{"id":35283,"date":"2025-06-24T09:22:15","date_gmt":"2025-06-24T09:22:15","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35283"},"modified":"2025-06-24T09:22:17","modified_gmt":"2025-06-24T09:22:17","slug":"determine-the-spontaneous-direction-at-ph","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/determine-the-spontaneous-direction-at-ph\/","title":{"rendered":"Determine the spontaneous direction at pH"},"content":{"rendered":"\n<p>5 [-(aq) I2O3(aq) + 6 H+(aq) \u00e2\u2020\u2019 3 [I-(s) + 9 H2O(l) (a) Determine the spontaneous direction at pH = 2.00 and [I2O3] = 0.100 M. (b) Repeat the calculation at pH = 1.00. (c) At what pH is this redox reaction at equilibrium at these concentrations of [I2O3] and [I-]? E\u00c2\u00b0(I2O3\/I-) = 0.5355 V, E\u00c2\u00b0(I2O3\/H+) = 1.195 V, E = E\u00c2\u00b0 &#8211; (0.0592\/n) log Q E = 0 at equilibrium<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this redox problem, we use the Nernst equation:E=E\u2218\u22120.0592nlog\u2061QE = E^\\circ &#8211; \\frac{0.0592}{n} \\log QE=E\u2218\u2212n0.0592\u200blogQ<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Balanced half-reaction:<br>I2O3(aq)+6H+(aq)+6e\u2212\u21922I\u2212(s)+3H2O(l)\\text{I}_2\\text{O}_3(aq) + 6H^+(aq) + 6e^- \\rightarrow 2I^-(s) + 3H_2O(l)I2\u200bO3\u200b(aq)+6H+(aq)+6e\u2212\u21922I\u2212(s)+3H2\u200bO(l)<\/li>\n\n\n\n<li>E\u2218=0.5355\u00a0VE^\\circ = 0.5355 \\text{ V}E\u2218=0.5355\u00a0V (for the I\u2082O\u2083\/I\u207b couple)<\/li>\n\n\n\n<li>n=6n = 6n=6<\/li>\n\n\n\n<li>[I2O3]=0.100\u00a0M[I_2O_3] = 0.100 \\text{ M}[I2\u200bO3\u200b]=0.100\u00a0M<\/li>\n\n\n\n<li>Solid I\u207b is pure, so activity = 1<\/li>\n\n\n\n<li>Water is a liquid, so also not included in Q<\/li>\n\n\n\n<li>Q=1[I2O3][H+]6Q = \\frac{1}{[I_2O_3][H^+]^6}Q=[I2\u200bO3\u200b][H+]61\u200b<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(a) At pH = 2.00<\/h3>\n\n\n\n<p>[H+]=10\u22122=0.010[H^+] = 10^{-2} = 0.010[H+]=10\u22122=0.010Q=1(0.100)(0.010)6=1(0.100)(1\u00d710\u221212)=1\u00d71013Q = \\frac{1}{(0.100)(0.010)^6} = \\frac{1}{(0.100)(1 \\times 10^{-12})} = 1 \\times 10^{13}Q=(0.100)(0.010)61\u200b=(0.100)(1\u00d710\u221212)1\u200b=1\u00d71013E=0.5355\u22120.05926log\u2061(1\u00d71013)=0.5355\u22120.05926\u22c513E = 0.5355 &#8211; \\frac{0.0592}{6} \\log(1 \\times 10^{13}) = 0.5355 &#8211; \\frac{0.0592}{6} \\cdot 13E=0.5355\u221260.0592\u200blog(1\u00d71013)=0.5355\u221260.0592\u200b\u22c513E=0.5355\u2212(0.0592\u22c52.1667)=0.5355\u22120.1282=0.4073&nbsp;VE = 0.5355 &#8211; (0.0592 \\cdot 2.1667) = 0.5355 &#8211; 0.1282 = 0.4073 \\text{ V}E=0.5355\u2212(0.0592\u22c52.1667)=0.5355\u22120.1282=0.4073&nbsp;V<\/p>\n\n\n\n<p>Since E &gt; 0, the reaction is spontaneous in the forward direction.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(b) At pH = 1.00<\/h3>\n\n\n\n<p>[H+]=0.10[H^+] = 0.10[H+]=0.10Q=1(0.100)(0.10)6=1(0.100)(1\u00d710\u22126)=1\u00d7107Q = \\frac{1}{(0.100)(0.10)^6} = \\frac{1}{(0.100)(1 \\times 10^{-6})} = 1 \\times 10^7Q=(0.100)(0.10)61\u200b=(0.100)(1\u00d710\u22126)1\u200b=1\u00d7107E=0.5355\u22120.05926\u22c5log\u2061(1\u00d7107)=0.5355\u2212(0.0592\u22c51.1667)=0.5355\u22120.0691=0.4664&nbsp;VE = 0.5355 &#8211; \\frac{0.0592}{6} \\cdot \\log(1 \\times 10^7) = 0.5355 &#8211; (0.0592 \\cdot 1.1667) = 0.5355 &#8211; 0.0691 = 0.4664 \\text{ V}E=0.5355\u221260.0592\u200b\u22c5log(1\u00d7107)=0.5355\u2212(0.0592\u22c51.1667)=0.5355\u22120.0691=0.4664&nbsp;V<\/p>\n\n\n\n<p>Again, E &gt; 0, so the reaction is spontaneous in the forward direction.