A) Zn acts as an oxidising agent when it reacts with \[HN{{O}_{3}}\]
B) \[HN{{O}_{3}}\] is weaker acid than \[{{H}_{2}}S{{O}_{4}}\]and \[HCl\]
C) In electrochemical series, Zn is above hydrogen
D) \[NO_{3}^{-}\] is reduced in preference to hydronium ion
Correct Answer: D
Solution :
[d] Zinc gives \[{{H}_{2}}\], gas with dil \[{{H}_{2}}S{{O}_{4}}/HCl\] but not with \[HN{{O}_{3}}\]because in \[HN{{O}_{3}},N{{O}_{3}}^{-}\] ion is reduced and give \[N{{H}_{4}}N{{O}_{3}},{{N}_{2}}O,NO\]and \[N{{O}_{2}}\](based upon the concentration of\[HN{{O}_{3}}\]) \[[Zn+\underset{left( nearly\text{ }6% \right)}{\mathop{2HN{{O}_{3}}}}\,\xrightarrow{{}}Zn{{\left( N{{O}_{3}} \right)}_{2}}+2H]\times 4\] \[HN{{O}_{3}}+8H\xrightarrow{{}}N{{H}_{3}}+3{{H}_{2}}O\] \[N{{H}_{3}}+HN{{O}_{3}}\xrightarrow{{}}N{{H}_{4}}N{{O}_{3}}\] \[4Zn+10HN{{O}_{3}}\xrightarrow{{}}\] \[4Zn{{\left( N{{O}_{3}} \right)}_{2}}+N{{H}_{4}}N{{O}_{3}}+3{{H}_{2}}O\] Zn is on the top position of hydrogen in electrochemical series. So Zn displaces \[{{H}_{2}}\]from dilute \[{{H}_{2}}S{{O}_{4}}\]and \[HCl\]with liberation of \[{{H}_{2}}\]. \[Zn+{{H}_{2}}S{{O}_{4}}\to ZnS{{O}_{4}}+{{H}_{2}}\]
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