$Zn+{{H}_{2}}S{{O}_{4}}(dil)\to ZnS{{O}_{4}}+{{H}_{2}}$
(a)- Write the ionic equation for the reaction.
(b)- Express the ionic equation in the form of two half-reaction.

Hint: Zinc is the element of group 12 of the d-block and it found as Zn in the elemental state and the formula of sulfuric acid is ${{H}_{2}}S{{O}_{4}}$. In the ionic form, the total charge on the reactant should be zero and the total charge on the product side should also be zero. The two half cell reactions are oxidation half-reaction and reduction half-reaction.

Complete step by step answer:
Zinc is the element of group 12 of the d-block and it found as Zn in the elemental state and the formula of sulfuric acid is ${{H}_{2}}S{{O}_{4}}$. When they react with each other there is a single displacement reaction i.e., the hydrogen atom will be replaced with the zinc element.
So, the reaction will be zinc and will react with sulfuric acid to form zinc sulfide and hydrogen gas. The reaction is given below:
$Zn+{{H}_{2}}S{{O}_{4}}(dil)\to ZnS{{O}_{4}}+{{H}_{2}}$

(a)- In the ionic form, the total charge on the reactant should be zero and the total charge on the product side should also be zero. On the reactant side, the sulfuric acid will split into ions and on the product side, the zinc sulfate will split into ions. The reaction is given below:
$Zn+2{{H}^{+}}+SO_{4}^{2-}\to Z{{n}^{2+}}+SO_{4}^{2-}+{{H}_{2}}$

(b)- The two half cell reactions are oxidation half-reaction and reduction half-reaction. Oxidation half-reaction is when electrons are released and the ions are formed, so the oxidation half-reaction is given below:
$Zn\to Z{{n}^{2+}}+2{{e}^{-}}$
Reduction half-reaction is when the electrons are consumed for the ions to convert into the elemental state, the reaction is given below:
$2{{H}^{+}}+2{{e}^{-}}\to {{H}_{2}}$

Note: For writing the ionic reaction and the two half cell reactions, the main reaction must be balanced first otherwise the number of electrons in the reaction will be incorrect.