{"id":148913,"date":"2024-02-27T17:45:53","date_gmt":"2024-02-27T12:15:53","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=148913"},"modified":"2024-02-27T17:45:53","modified_gmt":"2024-02-27T12:15:53","slug":"calculate-emf-of-the-following-cell-at-25-c","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/quiz\/calculate-emf-of-the-following-cell-at-25-c","title":{"rendered":"Calculate emf of the following cell at 25 \u00b0C:…"},"content":{"rendered":"

Calculate emf of the following cell at 25 \u00b0C:<\/h2>\n

\"\"<\/p>\n

Ans.<\/h3>\n
    \n
  1. \n

    Sn\/Sn2+ (0.001 M) and H+\/H2(g) (1 bar) | Pt(s)<\/h3>\n<\/li>\n<\/ol>\n

    The overall cell reaction is:<\/h3>\n

    Sn\u2192Sn2++2e\u2212<\/span>Sn<\/span><\/span>\u2192<\/span><\/span>Sn<\/span>2+<\/span><\/span><\/span><\/span><\/span><\/span><\/span>+<\/span><\/span>2<\/span>e<\/span>\u2212<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

    From the given data:<\/h3>\n
      \n
    • \n

      \ufffd\u2218(Sn2+\/Sn)=\u22120.14<\/span>E<\/span>\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span>(<\/span>Sn<\/span>2+<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\/<\/span>Sn<\/span><\/span>)<\/span>=<\/span><\/span>\u2212<\/span>0.14<\/span><\/span><\/span><\/span><\/span> V<\/h3>\n<\/li>\n
    • \n

      \ufffd\u2218(H+\/H2)=0.00<\/span>E<\/span>\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span>(<\/span>H<\/span>+<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\/<\/span>H<\/span>2<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>)<\/span>=<\/span><\/span>0.00<\/span><\/span><\/span><\/span><\/span> V<\/h3>\n<\/li>\n
    • \n

      pH=\u2212log\u2061(0.01)=2<\/span>pH<\/span><\/span>=<\/span><\/span>\u2212<\/span>log<\/span>(<\/span>0.01<\/span>)<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span><\/h3>\n<\/li>\n<\/ul>\n

      The reaction quotient (\ufffd<\/span>Q<\/span><\/span><\/span><\/span><\/span>) for this cell is:<\/h3>\n

      \ufffd=[H+]\ufffdH2=0.011=0.01<\/span>Q<\/span>=<\/span><\/span>P<\/span>H<\/span>2\u200b<\/span><\/span>\u200b<\/span><\/span><\/span><\/span>[<\/span>H<\/span>+<\/span><\/span><\/span><\/span>]<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>1<\/span><\/span>0.01<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>=<\/span><\/span>0.01<\/span><\/span><\/span><\/span><\/span><\/h3>\n

      Now, let’s calculate the emf (\ufffd<\/span>E<\/span><\/span><\/span><\/span><\/span>) of the cell using the Nernst equation:<\/h3>\n

      \ufffd=\ufffd\u2218\u2212\ufffd\ufffd\ufffd\ufffdln\u2061(\ufffd)<\/span>E<\/span>=<\/span><\/span>E<\/span>\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span>\u2212<\/span><\/span>n<\/span>F<\/span><\/span><\/span>RT<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>ln<\/span>(<\/span>Q<\/span>)<\/span><\/span><\/span><\/span><\/span><\/h3>\n

      Since the cell reaction involves the transfer of 2 electrons (\ufffd=2<\/span>n<\/span>=<\/span><\/span>2<\/span><\/span><\/span><\/span><\/span>), we have:<\/h3>\n

      \ufffd=\u22120.14\u2212(8.314\u00d7298)(2\u00d796485)\u00d7ln\u2061(0.01)<\/span>E<\/span>=<\/span><\/span>\u2212<\/span>0.14<\/span>\u2212<\/span><\/span>(<\/span>2\u00d7<\/span>96485)<\/span><\/span><\/span>(<\/span>8.314\u00d7<\/span>298)<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>ln<\/span>(<\/span>0.01<\/span>)<\/span><\/span><\/span><\/span><\/span><\/h3>\n

      Now, let’s compute the value:<\/h3>\n

      \ufffd=\u22120.14\u2212(2475.972)(192970)\u00d7ln\u2061(0.01)<\/span>E<\/span>=<\/span><\/span>\u2212<\/span>0.14<\/span>\u2212<\/span><\/span>(<\/span>192970)<\/span><\/span><\/span>(<\/span>2475.972)<\/span><\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>ln<\/span>(<\/span>0.01<\/span>)<\/span><\/span><\/span><\/span><\/span> \ufffd=\u22120.14\u22122475.972192970\u00d7(\u22124.605)<\/span>E<\/span>=<\/span><\/span>\u2212<\/span>0.14<\/span>\u2212<\/span><\/span>192970<\/span><\/span>2475.972<\/span><\/span><\/span>\u200b<\/span><\/span><\/span><\/span><\/span>\u00d7<\/span><\/span>(<\/span>\u2212<\/span>4.605<\/span>)<\/span><\/span><\/span><\/span><\/span> \ufffd=\u22120.14+0.005914<\/span>E<\/span>=<\/span><\/span>\u2212<\/span>0.14<\/span>+<\/span><\/span>0.005914<\/span><\/span><\/span><\/span><\/span> \ufffd=\u22120.134086\u00a0V<\/span>E<\/span>=<\/span><\/span>\u2212<\/span>0.134086<\/span>\u00a0V<\/span><\/span><\/span><\/span><\/span><\/span><\/h3>\n

      Therefore, the emf of the given cell at 25\u00b0C is approximately -0.134 V.<\/h3>\n","protected":false},"excerpt":{"rendered":"

      Calculate emf of the following cell at 25 \u00b0C: Ans. Sn\/Sn2+ (0.001 M) and H+\/H2(g) (1 bar) | Pt(s) The overall cell reaction is: Sn\u2192Sn2++2e\u2212Sn\u2192Sn2++2e\u2212 From the given data: \ufffd\u2218(Sn2+\/Sn)=\u22120.14E\u2218(Sn2+\/Sn)=\u22120.14 V \ufffd\u2218(H+\/H2)=0.00E\u2218(H+\/H2\u200b)=0.00 V pH=\u2212log\u2061(0.01)=2pH=\u2212log(0.01)=2 The reaction quotient (\ufffdQ) for this cell is: \ufffd=[H+]\ufffdH2=0.011=0.01Q=PH2\u200b\u200b[H+]\u200b=10.01\u200b=0.01 Now, let’s calculate the emf (\ufffdE) of the cell using the Nernst […]<\/p>\n","protected":false},"author":21830,"featured_media":114041,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[12119],"tags":[],"class_list":{"0":"post-148913","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-quiz"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148913","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/users\/21830"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/comments?post=148913"}],"version-history":[{"count":1,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148913\/revisions"}],"predecessor-version":[{"id":148914,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/posts\/148913\/revisions\/148914"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media\/114041"}],"wp:attachment":[{"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/media?parent=148913"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/categories?post=148913"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.mapsofindia.com\/my-india\/wp-json\/wp\/v2\/tags?post=148913"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}