Question

Balance the following reaction:
\[{\text{Cu}} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{CuS}}{{\text{O}}_4} + {\text{S}}{{\text{O}}_2} \uparrow + {{\text{H}}_2}{\text{O}}\] .

Answer 1

Hint: If the number of each element involved in the reaction is not the same on both sides, such chemical equation is said to be unbalanced. For balancing a chemical equation, we need to equalize the number of atoms of each element in order to obey the law of conservation of mass.

Complete step by step answer:
The substance that undergoes a chemical change in the reaction is known as the reactants and new substance formed during the reaction is known as product. A chemical equation represents a chemical reaction.
 To balance the chemical equation by oxidation number method:
-Write the skeleton equation representing the chemical change.
-Assign the oxidation number to find out which atoms are undergoing oxidation and reduction, write a separate equation for the atoms undergoing oxidation and reduction.
-Find the change in oxidation number in each equation. Make the change equal in both the equation by multiplying with suitable integers. Add both the equations.
-Now balance those elements which are not undergoing oxidation or reduction followed by balancing of hydrogen and oxygen with the help of water.
-To balance the given reaction in question:
\[{\text{Cu}} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{CuS}}{{\text{O}}_4} + {\text{S}}{{\text{O}}_2} \uparrow + {{\text{H}}_2}{\text{O}}\]
\[{\text{Cu}}\left( 0 \right) + {{\text{H}}_2}{\text{S}}{{\text{O}}_4}\left( { + 6} \right) \to {\text{CuS}}{{\text{O}}_4}\left( { + 2} \right) + {\text{S}}{{\text{O}}_2}\left( { + 4} \right) + {{\text{H}}_2}{\text{O}}\] , as oxidation state of copper changes from zero to \[ + 2\] which means oxidation of copper taking place. Oxidation state of sulphur changes from \[ + 6\] to \[ + 4\] which means reduction of sulphur taking place. Oxidation half reaction can be written as: \[{\text{Cu}} \to {\text{CuS}}{{\text{O}}_4}\] and reduction half can be written as: \[{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{S}}{{\text{O}}_2}\] .
As in both the change of oxidation state is by 2. There is no need to add integers. Now, adding both the equations: \[{\text{Cu}} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{CuS}}{{\text{O}}_4} + {\text{S}}{{\text{O}}_2}\].
First balance sulphur atom by adding a coefficient 2 to sulphuric acid. Now balance hydrogen and oxygen by adding two water molecules on the product side to have 4 hydrogen atoms on each side and 8 oxygen atoms on both sides. Balanced reaction will be: \[{\text{Cu}} + 2{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{CuS}}{{\text{O}}_4} + {\text{S}}{{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}}\] .

Note:
The total mass of the elements of products of chemical reaction has to be equal to the total mass of the elements of the reactants. So the number of atoms of each element remains the same before and after the reaction.