20) Identify the location of reduction in an electrochemical cell. A) the socket B) the salt bridge C) the anode D) the electrode E) the cathode 21) The value of AG\u00b0 at 141.0\u00b0C for the formation of phosphorous trichloride from its constituent elements, P2 (g) + 3Cl2 (g) 2PC13 (g) is kJ\/mol. At 25.0\u00b0C for this reaction, AH is -720.5 kJ\/mol, AG\u00b0 is -642.9 kJ\/mol, and AS is -263.7 J\/K. A) -829.7 D) 1.08 \u00d7 105 B) 3.65 x 104 22) Phosphorous and chlorine gases combine to produce phosphorous trichloride: P2 (g) + 3Cl2 (g) 2PC13 (g) B) -3.88 \u00d7 103 X C) -7.28 \u00d7 103 D) -708.4 E) -44.2 C) -612.3 ?>> AGO at 298 K for this reaction is -642.9 kJ\/mol. The value of AG at 298 K for a reaction mixture that consists of 1.5 atm P2, 1.6 atm Cl2, and 0.65 atm PCl3 is A) -649.5 E) -683.3<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s go through each of the questions systematically:<\/p>\n\n\n\n The reduction process in an electrochemical cell occurs at the cathode<\/strong>. The cathode is where electrons are gained by the species involved in the reduction reaction. Therefore, the correct answer is:<\/p>\n\n\n\n E) the cathode<\/strong><\/p>\n\n\n\n In electrochemical cells, oxidation happens at the anode (where electrons are lost), and reduction occurs at the cathode (where electrons are gained). This is a fundamental concept of redox reactions.<\/p>\n\n\n\n We are given the following data:<\/p>\n\n\n\n We need to calculate the standard Gibbs free energy (\u0394G\u00b0) at 141.0\u00b0C (which is 414.15 K).<\/p>\n\n\n\n The formula to calculate \u0394G\u00b0 is:<\/p>\n\n\n\n [ First, convert \u0394S\u00b0 into kJ\/K (because \u0394H\u00b0 is in kJ\/mol):<\/p>\n\n\n\n [ Now, substitute the values into the formula:<\/p>\n\n\n\n [ Calculating this:<\/p>\n\n\n\n [ However, the closest value is not directly calculated in this way, and typically, we’d need to apply more specific temperature dependencies or data. But based on typical answer choices, the closest approximation can be:<\/p>\n\n\n\n C) -612.3<\/strong><\/p>\n\n\n\n To calculate \u0394G under non-standard conditions, we use the formula:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n [ Substitute the given values:<\/p>\n\n\n\n [ First, calculate Q:<\/p>\n\n\n\n [ Now, use this to calculate \u0394G:<\/p>\n\n\n\n [ [ Since ( \\ln 0.0688 \\approx -2.673 ):<\/p>\n\n\n\n [ So, the correct answer is:<\/p>\n\n\n\n A) -649.5<\/strong><\/p>\n\n\n\n 20) Identify the location of reduction in an electrochemical cell. A) the socket B) the salt bridge C) the anode D) the electrode E) the cathode 21) The value of AG\u00b0 at 141.0\u00b0C for the formation of phosphorous trichloride from its constituent elements, P2 (g) + 3Cl2 (g) 2PC13 (g) is kJ\/mol. At 25.0\u00b0C for […]<\/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-188790","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188790","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=188790"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/188790\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=188790"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=188790"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=188790"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}20) Identify the location of reduction in an electrochemical cell.<\/h3>\n\n\n\n
Explanation:<\/h4>\n\n\n\n
21) The value of \u0394G\u00b0 at 141.0\u00b0C for the formation of phosphorus trichloride from its constituent elements, P\u2082 (g) + 3Cl\u2082 (g) \u2192 2PCl\u2083 (g) is kJ\/mol.<\/h3>\n\n\n\n
\n
\\Delta G^\\circ = \\Delta H^\\circ – T \\Delta S^\\circ
]<\/p>\n\n\n\n
\\Delta S^\\circ = -263.7 \\, \\text{J\/K} = -0.2637 \\, \\text{kJ\/K}
]<\/p>\n\n\n\n
\\Delta G^\\circ = -720.5 – (414.15 \\times -0.2637)
]<\/p>\n\n\n\n
\\Delta G^\\circ = -720.5 + 109.4 = -611.1 \\, \\text{kJ\/mol}
]<\/p>\n\n\n\n22) Given the reaction: P\u2082 (g) + 3Cl\u2082 (g) \u2192 2PCl\u2083 (g), and we know that at 298 K, \u0394G\u00b0 = -642.9 kJ\/mol, we need to calculate \u0394G for a mixture with partial pressures:<\/h3>\n\n\n\n
\n
\\Delta G = \\Delta G^\\circ + RT \\ln Q
]<\/p>\n\n\n\n\n
Q = \\frac{{[PCl_3]^2}}{{[P_2] \\cdot [Cl_2]^3}}
]<\/p>\n\n\n\n
Q = \\frac{{(0.65)^2}}{{(1.5) \\cdot (1.6)^3}}
]<\/p>\n\n\n\n
Q = \\frac{{0.4225}}{{1.5 \\cdot 4.096}} = \\frac{{0.4225}}{{6.144}} = 0.0688
]<\/p>\n\n\n\n
\\Delta G = -642.9 + (8.314 \\times 10^{-3} \\times 298 \\times \\ln 0.0688)
]<\/p>\n\n\n\n
\\Delta G = -642.9 + (2.478 \\times \\ln 0.0688)
]<\/p>\n\n\n\n
\\Delta G = -642.9 + (2.478 \\times -2.673) = -642.9 – 6.62 = -649.5 \\, \\text{kJ\/mol}
]<\/p>\n\n\n\nFinal Answer Summary:<\/h3>\n\n\n\n
\n