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Karnataka Board PUCPUC Science Class 11

Balance the following ionic equations. CrX2OX72−+HX++IX−⟶CrX3++IX2+HX2O

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Question

Balance the following ionic equations.

\[\ce{Cr2O^{2-}7 + H^{+} + I- -> Cr^{3+} + I2 + H2O}\]

Short/Brief Note
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Solution

Dividing the equation into two half-reactions:

Oxidation half-reaction: \[\ce{I- -> I2}\]

Reduction half-reaction: \[\ce{CrO7^{2-} -> Cr^{3+}}\]

Balancing oxidation and reduction half-reactions separately as:

Oxidation half-reaction:

\[\ce{I- -> I2}\]

\[\ce{2I- -> I2}\]

\[\ce{2I- -> I2 + 2e-}\]  ......(i)

Reduction half-reaction:

\[\ce{CrO^{2-}7 -> Cr^{3+}}\]

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

\[\ce{Cr2O^{2-}7 + 6e- -> 2Cr^{3+}}\]

\[\ce{Cr2O^{2-}7 + 14H^{+} + 6e- -> \underset{(acidic medium)}{2Cr^{3+} + 7H2O}}\]  .....(ii)

To balance the electrons, multiply equation (i) by 3 and add to equation (ii)

\[\ce{Cr2O^{2-}7 + 14H^{+} + 6I- -> 2Cr^{3+} + 3I2 + 7H2O}\]

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Chapter 8: Redox Reactions - Multiple Choice Questions (Type - I) [Page 108]

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NCERT Exemplar Chemistry Exemplar [English] Class 11
Chapter 8 Redox Reactions
Multiple Choice Questions (Type - I) | Q 26.(i) | Page 108

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