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Balance the following equation by ion electron method in acidic mediuma) Cr2O7²- +C2H4O→ C2H4O2 +Cr³+​

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Explanation:

Cr2O72- + C2H2O- → C2H2O + Cr3+ ...(i) Here Cr2O72- → 2Cr3+ i.e., Cr2O72- → 2Cr3+ + 7H2O 6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O [H2O + C2H4O- → C2H4O2 + 2H+ + 3E-] x 2 ...(ii) 10H+ + Cr2O72- + 2C2H4O- → 2C2H4O2 + 2Cr3+ + 5H2ORead more on Sarthaks.com - https://www.sarthaks.com/608497/cr2o7-2-c2h4o-c2h4o2-cr-3-in-acid-solution-by-ion-electron-method


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