{"id":4412,"date":"2026-04-07T10:43:58","date_gmt":"2026-04-07T10:43:58","guid":{"rendered":"https:\/\/ksquareinstitute.in\/blog\/?p=4412"},"modified":"2026-04-07T10:43:59","modified_gmt":"2026-04-07T10:43:59","slug":"top-5-redox-reactions-questions","status":"publish","type":"post","link":"https:\/\/ksquareinstitute.in\/blog\/top-5-redox-reactions-questions\/","title":{"rendered":"Top 5 Redox Reactions Questions for NEET (Repeated Problems with Solutions)"},"content":{"rendered":"\n

Top Redox Reactions Questions for NEET<\/h2>\n\n\n\n

Redox reactions are one of the most fundamental and high-weightage topics in NEET Chemistry. Questions from oxidation number, balancing redox equations, and equivalent concept are asked almost every year. If you master the Top 5 Redox Reactions Questions<\/strong>, you can confidently solve both conceptual and numerical problems in the exam.<\/p>\n\n\n\n

In this article, we will go through the Top 5 Redox Reactions Questions<\/strong> that are frequently repeated in NEET PYQs. Along with step-by-step solutions, you will also learn important tricks to solve questions quickly and accurately.<\/p>\n\n\n\n

\"Top<\/figure>\n\n\n\n

Question 1: Oxidation Number Calculation<\/h2>\n\n\n\n

Find the oxidation number of chromium in K<\/mi>2<\/mn><\/msub>C<\/mi>r<\/mi>2<\/mn><\/msub>O<\/mi>7<\/mn><\/msub><\/mrow>K_2Cr_2O_7<\/annotation><\/semantics><\/math>K2\u200bCr2\u200bO7\u200b.<\/p>\n\n\n\n

Detailed Explanation<\/h3>\n\n\n\n

Let the oxidation number of Cr = x<\/mi><\/mrow>x<\/annotation><\/semantics><\/math>x<\/p>\n\n\n\n

We know:<\/p>\n\n\n\n

    \n
  • K = +1<\/li>\n\n\n\n
  • O = -2<\/li>\n<\/ul>\n\n\n\n

    Applying the rule:2<\/mn>(<\/mo>+<\/mo>1<\/mn>)<\/mo>+<\/mo>2<\/mn>x<\/mi>+<\/mo>7<\/mn>(<\/mo>\u2212<\/mo>2<\/mn>)<\/mo>=<\/mo>0<\/mn><\/mrow>2(+1) + 2x + 7(-2) = 0<\/annotation><\/semantics><\/math>2(+1)+2x+7(\u22122)=0 2<\/mn>+<\/mo>2<\/mn>x<\/mi>\u2212<\/mo>14<\/mn>=<\/mo>0<\/mn><\/mrow>2 + 2x – 14 = 0<\/annotation><\/semantics><\/math>2+2x\u221214=0 2<\/mn>x<\/mi>\u2212<\/mo>12<\/mn>=<\/mo>0<\/mn>\u21d2<\/mo>x<\/mi>=<\/mo>+<\/mo>6<\/mn><\/mrow>2x – 12 = 0 \\Rightarrow x = +6<\/annotation><\/semantics><\/math>2x\u221212=0\u21d2x=+6<\/p>\n\n\n\n

    Final Answer:<\/h3>\n\n\n\n

    Oxidation number of Cr = +6<\/p>\n\n\n\n

    This is one of the most basic yet frequently asked problems in the Top 5 Redox Reactions Questions<\/strong>.<\/p>\n\n\n\n

    Question 2: Identification of Oxidation and Reduction<\/h2>\n\n\n\n

    Identify which species is oxidized and which is reduced:Z<\/mi>n<\/mi>+<\/mo>C<\/mi>u<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>\u2192<\/mo>Z<\/mi>n<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>+<\/mo>C<\/mi>u<\/mi><\/mrow>Zn + Cu^{2+} \\rightarrow Zn^{2+} + Cu<\/annotation><\/semantics><\/math>Zn+Cu2+\u2192Zn2++Cu<\/p>\n\n\n\n

    Detailed Explanation<\/h3>\n\n\n\n
      \n
    • Zn \u2192 Zn\u00b2\u207a (loss of electrons) \u2192 Oxidation<\/li>\n\n\n\n
    • Cu\u00b2\u207a \u2192 Cu (gain of electrons) \u2192 Reduction<\/li>\n<\/ul>\n\n\n\n

