question_answer

Zinc reacts with dilute \[{{H}_{2}}S{{O}_{4}}\] to give \[{{H}_{2}}\]. It also reacts with cone. \[{{H}_{2}}S{{O}_{4}}\]to give \[S{{O}_{2}}\]. How will you account for this difference?

Answer:

                Dilute \[{{H}_{2}}S{{O}_{4}}\] behaves as a proton acid and zinc metal displaces hydrogen from the acid. \[Zn+{{H}_{2}}S{{O}_{4}}\text{(dil)}\xrightarrow{}\,ZnS{{O}_{4}}\,+{{H}_{2}}\] Concentrated \[{{H}_{2}}S{{O}_{4}}\] behaves as a strong oxidising agent and itself is reduced to \[S{{O}_{2}}\]. \[Zn+2{{H}_{2}}S{{O}_{4}}\text{(conc}\text{.)}\xrightarrow{\,}\,ZnS{{O}_{4}}+S{{O}_{2}}+2{{H}_{2}}O\]