions to form $H_{2}O$. Then we cancel $H_{2}O$ which is common to both sides and which can be eliminated thus finally $O{H}^{-}$ is left on one side and the equation is balanced. Then we can verify whether the equation is balanced or not by checking the balance of charge on both sides.

formula used for the n-factor calculation,
$n_f=(|O.S{.}_{Product}-O.S{.}_{Reactant}|\times \text{number of atoms}$

Taking the given equation and following the above mentioned steps:
$H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to Cl{O}_2^{-}\left(aq\right)+O_2\left(g\right)$

Oxidation state of O in $H_2{O}_2=-1$
Oxidation state of Oin $O_2=0$
Oxidation state of Cl in $C{l}_2{O}_7=+7$
Oxidation state of Cl in $Cl{O}_2^{-}=+3$

Clearly, $H_2{O}_2$ is undergoing oxidation and $C{l}_2{O}_7$ is undergoing reduction.
using the formula of n-factor given above,
$n_f$ of $H_2{O}_2=1$
$n_f$ of $C{l}_2{O}_7=4$

Cross multiplying these with $n_f$ of each other.
we get,
$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to Cl{O}_2^{-}\left(aq\right)+O_2\left(g\right)$

Balancing the main elements on both sides, generally oxygen is not balanced in this step but we will balance as it is undergoing oxidation in this reaction.

$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)$

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

LHS of the equation contains 15 oxygens while RHS contains only 12 oxygens, so adding 3 $H_{2}O$ to RHS.
$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+3H_{2}O$
adding $H^{+}$ to balance hydrogen,

As LHS of the equation contains 8 hydrogens while RHS contains 6 hydrogens, so adding 2$H^{+}$ to RHS.
$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+3H_{2}O+2H^{+}$

Now adding $O{H}^{-}$ to both sides to combine with $H^{+}$ and make it $H_{2}O$,
$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)+2O{H}^{-}\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+3H_{2}O+2H^{+}+2O{H}^{-}$

$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)+2O{H}^{-}\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+5H_{2}O$

This is the final balanced equation.

We can also see that charge on both sides is -2. which also indicates that the equation is balanced now.

$4H_2{O}_2\left(aq\right)+C{l}_2{O}_7\left(aq\right)+2O{H}^{-}\to 2Cl{O}_2^{-}\left(aq\right)+4O_2\left(g\right)+5H_{2}O$

So, 2, 5 moles of $O{H}^{-}$ and $H_{2}O$ respectively.