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">(c) At equilibrium, E = 0<\/h3>\n\n\n\n<p>Set up the Nernst equation and solve for [H\u207a]:0=0.5355\u22120.05926log\u2061Q0 = 0.5355 &#8211; \\frac{0.0592}{6} \\log Q0=0.5355\u221260.0592\u200blogQlog\u2061Q=0.5355\u22c560.0592=54.27\\log Q = \\frac{0.5355 \\cdot 6}{0.0592} = 54.27logQ=0.05920.5355\u22c56\u200b=54.27Q=1054.27Q = 10^{54.27}Q=1054.27<\/p>\n\n\n\n<p>Recall that Q=1(0.100)[H+]6Q = \\frac{1}{(0.100)[H^+]^6}Q=(0.100)[H+]61\u200b(0.100)[H+]6=10\u221254.27\u21d2[H+]6=10\u221254.27\/0.100=10\u221253.27(0.100)[H^+]^6 = 10^{-54.27} \\Rightarrow [H^+]^6 = 10^{-54.27} \/ 0.100 = 10^{-53.27}(0.100)[H+]6=10\u221254.27\u21d2[H+]6=10\u221254.27\/0.100=10\u221253.27log\u2061[H+]=\u221253.276=\u22128.88\u21d2pH=8.88\\log [H^+] = \\frac{-53.27}{6} = -8.88 \\Rightarrow \\text{pH} = 8.88log[H+]=6\u221253.27\u200b=\u22128.88\u21d2pH=8.88<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<p><strong>(a)<\/strong> Reaction is spontaneous forward, E = 0.407 V<br><strong>(b)<\/strong> Reaction is spontaneous forward, E = 0.466 V<br><strong>(c)<\/strong> Reaction is at equilibrium when <strong>pH = 8.88<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation<\/h3>\n\n\n\n<p>This redox reaction involves the reduction of iodic oxide (I\u2082O\u2083) to iodide ion (I\u207b) in acidic solution. The spontaneity of a redox reaction can be assessed by calculating the cell potential (E) using the Nernst equation. A positive E indicates a spontaneous reaction in the forward direction.<\/p>\n\n\n\n<p>To apply the Nernst equation, we first identify the half-reaction and its standard reduction potential. The balanced reaction shows that 6 electrons are transferred, and water is formed from hydrogen ions. The reaction quotient (Q) includes only aqueous species, so solids like iodide and liquids like water are excluded.<\/p>\n\n\n\n<p>In part (a), at pH 2.00, the hydrogen ion concentration is 0.010 M. Substituting values into the Nernst equation, we find a positive E of 0.407 V, indicating that the forward reaction is spontaneous. In part (b), at a more acidic pH of 1.00, the higher hydrogen ion concentration further favors the forward direction, giving a slightly larger E value of 0.466 V.<\/p>\n\n\n\n<p>In part (c), we find the pH where the reaction is at equilibrium by setting E = 0 and solving for [H\u207a]. This gives a very low hydrogen ion concentration, corresponding to a pH of 8.88. At this point, the forward and reverse reactions occur at the same rate, and no net change occurs in concentrations.<\/p>\n\n\n\n<p>This problem highlights how pH affects redox equilibria. As pH increases (fewer H\u207a ions), the forward redox reaction becomes less favorable, and equilibrium is reached at basic pH.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-661.jpeg\" alt=\"\" class=\"wp-image-35284\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-661.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-661-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-661-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>5 [-(aq) I2O3(aq) + 6 H+(aq) \u00e2\u2020\u2019 3 [I-(s) + 9 H2O(l) (a) Determine the spontaneous direction at pH = 2.00 and [I2O3] = 0.100 M. (b) Repeat the calculation at pH = 1.00. (c) At what pH is this redox reaction at equilibrium at these concentrations of [I2O3] and [I-]? E\u00c2\u00b0(I2O3\/I-) = 0.5355 V, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-35283","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35283","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=35283"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35283\/revisions"}],"predecessor-version":[{"id":35285,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35283\/revisions\/35285"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35283"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35283"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35283"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}