{"id":202616,"date":"2025-03-17T17:54:11","date_gmt":"2025-03-17T17:54:11","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=202616"},"modified":"2025-03-17T17:54:14","modified_gmt":"2025-03-17T17:54:14","slug":"complete-the-statements-about-the-redox-reaction-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/03\/17\/complete-the-statements-about-the-redox-reaction-below\/","title":{"rendered":"Complete the statements about the redox reaction below"},"content":{"rendered":"\n<p>Complete the statements about the redox reaction below. Al*(aq) + 3Ag(s) + 3Cl(aq) + Al(s) + 3AgCl(s) Half-reactions: AP+(aq) + 3e&#8221; &#8211; Al(s) E\u00b0 = -1.66 V AgCl(s) + e &#8211; Ag(s) + Cl(aq) E = +0.22 V AgCl(s) 0 is being oxidized. 2 is being reduced electrolytic is acting as an reducing agent. Al(s) Ag(s) is acting as an oxidizing agent drag) voltaic AP(aq) moles of electrons are transferred. 3 The reaction is an) cell<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Completion of the Statements:<\/strong><\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Al(s)<\/strong> is being oxidized.<\/li>\n\n\n\n<li><strong>AgCl(s)<\/strong> is being reduced.<\/li>\n\n\n\n<li><strong>Al(s)<\/strong> is acting as a reducing agent.<\/li>\n\n\n\n<li><strong>Ag\u207a(aq)<\/strong> is acting as an oxidizing agent.<\/li>\n\n\n\n<li><strong>3<\/strong> moles of electrons are transferred.<\/li>\n\n\n\n<li>The reaction is a <strong>voltaic<\/strong> cell.<\/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>Explanation (300 Words)<\/strong><\/h3>\n\n\n\n<p>A redox reaction involves the transfer of electrons between two species, where one undergoes oxidation (losing electrons) and the other undergoes reduction (gaining electrons).<\/p>\n\n\n\n<p>In this reaction, <strong>Al(s)<\/strong> (solid aluminum) is oxidized to <strong>Al\u00b3\u207a(aq)<\/strong> by losing three electrons: Al(s)\u2192Al\u00b3\u207a(aq)+3e\u2212\\text{Al(s)} \\rightarrow \\text{Al\u00b3\u207a(aq)} + 3e\u207b<\/p>\n\n\n\n<p>The standard reduction potential of aluminum is <strong>-1.66 V<\/strong>, which means it has a strong tendency to lose electrons and act as a reducing agent.<\/p>\n\n\n\n<p>Meanwhile, silver chloride (<strong>AgCl(s)<\/strong>) undergoes reduction, as silver ions <strong>(Ag\u207a(aq))<\/strong> gain electrons to form solid silver: AgCl(s)+e\u2212\u2192Ag(s)+Cl\u207b(aq)\\text{AgCl(s)} + e\u207b \\rightarrow \\text{Ag(s)} + \\text{Cl\u207b(aq)}<\/p>\n\n\n\n<p>The reduction potential for this half-reaction is <strong>+0.22 V<\/strong>, indicating that <strong>Ag\u207a<\/strong> prefers to gain electrons and be reduced. Since <strong>Ag\u207a(aq)<\/strong> is gaining electrons, it is the oxidizing agent.<\/p>\n\n\n\n<p>The number of electrons exchanged per aluminum atom is <strong>3<\/strong>, meaning that <strong>3 moles of electrons<\/strong> are transferred in this reaction.<\/p>\n\n\n\n<p>Since the overall cell potential is <strong>positive<\/strong> (calculated as <strong>E\u00b0 = 0.22 V &#8211; (-1.66 V) = +1.88 V<\/strong>), the reaction is spontaneous and can function as a <strong>voltaic cell<\/strong> (not electrolytic). A voltaic cell generates electrical energy from a spontaneous chemical reaction.<\/p>\n\n\n\n<p>In summary:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Aluminum<\/strong> is oxidized, acting as the reducing agent.<\/li>\n\n\n\n<li><strong>Silver ions<\/strong> are reduced, acting as the oxidizing agent.<\/li>\n\n\n\n<li>The reaction is <strong>spontaneous<\/strong> and occurs in a voltaic cell.<\/li>\n\n\n\n<li><\/li>\n<\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Complete the statements about the redox reaction below. Al*(aq) + 3Ag(s) + 3Cl(aq) + Al(s) + 3AgCl(s) Half-reactions: AP+(aq) + 3e&#8221; &#8211; Al(s) E\u00b0 = -1.66 V AgCl(s) + e &#8211; Ag(s) + Cl(aq) E = +0.22 V AgCl(s) 0 is being oxidized. 2 is being reduced electrolytic is acting as an reducing agent. Al(s) [&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-202616","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202616","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=202616"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/202616\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=202616"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=202616"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=202616"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}