{"id":35287,"date":"2025-06-24T09:25:23","date_gmt":"2025-06-24T09:25:23","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=35287"},"modified":"2025-06-24T09:25:24","modified_gmt":"2025-06-24T09:25:24","slug":"given-these-balanced-half-reactions","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/given-these-balanced-half-reactions\/","title":{"rendered":"Given these balanced half-reactions"},"content":{"rendered":"\n
Given these balanced half-reactions: 3H2O + I2O3 + 6 Cl- -> 2 IO3- + 6 HCl Enter the overall balanced redox reaction in an acidic solution. Do not include phases in your answer. Enter Your answer: 45 Chemical equation: 3 H2O + I2O3 + 6 Cl- -> 2 IO3- + 6 HCl<\/pre>\n\n\n\n
\"\"<\/figure>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n
The correct overall balanced redox reaction in acidic solution is:<\/p>\n\n\n\n
3 H\u2082O + I\u2082O\u2083 + 6 Cl\u207b \u2192 2 IO\u2083\u207b + 6 H\u207a + 6 Cl\u207b<\/strong><\/p>\n\n\n\n
Let\u2019s unpack this.<\/p>\n\n\n\n
Step-by-step explanation:<\/h3>\n\n\n\n
This problem involves a redox reaction between iodine trioxide (I\u2082O\u2083) and chloride ions (Cl\u207b) in acidic medium.<\/p>\n\n\n\n
1. Break the overall process into half-reactions:<\/strong><\/h4>\n\n\n\n
We identify two half-reactions:<\/p>\n\n\n\n
    \n
  • Oxidation half-reaction (Cl\u207b to Cl\u2082):<\/strong> But here, Cl\u207b is not oxidized<\/strong>, it remains Cl\u207b in the product. So it is a spectator.<\/li>\n\n\n\n
  • Reduction half-reaction (I in I\u2082O\u2083 to IO\u2083\u207b):<\/strong> To determine how iodine is transformed, consider the change in oxidation states.<\/li>\n<\/ul>\n\n\n\n
    Each iodine in I\u2082O\u2083 has an average oxidation state of +3. It is converted to IO\u2083\u207b, where iodine is +5. Thus, iodine is oxidized<\/strong>, which might seem counterintuitive since it usually acts as an oxidizing agent. But since Cl\u207b does not change, this process is simply:<\/p>\n\n\n\n
    I\u2082O\u2083 + 3 H\u2082O \u2192 2 IO\u2083\u207b + 6 H\u207a + 6 e\u207b<\/strong><\/p>\n\n\n\n
    We\u2019re missing an explicit reduction half-reaction because Cl\u207b appears on both sides. It’s not undergoing oxidation or reduction in this case but is present to balance charge as part of HCl.<\/p>\n\n\n\n
    2. Balance charges and combine:<\/strong><\/h4>\n\n\n\n
    To provide chloride ions for 6 HCl molecules, we include 6 Cl\u207b on the left-hand side. HCl is written as a product to represent hydrogen ions pairing with Cl\u207b:<\/p>\n\n\n\n
    3 H\u2082O + I\u2082O\u2083 + 6 Cl\u207b \u2192 2 IO\u2083\u207b + 6 HCl<\/strong><\/p>\n\n\n\n
    However, this representation combines neutralization directly. To clarify acidic conditions and charge balance separately, one might instead write:<\/p>\n\n\n\n
    3 H\u2082O + I\u2082O\u2083 + 6 Cl\u207b \u2192 2 IO\u2083\u207b + 6 H\u207a + 6 Cl\u207b<\/strong><\/p>\n\n\n\n
    Either format is acceptable depending on whether the goal is to show HCl as product or to focus on the redox process under acidic conditions.<\/p>\n\n\n\n
    \"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"
    Given these balanced half-reactions: 3H2O + I2O3 + 6 Cl- -> 2 IO3- + 6 HCl Enter the overall balanced redox reaction in an acidic solution. Do not include phases in your answer. Enter Your answer: 45 Chemical equation: 3 H2O + I2O3 + 6 Cl- -> 2 IO3- + 6 HCl The Correct Answer […]<\/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-35287","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35287","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=35287"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35287\/revisions"}],"predecessor-version":[{"id":35290,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35287\/revisions\/35290"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35287"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35287"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35287"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}