      Final Answer:<\/h3>\n\n\n\n

      Zn is oxidized, and Cu\u00b2\u207a is reduced.<\/p>\n\n\n\n

      This is a conceptual question commonly included in the Top 5 Redox Reactions Questions<\/strong>.<\/p>\n\n\n\n

      Question 3: Balancing Redox Reaction (Ion-Electron Method)<\/h2>\n\n\n\n

      Balance the reaction in acidic medium:M<\/mi>n<\/mi>O<\/mi>4<\/mn>\u2212<\/mo><\/msubsup>+<\/mo>F<\/mi>e<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>\u2192<\/mo>M<\/mi>n<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>+<\/mo>F<\/mi>e<\/mi>3<\/mn>+<\/mo><\/mrow><\/msup><\/mrow>MnO_4^- + Fe^{2+} \\rightarrow Mn^{2+} + Fe^{3+}<\/annotation><\/semantics><\/math>MnO4\u2212\u200b+Fe2+\u2192Mn2++Fe3+<\/p>\n\n\n\n

      Detailed Explanation<\/h3>\n\n\n\n

      Step 1: Write half-reactions<\/p>\n\n\n\n

      Reduction:M<\/mi>n<\/mi>O<\/mi>4<\/mn>\u2212<\/mo><\/msubsup>\u2192<\/mo>M<\/mi>n<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup><\/mrow>MnO_4^- \\rightarrow Mn^{2+}<\/annotation><\/semantics><\/math>MnO4\u2212\u200b\u2192Mn2+<\/p>\n\n\n\n

      Oxidation:F<\/mi>e<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>\u2192<\/mo>F<\/mi>e<\/mi>3<\/mn>+<\/mo><\/mrow><\/msup><\/mrow>Fe^{2+} \\rightarrow Fe^{3+}<\/annotation><\/semantics><\/math>Fe2+\u2192Fe3+<\/p>\n\n\n\n

      Step 2: Balance oxygen and hydrogenM<\/mi>n<\/mi>O<\/mi>4<\/mn>\u2212<\/mo><\/msubsup>+<\/mo>8<\/mn>H<\/mi>+<\/mo><\/msup>\u2192<\/mo>M<\/mi>n<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>+<\/mo>4<\/mn>H<\/mi>2<\/mn><\/msub>O<\/mi><\/mrow>MnO_4^- + 8H^+ \\rightarrow Mn^{2+} + 4H_2O<\/annotation><\/semantics><\/math>MnO4\u2212\u200b+8H+\u2192Mn2++4H2\u200bO<\/p>\n\n\n\n

      Step 3: Balance electronsM<\/mi>n<\/mi>O<\/mi>4<\/mn>\u2212<\/mo><\/msubsup>+<\/mo>8<\/mn>H<\/mi>+<\/mo><\/msup>+<\/mo>5<\/mn>e<\/mi>\u2212<\/mo><\/msup>\u2192<\/mo>M<\/mi>n<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>+<\/mo>4<\/mn>H<\/mi>2<\/mn><\/msub>O<\/mi><\/mrow>MnO_4^- + 8H^+ + 5e^- \\rightarrow Mn^{2+} + 4H_2O<\/annotation><\/semantics><\/math>MnO4\u2212\u200b+8H++5e\u2212\u2192Mn2++4H2\u200bO F<\/mi>e<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>\u2192<\/mo>F<\/mi>e<\/mi>3<\/mn>+<\/mo><\/mrow><\/msup>+<\/mo>e<\/mi>\u2212<\/mo><\/msup><\/mrow>Fe^{2+} \\rightarrow Fe^{3+} + e^-<\/annotation><\/semantics><\/math>Fe2+\u2192Fe3++e\u2212<\/p>\n\n\n\n

      Step 4: Equalize electrons and combineM<\/mi>n<\/mi>O<\/mi>4<\/mn>\u2212<\/mo><\/msubsup>+<\/mo>8<\/mn>H<\/mi>+<\/mo><\/msup>+<\/mo>5<\/mn>F<\/mi>e<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>\u2192<\/mo>M<\/mi>n<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>+<\/mo>4<\/mn>H<\/mi>2<\/mn><\/msub>O<\/mi>+<\/mo>5<\/mn>F<\/mi>e<\/mi>3<\/mn>+<\/mo><\/mrow><\/msup><\/mrow>MnO_4^- + 8H^+ + 5Fe^{2+} \\rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}<\/annotation><\/semantics><\/math>MnO4\u2212\u200b+8H++5Fe2+\u2192Mn2++4H2\u200bO+5Fe3+<\/p>\n\n\n\n

