{"id":247868,"date":"2025-07-07T20:19:46","date_gmt":"2025-07-07T20:19:46","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=247868"},"modified":"2025-07-07T20:19:48","modified_gmt":"2025-07-07T20:19:48","slug":"the-cell-notation-of-a-standard-galvanic-voltaic-cell-containing-an-unknown-metal-electrode-x-is-shown-below","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/the-cell-notation-of-a-standard-galvanic-voltaic-cell-containing-an-unknown-metal-electrode-x-is-shown-below\/","title":{"rendered":"The cell notation of a standard galvanic (voltaic) cell containing an unknown metal electrode X is shown below."},"content":{"rendered":"\n

The cell notation of a standard galvanic (voltaic) cell containing an unknown metal electrode X is shown below. X(s) | X3+ (1 mol\u00e2\u02c6\u2122dm-3) || Pb2+ (1 mol\u00e2\u02c6\u2122dm-3) | Pb(s) a) Name the component of the cell represented by the double vertical lines (||) in the above cell notation and explain its function. b) Identify the oxidizing agent in the above cell. c) The initial reading on a voltmeter connected across the electrodes of the above cell is 1.53 V. Identify metal X by calculating the standard reduction potential of the unknown metal X. d) Write down the balanced equation for the net (overall) reaction taking place in this cell. Omit the spectator ions. e) How will the initial voltmeter reading be affected if the concentration of the electrolyte in the X(s)|X3+ (aq) half-cell is increased? Write down only INCREASES, DECREASES, or REMAINS THE SAME.<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

Name the component of the cell represented by the double vertical lines (||) in the above cell notation and explain its function.<\/h3>\n\n\n\n

The double vertical lines (||) represent the salt bridge<\/strong> or porous barrier<\/strong>. Its primary function is to maintain electrical neutrality by allowing the flow of ions between the two half-cells. This prevents the buildup of charge that would otherwise stop the reaction, ensuring the continued flow of electrons through the external circuit.<\/p>\n\n\n\n

\n\n\n\n

b) Identify the oxidizing agent in the above cell.<\/h3>\n\n\n\n

In the given cell notation, the oxidizing agent is Pb\u00b2\u207a<\/strong> (lead ion). The oxidizing agent is the species that gains electrons during the redox reaction. In this case, Pb\u00b2\u207a ions are reduced to Pb(s) at the cathode, meaning they accept electrons and thus act as the oxidizing agent.<\/p>\n\n\n\n

\n\n\n\n

c) The initial reading on a voltmeter connected across the electrodes of the above cell is 1.53 V. Identify metal X by calculating the standard reduction potential of the unknown metal X.<\/h3>\n\n\n\n

The cell potential can be calculated using the formula: Ecell=Ecathode\u2212EanodeE_{\\text{cell}} = E_{\\text{cathode}} – E_{\\text{anode}}Ecell\u200b=Ecathode\u200b\u2212Eanode\u200b<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • The cathode is where reduction occurs (Pb\u00b2\u207a \u2192 Pb),<\/li>\n\n\n\n
  • The anode is where oxidation occurs (X \u2192 X\u00b3\u207a).<\/li>\n<\/ul>\n\n\n\n

    From the given information:<\/p>\n\n\n\n

      \n
    • The standard reduction potential of Pb\u00b2\u207a\/Pb is +0.13 V (standard reduction potential for lead).<\/li>\n\n\n\n
    • The standard reduction potential for the unknown metal X (X\u00b3\u207a\/X) is to be determined.<\/li>\n<\/ul>\n\n\n\n

      The given total cell potential is 1.53 V, so: 1.53\u2009V=EPb\u00b2\u207a\/Pb\u2212EX\u00b3\u207a\/X1.53 \\, \\text{V} = E_{\\text{Pb\u00b2\u207a\/Pb}} – E_{\\text{X\u00b3\u207a\/X}}1.53V=EPb\u00b2\u207a\/Pb\u200b\u2212EX\u00b3\u207a\/X\u200b 1.53\u2009V=0.13\u2009V\u2212EX\u00b3\u207a\/X1.53 \\, \\text{V} = 0.13 \\, \\text{V} – E_{\\text{X\u00b3\u207a\/X}}1.53V=0.13V\u2212EX\u00b3\u207a\/X\u200b<\/p>\n\n\n\n

      Solving for EX\u00b3\u207a\/XE_{\\text{X\u00b3\u207a\/X}}EX\u00b3\u207a\/X\u200b: EX\u00b3\u207a\/X=0.13\u2009V\u22121.53\u2009VE_{\\text{X\u00b3\u207a\/X}} = 0.13 \\, \\text{V} – 1.53 \\, \\text{V}EX\u00b3\u207a\/X\u200b=0.13V\u22121.53V EX\u00b3\u207a\/X=\u22121.40\u2009VE_{\\text{X\u00b3\u207a\/X}} = -1.40 \\, \\text{V}EX\u00b3\u207a\/X\u200b=\u22121.40V<\/p>\n\n\n\n

