11th Class Chemistry Redox Reactions

  • question_answer 7) Why do the following reactions proceed differently? \[P{{b}_{3}}{{O}_{4}}+8HCl\to 3PbCl+C{{l}_{2}}+4{{H}_{2}}O\]and \[P{{b}_{3}}{{O}_{4}}+4HN{{O}_{3}}\to 2Pb{{(N{{O}_{3}})}_{2}}+Pb{{O}_{2}}+2{{H}_{2}}O.\]

    Answer:

    \[P{{b}_{3}}{{O}_{4}}\] is a mixture of \[PbO\] and \[Pb{{O}_{2}}\]in 2 : 1 ratio. \[PbO\]has stable oxidation state of +2 while in \[Pb{{O}_{2}}\], its oxidation state is +4. Thus, \[Pb{{O}_{2}}\] acts as an oxidising agent and oxidises \[HCl\]into\[C{{l}_{2}}\]. \[PbO\]on the other hand is a basic oxide which forms a salt \[(PbC{{l}_{2}})\] with \[HCl\] The reaction, \[P{{b}_{3}}{{O}_{4}}+8HCl\to 3PbC{{l}_{2}}+C{{l}_{2}}+4{{H}_{2}}O\] may be divided into two parts. \[2PbO+4HCl\to 2PbC{{l}_{2}}+2{{H}_{2}}O\](Acid base reaction) \[Pb{{O}_{2}}+4HCl\to PbC{{l}_{2}}+C{{l}_{2}}+2{{H}_{2}}O\](Redox reaction) In the second reaction, \[\text{HN}{{\text{O}}_{\text{3}}}\]acts as oxidising agent only once the reaction may not occur between \[Pb{{O}_{2}}\] and \[HN{{O}_{3}}\]. \[PbO\]being a basic oxide forms salt with\[HN{{O}_{3}}\]. \[2PbO+4HN{{O}_{3}}\to 2Pb(N{{O}_{3}})+2{{H}_{2}}O\] (\[Pb{{O}_{2}}\]remain unaffected.)


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