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Balance the following equations by the oxidation number method. FeX2++HX+ CrX2OX72−⟶CrX3++FeX3++HX2O

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प्रश्न

Balance the following equations by the oxidation number method.

\[\ce{Fe^{2+} + H^{+} + Cr2O^{2-}7 -> Cr^{3+} + Fe^{3+} + H2O}\]

टीपा लिहा
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उत्तर

We can balance the given equation by oxidation number method-

\[\ce{\overset{+2}{Fe}^{2+} + H^{+} + \overset{+6}{C}r2\overset{-2}{O^{2-}7} -> \overset{+3}{C}r^{3+} + Fe^{3+} + H2O}\]

(a) Balance the increase and decrease in O.N.

\[\ce{6\overset{+2}{Fe}^{2+} + H^{+} + \overset{+6}{C}r2\overset{-2}{O^{2-}7} -> \overset{+3}{C}r^{3+} + 6\overset{3+}{F}e^{3+} + H2O}\]

(b) Balancing H and O atoms by adding H+ and H2O molecules.

\[\ce{6\overset{+2}{Fe}^{2+} + 14H^{+} + \overset{+6}{C}r2O^{2-}7 -> 2Cr^{3+} + 6Fe^{3+} + 7H2O}\]

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पाठ 8: Redox Reactions - Multiple Choice Questions (Type - I) [पृष्ठ १०८]

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एनसीईआरटी एक्झांप्लर Chemistry Exemplar [English] Class 11
पाठ 8 Redox Reactions
Multiple Choice Questions (Type - I) | Q 24.(i) | पृष्ठ १०८

संबंधित प्रश्‍न

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