      Final Answer:<\/h3>\n\n\n\n

      M<\/mi>n<\/mi>O<\/mi>4<\/mn>\u2212<\/mo><\/msubsup>+<\/mo>8<\/mn>H<\/mi>+<\/mo><\/msup>+<\/mo>5<\/mn>F<\/mi>e<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>\u2192<\/mo>M<\/mi>n<\/mi>2<\/mn>+<\/mo><\/mrow><\/msup>+<\/mo>4<\/mn>H<\/mi>2<\/mn><\/msub>O<\/mi>+<\/mo>5<\/mn>F<\/mi>e<\/mi>3<\/mn>+<\/mo><\/mrow><\/msup><\/mrow>MnO_4^- + 8H^+ + 5Fe^{2+} \\rightarrow Mn^{2+} + 4H_2O + 5Fe^{3+}<\/annotation><\/semantics><\/math>MnO4\u2212\u200b+8H++5Fe2+\u2192Mn2++4H2\u200bO+5Fe3+<\/p>\n\n\n\n

      Balancing reactions is one of the most important parts of the Top 5 Redox Reactions Questions<\/strong>.<\/p>\n\n\n\n

      Question 4: Equivalent Weight Concept<\/h2>\n\n\n\n

      Find the equivalent weight of K<\/mi>M<\/mi>n<\/mi>O<\/mi>4<\/mn><\/msub><\/mrow>KMnO_4<\/annotation><\/semantics><\/math>KMnO4\u200b in acidic medium.<\/p>\n\n\n\n

      Detailed Explanation<\/h3>\n\n\n\n

      Equivalent weight = Molecular weight \/ n-factor<\/p>\n\n\n\n

      In acidic medium:<\/p>\n\n\n\n

        \n
      • Mn changes from +7 to +2<\/li>\n\n\n\n
      • Change in oxidation number = 5<\/li>\n<\/ul>\n\n\n\n

        Thus, n-factor = 5<\/p>\n\n\n\n

        Equivalent weight:158<\/mn>5<\/mn><\/mfrac>=<\/mo>31.6<\/mn><\/mrow>\\frac{158}{5} = 31.6<\/annotation><\/semantics><\/math>5158\u200b=31.6<\/p>\n\n\n\n

        Final Answer:<\/h3>\n\n\n\n

        Equivalent weight = 31.6<\/p>\n\n\n\n

        This concept is frequently tested in the Top 5 Redox Reactions Questions<\/strong>.<\/p>\n\n\n\n

        Question 5: Disproportionation Reaction<\/h2>\n\n\n\n

        Identify whether the following is a disproportionation reaction:2<\/mn>H<\/mi>2<\/mn><\/msub>O<\/mi>2<\/mn><\/msub>\u2192<\/mo>2<\/mn>H<\/mi>2<\/mn><\/msub>O<\/mi>+<\/mo>O<\/mi>2<\/mn><\/msub><\/mrow>2H_2O_2 \\rightarrow 2H_2O + O_2<\/annotation><\/semantics><\/math>2H2\u200bO2\u200b\u21922H2\u200bO+O2\u200b<\/p>\n\n\n\n

        Detailed Explanation<\/h3>\n\n\n\n

        In H<\/mi>2<\/mn><\/msub>O<\/mi>2<\/mn><\/msub><\/mrow>H_2O_2<\/annotation><\/semantics><\/math>H2\u200bO2\u200b, oxygen has oxidation number -1.<\/p>\n\n\n\n

        After reaction:<\/p>\n\n\n\n

          \n
        • In H<\/mi>2<\/mn><\/msub>O<\/mi><\/mrow>H_2O<\/annotation><\/semantics><\/math>H2\u200bO: O = -2 (reduction)<\/li>\n\n\n\n
        • In O<\/mi>2<\/mn><\/msub><\/mrow>O_2<\/annotation><\/semantics><\/math>O2\u200b: O = 0 (oxidation)<\/li>\n<\/ul>\n\n\n\n

