What quantity of Cu, in moles, can these electrons produce?<\/p>\n\n\n\n
Cu deposited: 0.3952 g<\/p>\n\n\n\n
6.219 \u00d7 10-3 mol Cu<\/p>\n\n\n\n
Charge passed: 1200 C<\/p>\n\n\n\n
1.2 \u00d7 10-2 mol e- passed<\/p>\n\n\n\n
The correct answer and explanation is :<\/strong><\/mark><\/p>\n\n\n\n To determine the quantity of Cu in moles that can be produced from the charge passed, we need to use the relationship between the amount of charge and the number of moles of electrons involved in the process. This relationship is governed by the Faraday constant and stoichiometry of the reaction.<\/p>\n\n\n\n We can use this information to calculate how many moles of copper (Cu) can be deposited based on the charge passed and the moles of electrons involved.<\/p>\n\n\n\n In the electrochemical reaction where copper is deposited, the half-reaction is:<\/p>\n\n\n\n [ This indicates that for every mole of Cu deposited, 2 moles of electrons are required.<\/p>\n\n\n\n From the problem, we are told that 1.2 \u00d7 10^-2 mol of electrons were passed. Since 2 moles of electrons are required to deposit 1 mole of Cu, we can set up the following proportion:<\/p>\n\n\n\n [ Using the moles of electrons passed (1.2 \u00d7 10^-2 mol e^-), we can calculate the moles of Cu:<\/p>\n\n\n\n [ Thus, the quantity of Cu that can be produced from the 1.2 \u00d7 10^-2 mol of electrons is 6.0 \u00d7 10^-3 mol of Cu.<\/p>\n\n\n\n We are given that 6.219 \u00d7 10^-3 mol of Cu were deposited for a certain amount of charge, and 1.2 \u00d7 10^-2 mol of electrons were passed. Our calculation aligns closely with this, confirming that the deposition of 6.0 \u00d7 10^-3 mol of Cu is correct, given the moles of electrons passed.<\/p>\n\n\n\n Hence, 6.0 \u00d7 10^-3 mol of Cu<\/strong> is the amount that can be produced from the charge passed.<\/p>\n","protected":false},"excerpt":{"rendered":" What quantity of Cu, in moles, can these electrons produce? Cu deposited: 0.3952 g 6.219 \u00d7 10-3 mol Cu Charge passed: 1200 C 1.2 \u00d7 10-2 mol e- passed The correct answer and explanation is : To determine the quantity of Cu in moles that can be produced from the charge passed, we need to […]<\/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-210721","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210721","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=210721"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/210721\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=210721"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=210721"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=210721"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Step 1: Identify the known values<\/h3>\n\n\n\n
\n
Step 2: Relationship between moles of Cu and moles of electrons<\/h3>\n\n\n\n
\\text{Cu}^{2+} + 2e^- \\rightarrow \\text{Cu (s)}
]<\/p>\n\n\n\nStep 3: Calculate the moles of Cu that can be produced<\/h3>\n\n\n\n
\\frac{\\text{moles of Cu}}{\\text{moles of electrons}} = \\frac{1 \\text{ mol Cu}}{2 \\text{ mol e}^-}
]<\/p>\n\n\n\n
\\text{moles of Cu} = \\frac{1.2 \\times 10^{-2} \\, \\text{mol e}^-}{2} = 6.0 \\times 10^{-3} \\, \\text{mol Cu}
]<\/p>\n\n\n\nStep 4: Confirm the result with the given data<\/h3>\n\n\n\n