How do you balance this redox reaction using oxidation number method?
\[A{{l}_{\left( s \right)}}\text{ }+\text{ }{{H}_{2}}S{{O}_{4\left( aq \right)\text{ }}}\to \text{ }A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3\left( aq \right)}}\text{ }+\text{ }{{H}_{2\left( g \right)}}\uparrow \]

Hint:We know the one approach to adjust redox responses is given by monitoring moving electrons by utilizing oxidation quantities of each and every one of the atoms. For Oxidation number change technique we have to start with an unbalanced skeleton condition.

Complete step-by-step answer:Let us start with calculating the oxidation number for each atom.
Step 1: When Sulfuric acid is a dilute Aluminium react with dilute Sulfuric acid \[\left( Al\text{ }{{H}_{2}}S{{O}_{4}} \right)\] in order to produce Aluminium sulfate and Hydrogen gas. The reaction is a Redox reaction. So the reaction is given by:
\[A{{l}_{\left( s \right)}}\text{ }+\text{ }{{H}_{2}}S{{O}_{4\left( aq \right)\text{ }}}\to \text{ }A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3\left( aq \right)}}\text{ }+\text{ }{{H}_{2\left( g \right)}}\uparrow \]
The above given reaction is Oxidation Reduction reaction because:
1.Al oxidation number in reactant is zero meanwhile Al on the left hand is \[+\text{ }3\] in aluminum sulfate.
2.Whereas the oxidation number of Hydrogen in reactants is \[+1\] whereas on the left side oxidation number is reduced to $0$ Thus this means there is change of electrons in between the reactants. That’s why the reaction is known as the Redox reaction.
Step 2: Next we have to identify what has to be oxidised as well as what is being reduced and the balance equation.
1.Al has to be oxidised \[0\text{ }to\text{ }+3\] and we need \[2Al\] on LHS as we have a $2$ on the right.
2.Thus, Hydrogen has to be reduced from\[+1\text{ }to\text{ }0\] Here we are with balanced $H$ and so it is given by:
\[2Al\text{ }+\text{ }{{H}_{2}}S{{O}_{4}}\text{ }\to \text{ }A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\text{ }+\text{ }{{H}_{2}}\uparrow \]
 Step 3: Now we have to calculate total increases in oxidation number which is due to oxidation taking place. We have oxidising \[2Al\text{ }\left( oxidation\text{ }state\text{ }0 \right)\text{ }to\text{
}2Al3+\text{ }\left( oxidation\text{ }state\text{ }+3 \right)\text{ }so\text{ }the\text{ }total\text{ }increase\text{ }is\text{ }+6.\]
Similarly we have to do same for reduction and now we get reduce
\[2\text{ }x\text{ }H+\text{ }\left( +1 \right)\text{ }to\text{ }2\text{ }x\text{ }H\text{ }\left( 0 \right)\text{ }so\text{ }total\text{ }decrease\text{ }is\text{ }-2\]
Step 4: Now we have to multiply oxidised or reduced so that total increase in a oxidation number = total decrease in oxidation number example we need to multiply terms involved in reduction of a hydrogen $\times 3$ to get total decrease of \[-6\] This mean we need to get the \[3{{H}_{2}}S{{O}_{4}}\] also we get \[3{{H}_{2}}\]
Final Equation: \[2Al\text{ }+\text{ }3{{H}_{2}}S{{O}_{4}}\text{ }\to \text{ }A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\text{ }+\text{ }3{{H}_{2}}\uparrow \]
Also the basic rule for balancing redox reaction using oxidation number rule:
1.Write the oxidation number of every component and distinguish components which are going through the change in an oxidation number. Identify oxidizing as a reducing agent.
2.By Multiplying equation of oxidizing as well as reducing agent by integer in order to adjust absolute increment in oxidation number. We have to balance all atom rather than $H$ as well as $O$

Note:The oxidation number technique and monitor electron picked up when substance is diminished and electron lost when substance is oxidized. Out oxidation numbers to every one of the particles in condition as well as compose numbers over the atom.