question_answer

Explain why zinc dissolves in dilute \[HCl\] to liberate \[{{H}_{2}}(g)\] but from concentrated \[{{H}_{2}}S{{O}_{4}},\] the gas evolved is\[S{{O}_{2}}\].

Answer:

                In dilute HCl, zinc reacts as follows: \[Zn+2{{H}^{+}}\,(aq)\xrightarrow{\,}\,Z{{n}^{2+}}(aq)+{{H}_{2}}(g)\] Therefore, \[{{H}_{2}}(g)\] is liberated. With concentrated\[{{H}_{2}}S{{O}_{4}},\] both \[{{H}^{+}}\] ions and \[SO_{4}^{2-}\] ions will be available. But \[SO_{4}^{2-}\] ion is a better oxidising agent than \[{{H}^{+}}\] ion. Therefore, the reaction will occur as follows: \[Zn(s)+SO_{4}^{2-}\,(aq)+4{{H}^{+}}(aq)\xrightarrow{\,}\]\[Z{{n}^{2+}}(aq)+S{{O}_{2}}(g)+2{{H}_{2}}O(aq)\]