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प्रश्न
Balance the following ionic equations.
\[\ce{Cr2O^{2-}7 + H^{+} + I- -> Cr^{3+} + I2 + H2O}\]
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उत्तर
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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