`Zn` gives `H_(2)` gas with `H_(2)SO_(4)` and `HCl` but not with `HNO_(3)` becauseA. Zn act as oxidising agent when react with `HNO_(3)`B. `HNO_(3)` is weaker acid than `H_(2)SO_(4)` and HClC. In electrochemical series `Zn` is placed above hydrogenD. `NO_(3)^(-)` is reduced in preference to hydronium ion

`Zn` gives `H_(2)` gas with `H_(2)SO_(4)` and `HCl` but not with `HNO_

1.	`Zn` gives `H_(2)` gas with `H_(2)SO_(4)` and `HCl` but not with `HNO_(3)` becauseA. Zn act as oxidising agent when react with `HNO_(3)`B. `HNO_(3)` is weaker acid than `H_(2)SO_(4)` and HClC. In electrochemical series `Zn` is placed above hydrogenD. `NO_(3)^(-)` is reduced in preference to hydronium ion
Answer» Correct Answer - D Zn have lower value of `E_("cell")^(@)` and easily gives oxidation. Zn is present above `H_(2)` in electrochemical series. So, it liberates hydrogen gas from dilu. `HCl//H_(2)SO_(4)`. But `HNO_(3)` is an oxidising agent. The hydrogen obtained in this reaction is converted into `H_(2)O` . In `HNO_(3), NO_(3)^(-)` ion is reduced and give `NH_(4)NO_(3), N_(2)O, NO` and `NO_(2)` (based upon the concentration of `HNO_(2)`). `[Zn+underset(("Nearly"6%))(2HNO_(3)) rarr Zn(NO_(3))_(2)+2H]xx4` `({:(HNO_(3)+8HrarrNH_(3)+3H_(2)O),(NH_(3)+HNO_(3)rarrNH_(4)NO_(3)):})/(4Zn+10HNO_(3)rarr4Zn(NO_(3))_(2)+NH_(4)NO_(3)+3H_(2)O)`

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