{"id":196850,"date":"2025-03-06T06:09:55","date_gmt":"2025-03-06T06:09:55","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=196850"},"modified":"2025-03-06T06:09:58","modified_gmt":"2025-03-06T06:09:58","slug":"the-solution-of-pbn32comno43-coomno2pb3o4no","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/06\/the-solution-of-pbn32comno43-coomno2pb3o4no\/","title":{"rendered":"The solution of Pb(N3)2+Co(MnO4)3 -> CoO+MnO2+Pb3O4+NO"},"content":{"rendered":"\n

The solution of Pb(N3)2+Co(MnO4)3 -> CoO+MnO2+Pb3O4+NO it is from O.P.Tondon oxidation n reduction Q.29 (k)<\/p>\n\n\n\n

The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n

Let’s balance the redox reaction step by step and provide a detailed explanation.<\/p>\n\n\n\n

Given Unbalanced Reaction:<\/h3>\n\n\n\n

[
\\text{Pb(N}_3\\text{)}_2 + \\text{Co(MnO}_4\\text{)}_3 \\rightarrow \\text{CoO} + \\text{MnO}_2 + \\text{Pb}_3\\text{O}_4 + \\text{NO}
]<\/p>\n\n\n\n

Step 1: Identify Oxidation and Reduction Half-Reactions<\/strong><\/h3>\n\n\n\n

Oxidation Half-Reaction (Lead)<\/strong><\/h4>\n\n\n\n
    \n
  • Lead in Pb(N\u2083)\u2082:<\/strong> Oxidation state = +2<\/strong> (Azide, N\u2083\u207b, is -1 per azide ion).<\/li>\n\n\n\n
  • Lead in Pb\u2083O\u2084:<\/strong> The oxidation states are +2 and +4<\/strong>, as Pb\u2083O\u2084 consists of 2 Pb\u00b2\u207a and 1 Pb\u2074\u207a<\/strong>.<\/li>\n\n\n\n
  • So, Pb\u00b2\u207a is partially oxidized to Pb\u2074\u207a.<\/li>\n<\/ul>\n\n\n\n

    Reduction Half-Reaction (Manganese)<\/strong><\/h4>\n\n\n\n
      \n
    • Mn in Co(MnO\u2084)\u2083:<\/strong> Mn in MnO\u2084\u207b has an oxidation state of +7<\/strong>.<\/li>\n\n\n\n
    • Mn in MnO\u2082:<\/strong> Oxidation state of Mn = +4<\/strong>.<\/li>\n\n\n\n
    • So, Mn\u2077\u207a is reduced to Mn\u2074\u207a.<\/li>\n<\/ul>\n\n\n\n

      Nitrogen Oxidation<\/strong><\/h4>\n\n\n\n
        \n
      • N in N\u2083\u207b (Azide ion)<\/strong> has an oxidation state of -1\/3 per nitrogen<\/strong>.<\/li>\n\n\n\n
      • N in NO<\/strong> has an oxidation state of +2<\/strong>.<\/li>\n\n\n\n
      • So, nitrogen is oxidized from -1\/3 to +2.<\/li>\n<\/ul>\n\n\n\n

        Step 2: Balance Half-Reactions<\/strong><\/h3>\n\n\n\n

        Oxidation Half-Reaction (Lead and Nitrogen)<\/strong><\/h4>\n\n\n\n

        [
        3 \\text{Pb}^{2+} + 2 \\text{H}_2\\text{O} \\rightarrow \\text{Pb}_3\\text{O}_4 + 4 \\text{H}^+ + 2 e^-
        ]
        [
        \\text{N}_3^- \\rightarrow \\text{NO}
        ]<\/p>\n\n\n\n

        Reduction Half-Reaction (Manganese)<\/strong><\/h4>\n\n\n\n

        [
        2 \\text{MnO}_4^- + 2 \\text{H}_2O + 2 e^- \\rightarrow 2 \\text{MnO}_2 + 4 \\text{OH}^-
        ]<\/p>\n\n\n\n

        Step 3: Balance the Overall Reaction<\/strong><\/h3>\n\n\n\n

        The balanced equation is:<\/p>\n\n\n\n

        [
        3 \\text{Pb(N}_3\\text{)}_2 + 2 \\text{Co(MnO}_4\\text{)}_3 + 2 \\text{H}_2O \\rightarrow 2 \\text{CoO} + 6 \\text{MnO}_2 + \\text{Pb}_3\\text{O}_4 + 6 \\text{NO}
        ]<\/p>\n\n\n\n

        \n\n\n\n

        Explanation (300 Words)<\/strong><\/h3>\n\n\n\n

        This reaction involves both oxidation and reduction processes.<\/p>\n\n\n\n

          \n
        1. Lead Oxidation:<\/strong>
          Lead starts as Pb\u00b2\u207a in Pb(N\u2083)\u2082<\/strong> and is partially oxidized to Pb\u00b3O\u2084<\/strong>, a mixed oxide where some Pb atoms reach +4 oxidation state<\/strong>. The oxidation occurs as Pb\u00b2\u207a loses electrons.<\/li>\n\n\n\n
        2. Manganese Reduction:<\/strong>
          MnO\u2084\u207b (permanganate) initially has Mn in the +7 oxidation state<\/strong>, which is reduced to MnO\u2082 (Mn in +4 state)<\/strong>. This reduction occurs by gaining electrons.<\/li>\n\n\n\n
        3. Azide (N\u2083\u207b) Oxidation:<\/strong>
          The azide ion (N\u2083\u207b) consists of nitrogen in an average oxidation state of -1\/3<\/strong>. It gets oxidized to NO (+2 oxidation state<\/strong>), releasing electrons in the process.<\/li>\n\n\n\n
        4. Cobalt Participation:<\/strong>
          Cobalt in Co(MnO\u2084)\u2083<\/strong> is already in a stable oxidation state, forming CoO<\/strong> as a byproduct.<\/li>\n\n\n\n
        5. Water Participation:<\/strong>
          Water molecules help balance the reaction by supplying oxygen to form Pb\u2083O\u2084 and MnO\u2082, ensuring mass and charge balance.<\/li>\n<\/ol>\n\n\n\n

          Thus, the reaction follows redox principles, where Mn\u2077\u207a and N\u2083\u207b undergo changes in oxidation states, leading to the final balanced equation.<\/p>\n\n\n\n

          \n\n\n\n
          \"\"<\/figure>\n\n\n\n

          I have provided an image illustrating the oxidation-reduction reaction, including oxidation states, electron transfers, and molecule involvement. <\/p>\n","protected":false},"excerpt":{"rendered":"

          The solution of Pb(N3)2+Co(MnO4)3 -> CoO+MnO2+Pb3O4+NO it is from O.P.Tondon oxidation n reduction Q.29 (k) The correct answer and explanation is : Let’s balance the redox reaction step by step and provide a detailed explanation. Given Unbalanced Reaction: [\\text{Pb(N}_3\\text{)}_2 + \\text{Co(MnO}_4\\text{)}_3 \\rightarrow \\text{CoO} + \\text{MnO}_2 + \\text{Pb}_3\\text{O}_4 + \\text{NO}] Step 1: Identify Oxidation and Reduction […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-196850","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196850","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=196850"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/196850\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=196850"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=196850"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=196850"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}