Balance the following ionic equation in acidic medium by the ion-electron method, showing both half-reactions

Exercise · Q10

\text{Cr}_2\text{O}_7^{2-} + \text{I}^{-} \to \text{Cr}^{3+} + \text{I}_2

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Reduction half: $\text{Cr}_2\text{O}_7^{2-}\to2\text{Cr}^{3+}$ (Cr already balanced); balance O with $7\text{H}_2\text{O}$ , H with $14\text{H}^{+}$ : $\text{Cr}_2\text{O}_7^{2-}+14\text{H}^{+}\to2\text{Cr}^{3+}+7\text{H}_2\text{O}$ ; balance charge ( $+12$ left vs $+6$ right) with $6e^{-}$ on the left: $\text{Cr}_2\text{O}_7^{2-}+14\text{H}^{+}+6e^{-}\to2\text{Cr}^{3+}+7\text{H}_2\text{O}$ .\n\nOxidation half: $2\text{I}^{-}\to\text{I}_2$ (I balanced 2=2); balance charge ( $-2$ left vs $0$ right) with $2e^{-}$ on the right: $2\text{I}^{-}\to\text{I}_2+2e^{-}$ .\n\nCombine: LCM of 6 and 2 is 6; the reduction half already has 6 electrons, multiply the oxidation half by 3: $6\text{I}^{-}\to3\text{I}_2+6e^{-}$ . Add and cancel the $6e^-$ : $\text{Cr}_2\text{O}_7^{2-} …

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Q.Balance the following ionic equation in acidic medium by the ion-electron method, showing both half-reactions separately: $\text{Cr}_2\text{O}_7^{2-} + \text{I}^{-} \to \text{Cr}^{3+} + \text{I}_2$ .

Concept understanding — Balancing Redox Reactions

Q.Balance the following ionic equation in acidic medium by the ion-electron method, showing both half-reactions separately: Cr2O72−+I−→Cr3++I2\text{Cr}_2\text{O}_7^{2-} + \text{I}^{-} \to \text{Cr}^{3+} + \text{I}_2Cr2​O72−​+I−→Cr3++I2​.

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Q.Balance the following ionic equation in acidic medium by the ion-electron method, showing both half-reactions separately: $\text{Cr}_2\text{O}_7^{2-} + \text{I}^{-} \to \text{Cr}^{3+} + \text{I}_2$ .

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