Balance the following ionic equation in acidic medium by the ion-electron (half-reaction) method, showing both half-reactions separately: $\text{MnO}_4^{-} + \text{C}_2\text{O}_4^{2-} \to \text{Mn}^{2+} + \text{CO}_2$.

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

Example · Example 9

\text{MnO}_4^{-} + \text{C}_2\text{O}_4^{2-} \to \text{Mn}^{2+} + \text{CO}_2

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Reduction half: $\text{MnO}_4^{-}\to\text{Mn}^{2+}$ ; balance O with $4\text{H}_2\text{O}$ , then H with $8\text{H}^+$ : $\text{MnO}_4^{-}+8\text{H}^{+}\to\text{Mn}^{2+}+4\text{H}_2\text{O}$ ; balance charge ( $+7$ left vs $+2$ right) with $5e^{-}$ added to the left: $\text{MnO}_4^{-}+8\text{H}^{+}+5e^{-}\to\text{Mn}^{2+}+4\text{H}_2\text{O}$ .\n\nOxidation half: $\text{C}_2\text{O}_4^{2-}\to2\text{CO}_2$ (C already balanced 2=2, O already balanced 4=4); balance charge ( $-2$ left vs $0$ right) with $2e^{-}$ added to the right: $\text{C}_2\text{O}_4^{2-}\to2\text{CO}_2+2e^{-}$ .\n\nCombine: LCM of 5 and 2 electrons is 10; multiply the reduction half by 2 and the oxidation half by 5, then add and cancel the $10e^-$ : $2\text{MnO}_4^{-} + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^{+} \to 2 …

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Q.Balance the following ionic equation in acidic medium by the ion-electron (half-reaction) method, showing both half-reactions separately: $\text{MnO}_4^{-} + \text{C}_2\text{O}_4^{2-} \to \text{Mn}^{2+} + \text{CO}_2$ .

Concept understanding — Balancing Redox Reactions

Q.Balance the following ionic equation in acidic medium by the ion-electron (half-reaction) method, showing both half-reactions separately: MnO4−+C2O42−→Mn2++CO2\text{MnO}_4^{-} + \text{C}_2\text{O}_4^{2-} \to \text{Mn}^{2+} + \text{CO}_2MnO4−​+C2​O42−​→Mn2++CO2​.

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Q.Balance the following ionic equation in acidic medium by the ion-electron (half-reaction) method, showing both half-reactions separately: $\text{MnO}_4^{-} + \text{C}_2\text{O}_4^{2-} \to \text{Mn}^{2+} + \text{CO}_2$ .

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