          Same element undergoes both oxidation and reduction.<\/p>\n\n\n\n

          Final Answer:<\/h3>\n\n\n\n

          Yes, it is a disproportionation reaction.<\/p>\n\n\n\n

          This is a classic concept included in the Top 5 Redox Reactions Questions<\/strong>.<\/p>\n\n\n\n

          Why These Top 5 Redox Reactions Questions Are Important<\/h2>\n\n\n\n

          The Top 5 Redox Reactions Questions<\/strong> represent the most important and repeated patterns in NEET. Questions are usually:<\/p>\n\n\n\n

            \n
          • Concept-based with simple calculations<\/li>\n\n\n\n
          • Focused on oxidation number and balancing<\/li>\n\n\n\n
          • Derived from standard NCERT examples<\/li>\n<\/ul>\n\n\n\n

            If you practice these Top 5 Redox Reactions Questions<\/strong>, you can solve most redox problems in NEET with ease.<\/p>\n\n\n\n

            Tricks to Solve Redox Questions Faster<\/h2>\n\n\n\n

            To master the Top 5 Redox Reactions Questions<\/strong>, follow these tricks:<\/p>\n\n\n\n

            Always remember basic oxidation number rules. Practice half-reaction method regularly. In equivalent weight questions, quickly identify n-factor. Look for disproportionation patterns where one element undergoes both oxidation and reduction.<\/p>\n\n\n\n

            These tricks will help you solve the Top 5 Redox Reactions Questions<\/strong> quickly during the exam.<\/p>\n\n\n\n

            Common Mistakes Students Make<\/h2>\n\n\n\n

            While practicing the Top 5 Redox Reactions Questions<\/strong>, students often make errors such as:<\/p>\n\n\n\n

            Incorrect oxidation number calculation. Skipping steps while balancing equations. Confusing oxidation with reduction. Miscalculating n-factor.<\/p>\n\n\n\n

            Avoiding these mistakes will significantly improve your accuracy.<\/p>\n\n\n\n

            FAQs on Top 5 Redox Reactions Questions<\/h2>\n\n\n\n

            Is redox reactions important for NEET?<\/h3>\n\n\n\n

            Yes, it is a fundamental topic and contributes at least 1\u20132 questions.<\/p>\n\n\n\n

            Are balancing questions difficult?<\/h3>\n\n\n\n

            No, with practice of the Top 5 Redox Reactions Questions<\/strong>, they become easy.<\/p>\n\n\n\n

            What is the most important concept in redox?<\/h3>\n\n\n\n

            Oxidation number and balancing reactions are the most important.<\/p>\n\n\n\n

            How to improve speed in redox?<\/h3>\n\n\n\n

            Practice regularly and use shortcut methods.<\/p>\n\n\n\n

            Conclusion<\/h2>\n\n\n\n

            The Top 5 Redox Reactions Questions<\/strong> covered in this article include the most repeated and important concepts for NEET. From oxidation number to balancing equations, these topics form the foundation of redox chemistry.<\/p>\n\n\n\n

            By consistently practicing these Top 5 Redox Reactions Questions<\/strong>, you can improve both speed and accuracy, ensuring better performance in NEET.<\/p>\n","protected":false},"excerpt":{"rendered":"

            Top Redox Reactions Questions for NEET Redox reactions are one of the most fundamental and high-weightage topics in NEET Chemistry. Questions from oxidation number, balancing redox equations, and equivalent concept are asked almost every year. If you master the Top 5 Redox Reactions Questions, you can confidently solve both conceptual and numerical problems in the […]<\/p>\n","protected":false},"author":1,"featured_media":4388,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[127,69],"tags":[757,758,759,756,754,755],"class_list":["post-4412","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-free-study-material","category-chemistry","tag-balancing-redox-equations","tag-equivalent-weight-chemistry","tag-neet-pyqs-chemistry","tag-oxidation-number-questions","tag-redox-reactions-neet","tag-top-5-redox-reactions-questions"],"blocksy_meta":{"page_structure_type":"type-1","styles_descriptor":{"styles":{"desktop":"","tablet":"","mobile":""},"google_fonts":[],"version":6}},"_links":{"self":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4412","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/comments?post=4412"}],"version-history":[{"count":1,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4412\/revisions"}],"predecessor-version":[{"id":4413,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/posts\/4412\/revisions\/4413"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/media\/4388"}],"wp:attachment":[{"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/media?parent=4412"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/categories?post=4412"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ksquareinstitute.in\/blog\/wp-json\/wp\/v2\/tags?post=4412"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}