Question

Balance the following equation in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
\[C{l_2}{O_7}(g) + {H_2}{O_2}(aq)\xrightarrow{{}}ClO_2^ - (aq) + {O_2}(g) + {H^ + }(aq)\]

Answer 1

Hint: We'll start by identifying the oxidising and reducing agents based on their oxidation numbers. To begin, we must understand the ion-electron technique. After that, we'll balance it. Following that, we'll go on to the oxidation number technique. And we'll strike a new balance with it. Then we'll compare the two balanced reactions to see if we're correct.

Complete Step by Step Solution:
Oxygen's oxidation number is increased from 1 to 0, whereas chlorine's oxidation number is dropped from \[ + 7\] to \[ + 3\] . We can deduce that \[C{l_2}{O_7}\] is the oxidising agent and \[{H_2}{O_2}\] is the reducing agent from this.
The half-reaction approach, commonly known as the ion-electron method, divides the redox reaction into two halves. The first is for oxidation, whereas the second is for reduction. Both are balanced independently before being merged for a complete balance reaction.

Ion electron method for balancing:
First, we'll write the oxidation component of the equation.
\[{H_2}{O_2}(aq)\xrightarrow{{}}{O_2}(g)\]
We add two electrons since the oxidation number differs by two.
\[{H_2}{O_2}(aq)\xrightarrow{{}}{O_2}(g) + 2{e^ - }\]
To make it charge-neutral, 2 additional hydroxide ions are added.
\[{H_2}{O_2}(aq) + 2O{H^ - }\xrightarrow{{}}{O_2}(g) + 2{e^ - }\]

We add two water molecules now that the hydrogen and oxygen molecules are more on one side.
\[{H_2}{O_2}(aq) + 2O{H^ - }\xrightarrow{{}}{O_2}(g) + 2{e^ - } + 2{H_2}O(aq)\]
Now we'll write the second half of the reduction.
\[C{l_2}{O_7}(g)\xrightarrow{{}}ClO_2^ - (aq)\]
We add 8 electrons to balance the oxidation number because the chlorine is already balanced.
\[C{l_2}{O_7}(g) + 8{e^ - }\xrightarrow{{}}ClO_2^ - (aq)\]
Six hydride ions are added to balance the charge.
\[C{l_2}{O_7}(g) + 8{e^ - }\xrightarrow{{}}ClO_2^ - (aq) + 6O{H^ - }(aq)\]
To remove the electrons, the oxidation half is multiplied by 4 and added to the reduction half.

The equation is balanced as follows:
\[C{l_2}{O_7}(g) + 4{H_2}{O_2}(aq) + 2O{H^ - }(aq)\xrightarrow{{}}ClO_2^ - + 4{O_2}(g) + 5{H_2}O(l)\]

In a redox reaction, the oxidation number approach is a manner of balancing the equation while watching the electrons. The theory is based on the transfer of electrons between charged atoms.

Oxidation number method:
The oxidation number of \[{H_2}{O_2}\] increases by 2. While \[C{l_2}{O_7}\] 's oxidation number decreases by 8.
\[{H_2}{O_2}\] and \[{O_2}\] are multiplied by four, while \[C{l_2}{O_7}\] is multiplied by two.
\[C{l_2}{O_7}(g) + 4{H_2}{O_2}(aq)\xrightarrow{{}}2ClO_2^ - (aq) + 4{O_2}(g)\]
Add 3 water molecules to balance the oxygen atoms, then 2 hydroxide ions and 2 additional water molecules to balance the hydrogen atoms. In the end, we get a balanced equation.
\[C{l_2}{O_7}(g) + 4{H_2}{O_2}(aq) + 2O{H^ - }(aq)\xrightarrow{{}}ClO_2^ - + 4{O_2}(g) + 5{H_2}O(l)\]
We can see that both methods result in the same equation.

Note: Because of the Law of Conservation of Mass, it is required to balance equations. One of the guiding principles is this. We can use the balanced reaction to figure out how much reactant we'll need to make a certain amount of product.