Question

Balance the following reaction by oxidation number method.
Reaction: \[C{r_2}{O_7}^{2 - } + S{O_{2(g)}} \to Cr_{(aq)}^{3 + } + SO_{4(aq)}^{2 - }\] [in acidic medium]

Answer 1

Hint: We are using the oxidation number method for balancing the given reaction. The oxidation number method involves finding the change in oxidation numbers of all atoms taking part in the reaction. We compare the total number of electrons lost and the total number of electrons gained during the reaction and make them equal.

Complete answer:
The oxidation number method is applied step by step as follows.
First, we write the correct chemical formulas of the reactant and the products.
Then we identify the atoms that change the oxidation number during the reaction.
Calculate the oxidation number of each atom for a given reaction and if numbers are not equal then make them equal by multiplying atoms with a simple whole number.
Remember the medium in which reaction is taking place if it is water then accordingly add hydroxyl and hydrogen ions if medium acidic add an appropriate number of hydrogen ions and in case of basic medium add an appropriate number of hydroxyl ions. The overall charge should be equal on both sides of the reaction.
For balancing the hydrogen and oxygen atoms, add water molecules to the other side of the reaction.
Let’s see the given reaction:
\[C{r_2}{O_7}^{2 - } + S{O_{2(g)}} \to Cr_{(aq)}^{3 + } + SO_{4(aq)}^{2 - }\]
Assigning oxidation number:
\[{[\mathop {C{r_2}}\limits^{ + 6} {\mathop O\limits^{ - 2} _7}]^{2 - }} + \mathop S\limits^{ + 4} {\mathop O\limits^{ - 2} _{2(g)}} \to [\mathop {Cr}\limits^{ + 3} ]_{(aq)}^{3 + } + [\mathop S\limits^{ + 6} {\mathop O\limits^{ - 2} _4}]_{(aq)}^{2 - }\]
Calculating the change in oxidation number and multiplying the atoms with a simple whole number to make them equal.
\[{[\mathop {C{r_2}}\limits^{ + 6} {\mathop O\limits^{ - 2} _7}]^{2 - }} + \mathop {3S}\limits^{ + 4} {\mathop O\limits^{ - 2} _{2(g)}} \to 2[\mathop {Cr}\limits^{ + 3} ]_{(aq)}^{3 + } + 3[\mathop S\limits^{ + 6} {\mathop O\limits^{ - 2} _4}]_{(aq)}^{2 - }\]
Now adding Hydrogen ions to balance the charge as the reaction is taking place in an acidic medium
\[C{r_2}{O_7}^{2 - } + 3S{O_{2(g)}} + 2H_{(aq)}^ + \to 2Cr_{(aq)}^{3 + } + 3SO_{4(aq)}^{2 - }\]
Adding water to balance the number of Hydrogen and Oxygen atoms
\[C{r_2}{O_7}^{2 - } + 3S{O_{2(g)}} + 2H_{(aq)}^ + \to 2Cr_{(aq)}^{3 + } + 3SO_{4(aq)}^{2 - } + {H_2}{O_{(l)}}\]
Hence a balanced reaction is obtained.

Note:
It is necessary to apply the oxidation number in the above-mentioned sequence. If any step is done before or after then the reaction obtained will not be balanced. Also carefully observe the change in oxidation state and charge of atoms as it can create confusion.