Balanced Redox Equation<\/h3>\n\n\n\n
Unbalanced reaction<\/strong>: We will balance the redox reaction in an acidic solution<\/strong> using the half-reaction method.<\/p>\n\n\n\n Oxidation (Arsenic changes from +3 in AsO\u2083\u00b3\u207b to +5 in AsO\u2084\u00b3\u207b):<\/strong> Reduction (Copper changes from +2 to 0):<\/strong> Already balanced for arsenic and copper.<\/p>\n\n\n\n AsO\u2083\u00b3\u207b has 3 oxygen atoms AsO\u2083\u00b3\u207b + H\u2082O \u2192 AsO\u2084\u00b3\u207b<\/p>\n\n\n\n 1 H\u2082O = 2 H atoms AsO\u2083\u00b3\u207b + H\u2082O \u2192 AsO\u2084\u00b3\u207b + 2 H\u207a<\/p>\n\n\n\n Left: AsO\u2083\u00b3\u207b (\u22123), H\u2082O (0) \u2192 total: \u22123 To balance charges: Oxidation half-reaction:<\/strong> Reduction half-reaction (Cu\u00b2\u207a gains 2 e\u207b):<\/strong> AsO\u2083\u00b3\u207b + H\u2082O + Cu\u00b2\u207a \u2192 AsO\u2084\u00b3\u207b + 2 H\u207a + Cu<\/p>\n\n\n\n AsO\u2083\u00b3\u207b + H\u2082O + Cu\u00b2\u207a \u2192 AsO\u2084\u00b3\u207b + 2 H\u207a + Cu<\/strong><\/p>\n\n\n\n This redox reaction involves arsenite (AsO\u2083\u00b3\u207b) and copper ions (Cu\u00b2\u207a). The key to balancing redox reactions is identifying the oxidation states of the elements involved and determining which species is oxidized and which is reduced.<\/p>\n\n\n\n In this reaction, arsenic in AsO\u2083\u00b3\u207b has an oxidation state of +3, while in AsO\u2084\u00b3\u207b it becomes +5. This increase in oxidation number indicates that arsenic is oxidized. On the other hand, copper goes from Cu\u00b2\u207a to elemental copper (Cu\u2070), meaning it is reduced.<\/p>\n\n\n\n We used the half-reaction method in acidic solution to balance the reaction. The oxidation half-reaction shows the conversion of arsenite to arsenate, balanced by adding water and hydrogen ions to match the oxygen and hydrogen atoms. Electrons are added to balance the charge. The reduction half-reaction shows Cu\u00b2\u207a gaining two electrons to form solid copper.<\/p>\n\n\n\n Combining the two balanced half-reactions cancels out the electrons, which confirms the conservation of charge and mass. Water and hydrogen ions are balanced appropriately for acidic conditions.<\/p>\n\n\n\n This balanced equation shows the fundamental concept of redox: the transfer of electrons. Arsenic loses electrons (oxidation), and copper gains them (reduction). Understanding how to separate and balance these processes helps in interpreting chemical reactions that involve electron exchange, which are common in electrochemistry, corrosion, and biological systems.<\/p>\n\n\n\n Write the balanced redox equation for AsO33- + Cu2+ AsO43- + Cu The Correct Answer and Explanation is: Balanced Redox Equation Unbalanced reaction:AsO\u2083\u00b3\u207b + Cu\u00b2\u207a \u2192 AsO\u2084\u00b3\u207b + Cu We will balance the redox reaction in an acidic solution using the half-reaction method. Step 1: Write oxidation and reduction half-reactions Oxidation (Arsenic changes from +3 […]<\/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-35671","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35671","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=35671"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35671\/revisions"}],"predecessor-version":[{"id":35676,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/35671\/revisions\/35676"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=35671"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=35671"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=35671"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
AsO\u2083\u00b3\u207b + Cu\u00b2\u207a \u2192 AsO\u2084\u00b3\u207b + Cu<\/p>\n\n\n\n
\n\n\n\nStep 1: Write oxidation and reduction half-reactions<\/h3>\n\n\n\n
AsO\u2083\u00b3\u207b \u2192 AsO\u2084\u00b3\u207b<\/p>\n\n\n\n
Cu\u00b2\u207a \u2192 Cu<\/p>\n\n\n\n
\n\n\n\nStep 2: Balance atoms other than O and H<\/h3>\n\n\n\n
\n\n\n\nStep 3: Balance oxygen using H\u2082O<\/h3>\n\n\n\n
AsO\u2084\u00b3\u207b has 4 oxygen atoms
We add 1 H\u2082O on the left side:<\/p>\n\n\n\n
\n\n\n\nStep 4: Balance hydrogen using H\u207a<\/h3>\n\n\n\n
Add 2 H\u207a on the right:<\/p>\n\n\n\n
\n\n\n\nStep 5: Balance charge using electrons<\/h3>\n\n\n\n
Right: AsO\u2084\u00b3\u207b (\u22123), 2 H\u207a (+2) \u2192 total: \u22121<\/p>\n\n\n\n
Add 2 e\u207b to the right<\/p>\n\n\n\n
AsO\u2083\u00b3\u207b + H\u2082O \u2192 AsO\u2084\u00b3\u207b + 2 H\u207a + 2 e\u207b<\/p>\n\n\n\n
Cu\u00b2\u207a + 2 e\u207b \u2192 Cu<\/p>\n\n\n\n
\n\n\n\nStep 6: Add half-reactions<\/h3>\n\n\n\n
\n\n\n\nFinal Balanced Redox Equation<\/h3>\n\n\n\n
\n\n\n\nExplanation<\/h3>\n\n\n\n
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<\/figure>\n","protected":false},"excerpt":{"rendered":"