      So, the standard reduction potential for X\u00b3\u207a\/X is -1.40 V<\/strong>.<\/p>\n\n\n\n

      From the standard reduction potential table, the metal with a reduction potential of -1.40 V is Aluminum (Al)<\/strong>. Therefore, the unknown metal X is Al<\/strong>.<\/p>\n\n\n\n

      \n\n\n\n

      d) Write down the balanced equation for the net (overall) reaction taking place in this cell. Omit the spectator ions.<\/h3>\n\n\n\n

      At the anode, the oxidation half-reaction is: X(s)\u2192X\u00b3\u207a(aq)+3e\u2212\\text{X(s)} \\rightarrow \\text{X\u00b3\u207a(aq)} + 3e\u207bX(s)\u2192X\u00b3\u207a(aq)+3e\u2212<\/p>\n\n\n\n

      At the cathode, the reduction half-reaction is: Pb\u00b2\u207a(aq)+2e\u2212\u2192Pb(s)\\text{Pb\u00b2\u207a(aq)} + 2e\u207b \\rightarrow \\text{Pb(s)}Pb\u00b2\u207a(aq)+2e\u2212\u2192Pb(s)<\/p>\n\n\n\n

      To balance the number of electrons, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3: 2(X(s)\u2192X\u00b3\u207a(aq)+3e\u2212)and3(Pb\u00b2\u207a(aq)+2e\u2212\u2192Pb(s))2(\\text{X(s)} \\rightarrow \\text{X\u00b3\u207a(aq)} + 3e\u207b) \\quad \\text{and} \\quad 3(\\text{Pb\u00b2\u207a(aq)} + 2e\u207b \\rightarrow \\text{Pb(s)})2(X(s)\u2192X\u00b3\u207a(aq)+3e\u2212)and3(Pb\u00b2\u207a(aq)+2e\u2212\u2192Pb(s))<\/p>\n\n\n\n

      This gives the overall balanced reaction: 2X(s)+3Pb\u00b2\u207a(aq)\u21922X\u00b3\u207a(aq)+3Pb(s)2\\text{X(s)} + 3\\text{Pb\u00b2\u207a(aq)} \\rightarrow 2\\text{X\u00b3\u207a(aq)} + 3\\text{Pb(s)}2X(s)+3Pb\u00b2\u207a(aq)\u21922X\u00b3\u207a(aq)+3Pb(s)<\/p>\n\n\n\n

      \n\n\n\n

      e) How will the initial voltmeter reading be affected if the concentration of the electrolyte in the X(s)|X\u00b3\u207a (aq) half-cell is increased? Write down only INCREASES, DECREASES, or REMAINS THE SAME.<\/h3>\n\n\n\n

      The initial voltmeter reading will increase<\/strong>. According to the Nernst equation, increasing the concentration of X\u00b3\u207a will shift the equilibrium towards more X(s) being oxidized (since the reaction is in the anode). This will increase the cell potential. Therefore, the voltmeter reading increases as the concentration of the electrolyte in the X(s)|X\u00b3\u207a half-cell increases.<\/p>\n\n\n\n

      \n\n\n\n

      In summary:<\/p>\n\n\n\n

        \n
      • Double vertical lines (||)<\/strong> represent the salt bridge<\/strong>, maintaining electrical neutrality.<\/li>\n\n\n\n
      • The oxidizing agent<\/strong> is Pb\u00b2\u207a<\/strong>.<\/li>\n\n\n\n
      • The unknown metal X<\/strong> is identified as Aluminum (Al)<\/strong>.<\/li>\n\n\n\n
      • The balanced equation<\/strong> is: 2X(s)+3Pb\u00b2\u207a(aq)\u21922X\u00b3\u207a(aq)+3Pb(s)2\\text{X(s)} + 3\\text{Pb\u00b2\u207a(aq)} \\rightarrow 2\\text{X\u00b3\u207a(aq)} + 3\\text{Pb(s)}2X(s)+3Pb\u00b2\u207a(aq)\u21922X\u00b3\u207a(aq)+3Pb(s).<\/li>\n\n\n\n
      • Increasing the concentration of the electrolyte in the X(s)|X\u00b3\u207a half-cell will cause the voltmeter reading to increase<\/strong>.<\/li>\n<\/ul>\n\n\n\n
        \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

        The cell notation of a standard galvanic (voltaic) cell containing an unknown metal electrode X is shown below. X(s) | X3+ (1 mol\u00e2\u02c6\u2122dm-3) || Pb2+ (1 mol\u00e2\u02c6\u2122dm-3) | Pb(s) a) Name the component of the cell represented by the double vertical lines (||) in the above cell notation and explain its function. b) Identify the […]<\/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-247868","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247868","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=247868"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247868\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=247868"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=247868"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=247868"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}