{"id":245328,"date":"2025-07-06T03:52:07","date_gmt":"2025-07-06T03:52:07","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=245328"},"modified":"2025-07-06T03:52:10","modified_gmt":"2025-07-06T03:52:10","slug":"name-name-lecture-pair-worksheet","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/06\/name-name-lecture-pair-worksheet\/","title":{"rendered":"Name Name Lecture Pair Worksheet"},"content":{"rendered":"\n<p>Name Name Lecture Pair Worksheet: Oxidation States Determine the oxidation number for nitrogen in each of the following molecules or ions. 1. N2 2. HNO3 3. NO2 4. N2O 5. NH3 For the following redox reaction, assign oxidation numbers to all elements and identify the substance reduced (oxidizing agent) and the substance oxidized (reducing agent). Write the oxidation half reaction and the reduction half reaction. Fe<br>O<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>3 CO<br>2 Fe + 3 CO<\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-215.png\" alt=\"\" class=\"wp-image-245329\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Of course. Here are the correct answers and a detailed explanation for the chemistry worksheet on oxidation states.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part 1: Oxidation Numbers for Nitrogen<\/strong><\/h3>\n\n\n\n<p>To determine the oxidation number, we use a set of rules. The key rules for this section are:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The oxidation number of an element in its elemental form (like N\u2082) is 0.<\/li>\n\n\n\n<li>The oxidation number of oxygen in a compound is usually -2.<\/li>\n\n\n\n<li>The oxidation number of hydrogen when bonded to a nonmetal is usually +1.<\/li>\n\n\n\n<li>The sum of all oxidation numbers in a neutral molecule is 0.<\/li>\n\n\n\n<li>The sum of all oxidation numbers in a polyatomic ion equals the charge of the ion.<\/li>\n<\/ul>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>N\u2082<\/strong>: Nitrogen is in its elemental form, so its oxidation number is\u00a0<strong>0<\/strong>.<\/li>\n\n\n\n<li><strong>HNO\u2083<\/strong>: This is a neutral molecule. Hydrogen (H) is +1, and each Oxygen (O) is -2. Let the oxidation number of Nitrogen (N) be &#8216;x&#8217;.<br>(+1) + (x) + 3(-2) = 0<br>1 + x &#8211; 6 = 0<br>x = +5. The oxidation number of nitrogen is\u00a0<strong>+5<\/strong>.<\/li>\n\n\n\n<li><strong>NO\u2082\u207b<\/strong>: This is an ion with a charge of -1. Each Oxygen (O) is -2. Let the oxidation number of Nitrogen (N) be &#8216;x&#8217;.<br>(x) + 2(-2) = -1<br>x &#8211; 4 = -1<br>x = +3. The oxidation number of nitrogen is\u00a0<strong>+3<\/strong>.<\/li>\n\n\n\n<li><strong>N\u2082O<\/strong>: This is a neutral molecule. Oxygen (O) is -2. There are two nitrogen atoms, so we&#8217;ll use &#8216;2x&#8217;.<br>2(x) + (-2) = 0<br>2x = 2<br>x = +1. The oxidation number of nitrogen is\u00a0<strong>+1<\/strong>.<\/li>\n\n\n\n<li><strong>NH\u2083<\/strong>: This is a neutral molecule. Each Hydrogen (H) is +1. Let the oxidation number of Nitrogen (N) be &#8216;x&#8217;.<br>(x) + 3(+1) = 0<br>x + 3 = 0<br>x = -3. The oxidation number of nitrogen is\u00a0<strong>-3<\/strong>.<\/li>\n<\/ol>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Part 2: Redox Reaction Analysis<\/strong><\/h3>\n\n\n\n<p><strong>Reaction:<\/strong>&nbsp;Fe\u2082O\u2083 + 3CO \u2192 2Fe + 3CO\u2082<\/p>\n\n\n\n<p><strong>1. Assign Oxidation Numbers:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In\u00a0<strong>Fe\u2082O\u2083<\/strong>: Oxygen is -2. To balance, the two Fe atoms must total +6, so each\u00a0<strong>Fe is +3<\/strong>.<\/li>\n\n\n\n<li>In\u00a0<strong>CO<\/strong>: Oxygen is -2, so\u00a0<strong>C is +2<\/strong>.<\/li>\n\n\n\n<li>In\u00a0<strong>Fe<\/strong>: It is an element in its pure form, so\u00a0<strong>Fe is 0<\/strong>.<\/li>\n\n\n\n<li>In\u00a0<strong>CO\u2082<\/strong>: Oxygen is -2. To balance,\u00a0<strong>C is +4<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p><strong>2. Identify Reduction and Oxidation:<\/strong><br>To remember the difference, use the mnemonic LEO the lion says GER:&nbsp;<strong>L<\/strong>oss of&nbsp;<strong>E<\/strong>lectrons is&nbsp;<strong>O<\/strong>xidation,&nbsp;<strong>G<\/strong>ain of&nbsp;<strong>E<\/strong>lectrons is&nbsp;<strong>R<\/strong>eduction.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Reduction:<\/strong>\u00a0Iron (Fe) goes from an oxidation state of +3 in Fe\u2082O\u2083 to 0 in Fe. Its oxidation number decreased, meaning it gained electrons.<\/li>\n\n\n\n<li><strong>Oxidation:<\/strong>\u00a0Carbon (C) goes from an oxidation state of +2 in CO to +4 in CO\u2082. Its oxidation number increased, meaning it lost electrons.<\/li>\n<\/ul>\n\n\n\n<p><strong>3. Identify Agents:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Substance Reduced (Oxidizing Agent):<\/strong>\u00a0The species that gets reduced (gains electrons) is the oxidizing agent because it causes the other substance to be oxidized. Here,\u00a0<strong>Fe\u2082O\u2083<\/strong>\u00a0is the oxidizing agent.<\/li>\n\n\n\n<li><strong>Substance Oxidized (Reducing Agent):<\/strong>\u00a0The species that gets oxidized (loses electrons) is the reducing agent because it causes the other substance to be reduced. Here,\u00a0<strong>CO<\/strong>\u00a0is the reducing agent.<\/li>\n<\/ul>\n\n\n\n<p><strong>4. Write Half-Reactions:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Oxidation half-reaction (shows electron loss):<\/strong><br>C\u207a\u00b2 \u2192 C\u207a\u2074 + 2e\u207b<br>For the whole molecule:\u00a0<strong>CO \u2192 CO\u2082 + 2e\u207b<\/strong><\/li>\n\n\n\n<li><strong>Reduction half-reaction (shows electron gain):<\/strong><br>Fe\u00b3\u207a + 3e\u207b \u2192 Fe\u2070<br>For the whole molecule:\u00a0<strong>Fe\u2082O\u2083 + 6e\u207b \u2192 2Fe<\/strong><\/li>\n<\/ul>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-630.jpeg\" alt=\"\" class=\"wp-image-245330\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Name Name Lecture Pair Worksheet: Oxidation States Determine the oxidation number for nitrogen in each of the following molecules or ions. 1. N2 2. HNO3 3. NO2 4. N2O 5. NH3 For the following redox reaction, assign oxidation numbers to all elements and identify the substance reduced (oxidizing agent) and the substance oxidized (reducing agent). [&hellip;]<\/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-245328","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245328","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=245328"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/245328\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=245328"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=245328"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=245328"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}