For the following electrochemical cell Sn(s)|Sn2 (aq, 0.0155 M)||Ag (aq, 3.50 M)|Ag(s) write the net cell equation. Phases are optional. Do not include the concentrations. Calculate the following values at 25.0 \u00b0C using standard potentials as needed<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n For the Sn<\/strong> electrode (oxidation half-reaction):<\/p>\n\n\n\n [ For the Ag<\/strong> electrode (reduction half-reaction):<\/p>\n\n\n\n [ From standard tables:<\/p>\n\n\n\n Since oxidation occurs at the anode and reduction at the cathode, the half-reaction at the anode involves the oxidation of Sn, and at the cathode, Ag\u207a is reduced to Ag.<\/p>\n\n\n\n The standard cell potential is calculated using:<\/p>\n\n\n\n [ Substituting values:<\/p>\n\n\n\n [ Thus, the standard cell potential is ( 0.94 \\, \\text{V} ).<\/p>\n\n\n\n Since the concentrations of ions in the cell are non-standard, we need to use the Nernst equation<\/strong> to calculate the actual cell potential at the given conditions:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n [ From the problem:<\/p>\n\n\n\n Thus:<\/p>\n\n\n\n [ Now, we can calculate ( E ):<\/p>\n\n\n\n [ [ [ [ So, the actual cell potential is ( E = 1.02 \\, \\text{V} ) at 25\u00b0C.<\/p>\n\n\n\n The overall cell reaction is the sum of the half-reactions:<\/p>\n\n\n\n To balance the number of electrons, multiply the reduction half-reaction by 2:<\/p>\n\n\n\n [ Thus, the net reaction is:<\/p>\n\n\n\n [ The overall cell reaction for the given electrochemical cell is:<\/p>\n\n\n\n [ The standard cell potential is ( 0.94 \\, \\text{V} ), and the actual cell potential, calculated using the Nernst equation, is ( 1.02 \\, \\text{V} ) at 25\u00b0C.<\/p>\n","protected":false},"excerpt":{"rendered":" For the following electrochemical cell Sn(s)|Sn2 (aq, 0.0155 M)||Ag (aq, 3.50 M)|Ag(s) write the net cell equation. Phases are optional. Do not include the concentrations. Calculate the following values at 25.0 \u00b0C using standard potentials as needed The Correct Answer and Explanation is : Electrochemical Cell: Sn(s) | Sn\u00b2\u207a(aq) || Ag\u207a(aq) | Ag(s) Step 1: […]<\/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-187068","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187068","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=187068"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/187068\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=187068"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=187068"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=187068"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}Electrochemical Cell: Sn(s) | Sn\u00b2\u207a(aq) || Ag\u207a(aq) | Ag(s)<\/h3>\n\n\n\n
Step 1: Write Half-Cell Reactions<\/h4>\n\n\n\n
\\text{Sn(s)} \\rightarrow \\text{Sn}^{2+}(aq) + 2e^-
]<\/p>\n\n\n\n
\\text{Ag}^+(aq) + e^- \\rightarrow \\text{Ag(s)}
]<\/p>\n\n\n\nStep 2: Determine the Standard Reduction Potentials<\/h4>\n\n\n\n
\n
Step 3: Calculate the Cell Potential (( E^\\circ_{\\text{cell}} ))<\/h4>\n\n\n\n
E^\\circ_{\\text{cell}} = E^\\circ_{\\text{cathode}} – E^\\circ_{\\text{anode}}
]<\/p>\n\n\n\n
E^\\circ_{\\text{cell}} = 0.80 \\, \\text{V} – (-0.14 \\, \\text{V}) = 0.80 \\, \\text{V} + 0.14 \\, \\text{V} = 0.94 \\, \\text{V}
]<\/p>\n\n\n\nStep 4: Calculate the Nernst Equation<\/h4>\n\n\n\n
E = E^\\circ_{\\text{cell}} – \\frac{0.0592}{n} \\log Q
]<\/p>\n\n\n\n\n
Q = \\frac{[\\text{Sn}^{2+}]{\\text{cathode}}}{[\\text{Ag}^+]<\/em>{\\text{anode}}}
]<\/p>\n\n\n\n\n
Q = \\frac{0.0155}{3.50} = 0.00443
]<\/p>\n\n\n\n
E = 0.94 \\, \\text{V} – \\frac{0.0592}{2} \\log(0.00443)
]<\/p>\n\n\n\n
E = 0.94 \\, \\text{V} – \\frac{0.0592}{2} (-2.653)
]<\/p>\n\n\n\n
E = 0.94 \\, \\text{V} + 0.0786 \\, \\text{V}
]<\/p>\n\n\n\n
E = 1.0186 \\, \\text{V}
]<\/p>\n\n\n\nStep 5: Write the Net Cell Reaction<\/h4>\n\n\n\n
\n
\\text{2Ag}^+(aq) + 2e^- \\rightarrow \\text{2Ag(s)}
]<\/p>\n\n\n\n
\\text{Sn(s)} + 2\\text{Ag}^+(aq) \\rightarrow \\text{Sn}^{2+}(aq) + 2\\text{Ag(s)}
]<\/p>\n\n\n\nConclusion<\/h3>\n\n\n\n
\\text{Sn(s)} + 2\\text{Ag}^+(aq) \\rightarrow \\text{Sn}^{2+}(aq) + 2\\text{Ag(s)}
]<\/p>\n\n\n\n