What are the oxidation number of nitrogen in the reactants and products of this reaction? NH4+(aq)+2H2O(l) \u2014> 4H+(aq)+NO2^-(aq)+6e^- Find NH4^+ and NO2^-Then answer\u2026..A voltaic cell with an alkaline aqueous electrolyte is based on the reaction between MnO4^-(aq) and Cd(s), producing MnO2(s) and Cd^2+(aq). Write half-reactions for the anode and cathode and then write a balanced cell reaction. Please include the states of matter in the equations.<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s break this into two parts:<\/p>\n\n\n\n Given the reaction: We are tasked with finding the oxidation numbers of nitrogen in both the reactants (NH\u2084\u207a) and products (NO\u2082\u207b).<\/p>\n\n\n\n So, the equation for the oxidation number of nitrogen in NH\u2084\u207a is:<\/p>\n\n\n\n [ [ [ Thus, the oxidation number of nitrogen in NH\u2084\u207a is -3<\/strong>.<\/p>\n\n\n\n We can set up the equation for nitrogen’s oxidation number:<\/p>\n\n\n\n [ [ [ [ Thus, the oxidation number of nitrogen in NO\u2082\u207b is +3<\/strong>.<\/p>\n\n\n\n In this voltaic cell, the reaction is between the permanganate ion (MnO\u2084\u207b) and solid cadmium (Cd), producing manganese dioxide (MnO\u2082) and cadmium ions (Cd\u00b2\u207a).<\/p>\n\n\n\n To balance the cell reaction, we need to balance the number of electrons involved in each half-reaction. The anode half-reaction involves 2 electrons, while the cathode half-reaction involves 5 electrons. To equalize the electrons, multiply the anode half-reaction by 5 and the cathode half-reaction by 2:<\/p>\n\n\n\n Now add both half-reactions:<\/p>\n\n\n\n [ This is the overall balanced cell reaction.<\/p>\n\n\n\n What are the oxidation number of nitrogen in the reactants and products of this reaction? NH4+(aq)+2H2O(l) \u2014> 4H+(aq)+NO2^-(aq)+6e^- Find NH4^+ and NO2^-Then answer\u2026..A voltaic cell with an alkaline aqueous electrolyte is based on the reaction between MnO4^-(aq) and Cd(s), producing MnO2(s) and Cd^2+(aq). Write half-reactions for the anode and cathode and then write a balanced […]<\/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-197513","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197513","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=197513"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/197513\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=197513"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=197513"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=197513"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}1. Oxidation Numbers in the Reaction:<\/strong><\/h3>\n\n\n\n
[ \\text{NH}_4^+(aq) + 2\\text{H}_2\\text{O}(l) \\rightarrow 4\\text{H}^+(aq) + \\text{NO}_2^-(aq) + 6e^- ]<\/p>\n\n\n\nIn NH\u2084\u207a (Ammonium ion):<\/strong><\/h4>\n\n\n\n
\n
\\text{Oxidation number of N} + 4(\\text{Oxidation number of H}) = +1
]<\/p>\n\n\n\n
\\text{Oxidation number of N} + 4(1) = +1
]<\/p>\n\n\n\n
\\text{Oxidation number of N} = +1 – 4 = -3
]<\/p>\n\n\n\nIn NO\u2082\u207b (Nitrite ion):<\/strong><\/h4>\n\n\n\n
\n
\\text{Oxidation number of N} + 2(\\text{Oxidation number of O}) = -1
]<\/p>\n\n\n\n
\\text{Oxidation number of N} + 2(-2) = -1
]<\/p>\n\n\n\n
\\text{Oxidation number of N} – 4 = -1
]<\/p>\n\n\n\n
\\text{Oxidation number of N} = -1 + 4 = +3
]<\/p>\n\n\n\nSummary:<\/strong><\/h4>\n\n\n\n
\n
\n\n\n\n2. Voltaic Cell Based on MnO\u2084\u207b and Cd(s):<\/strong><\/h3>\n\n\n\n
Half-Reactions:<\/strong><\/h4>\n\n\n\n
\n
The cadmium metal is oxidized at the anode. The half-reaction is:
[
\\text{Cd}(s) \\rightarrow \\text{Cd}^{2+}(aq) + 2e^-
] Cadmium (Cd) loses two electrons to form Cd\u00b2\u207a. The oxidation number of cadmium increases from 0 to +2.<\/li>\n\n\n\n
The permanganate ion (MnO\u2084\u207b) is reduced at the cathode. The half-reaction is:
[
\\text{MnO}_4^-(aq) + 8\\text{H}^+(aq) + 5e^- \\rightarrow \\text{MnO}_2(s) + 4\\text{H}_2\\text{O}(l)
] Here, the MnO\u2084\u207b ion gains electrons (reduction) to form MnO\u2082, where the oxidation state of manganese decreases from +7 to +4.<\/li>\n<\/ul>\n\n\n\nBalanced Cell Reaction:<\/strong><\/h4>\n\n\n\n
\n
[
5\\text{Cd}(s) \\rightarrow 5\\text{Cd}^{2+}(aq) + 10e^-
]<\/li>\n\n\n\n
[
2\\text{MnO}_4^-(aq) + 16\\text{H}^+(aq) + 10e^- \\rightarrow 2\\text{MnO}_2(s) + 8\\text{H}_2\\text{O}(l)
]<\/li>\n<\/ul>\n\n\n\n
5\\text{Cd}(s) + 2\\text{MnO}_4^-(aq) + 16\\text{H}^+(aq) \\rightarrow 5\\text{Cd}^{2+}(aq) + 2\\text{MnO}_2(s) + 8\\text{H}_2\\text{O}(l)
]<\/p>\n\n\n\n
\n\n\n\nExplanation:<\/strong><\/h3>\n\n\n\n
\n