A) Zn act as oxidising agent when react with \[HN{{O}_{3}}\]
B) \[HN{{O}_{3}}\] is weaker acid then \[{{H}_{2}}S{{O}_{4}}\] and HCI
C) In electrochemical series Zn is above hydrogen
D) \[NO_{3}^{-}\] is reduced in preference to hydronium ion
Correct Answer: D
Solution :
Zn is present above \[{{H}_{2}}\] in electrochemical series. So, it liberates hydrogen gas from dilute \[HCl/{{H}_{2}}S{{O}_{4}}\]. But \[HN{{O}_{3}}\] is oxidising agent. The hydrogen obtained in this reaction is converted into \[{{H}_{2}}O\] in \[HN{{O}_{3}},\] \[NO_{3}^{-}\] ion is reduced and give \[N{{H}_{4}}N{{O}_{3}},\,{{N}_{2}}O,\,NO\] and \[N{{O}_{2}}\] (based upon the concentration of \[HN{{O}_{3}}\]) \[[Zn+\underset{\text{(nearly 6 }\!\!%\!\!\text{ )}}{\mathop{2HN{{O}_{3}}}}\,\xrightarrow{{}}Zn{{(N{{O}_{3}})}_{2}}+2H]\times 4\] \[HN{{O}_{3}}+8H\xrightarrow{{}}N{{H}_{3}}+3{{H}_{2}}O\] \[\frac{N{{H}_{3}}+HN{{O}_{3}}\xrightarrow{{}}N{{H}_{4}}+3{{H}_{2}}O}{4Zn+10HN{{O}_{3}}\xrightarrow{{}}4Zn{{(N{{O}_{3}})}_{2}}+N{{H}_{4}}N{{O}_{3}}+3{{H}_{2}}O}\]
You need to login to perform this action.
You will be redirected in
3 sec
