Different Oxidation States of Iron Comparison Standards Solution FeSO4) FeCl(oa) Iron ion present (+2\/+3) Interpret how the KSCN allows you to better tell which iron ion you have. Reaction 1 Balance each half-reaction including electrons, then balance the overall net ionic equation. Give the oxidation state for each atom in the equation. Ox: Fe? Red: NO, Net: What ion\/molecule contains the element that gets oxidized? What is the oxidizing agent in the above reaction?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The oxidation reaction involves iron (Fe\u00b2\u207a) being oxidized to Fe\u00b3\u207a by losing one electron.<\/p>\n\n\n\n [ In the reduction half-reaction, nitrate ion (NO\u2083\u207b) is reduced to nitrogen monoxide (NO) by gaining electrons. To balance the equation, we need to account for the hydrogen ions (H\u207a) that balance the overall charges and atoms.<\/p>\n\n\n\n [ The overall reaction involves combining the two half-reactions, ensuring that electrons are balanced. The number of electrons lost in the oxidation half-reaction is 1, and in the reduction half-reaction is 3. To balance this, we multiply the oxidation reaction by 3 and the reduction reaction by 1.<\/p>\n\n\n\n [ KSCN (Potassium thiocyanate) helps in determining the iron oxidation state because thiocyanate ions react with Fe\u00b3\u207a ions to form a reddish-brown complex, while Fe\u00b2\u207a does not form this complex. This visual test can distinguish between Fe\u00b2\u207a and Fe\u00b3\u207a ions in a solution. The formation of the red complex indicates the presence of Fe\u00b3\u207a, whereas the absence of the color change suggests Fe\u00b2\u207a. Therefore, KSCN serves as a qualitative indicator of the iron oxidation state in a given solution.<\/p>\n","protected":false},"excerpt":{"rendered":" Different Oxidation States of Iron Comparison Standards Solution FeSO4) FeCl(oa) Iron ion present (+2\/+3) Interpret how the KSCN allows you to better tell which iron ion you have. Reaction 1 Balance each half-reaction including electrons, then balance the overall net ionic equation. Give the oxidation state for each atom in the equation. Ox: Fe? Red: […]<\/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-186266","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186266","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=186266"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186266\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186266"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186266"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186266"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Balanced Half-Reactions and Overall Net Ionic Equation<\/h3>\n\n\n\n
Oxidation Half-Reaction (Fe\u00b2\u207a to Fe\u00b3\u207a):<\/h4>\n\n\n\n
\\text{Fe}^{2+} \\rightarrow \\text{Fe}^{3+} + e^-
]<\/p>\n\n\n\nReduction Half-Reaction (NO\u2083\u207b to NO):<\/h4>\n\n\n\n
\\text{NO}_3^- + 4\\text{H}^+ + 3e^- \\rightarrow \\text{NO} + 2\\text{H}_2\\text{O}
]<\/p>\n\n\n\nOverall Net Ionic Equation:<\/h4>\n\n\n\n
3 \\text{Fe}^{2+} + \\text{NO}_3^- + 12\\text{H}^+ \\rightarrow 3\\text{Fe}^{3+} + \\text{NO} + 6\\text{H}_2\\text{O}
]<\/p>\n\n\n\nOxidation States and Key Points:<\/h3>\n\n\n\n
\n
Role of KSCN:<\/h3>\n\n\n\n