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\[ Cr _{2} O _{7}^{2-}+ C _{2} O _{4}^{2-} \stackrel{ H ^{+}}{\longrightarrow} Cr ^{3+}+ CO _{2} \] Balance by onidation method |
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Answer» Cr2O72- + C2O4-2 \(\xrightarrow{H^{+}}\) Cr+3 + Co2 hall cell reaction Cr2 O72- + 6e → 2cr+3 - (reduction) C2O4-2 → 2 Co2 + 2e- (oxidation) Reduction half reaction. balancing oxygen atom - Cr2O7-2 + 6e → 2Cr+3 + 7H2O balancing Hydrogenation Cr2O72- + 6e- + 14 H+ → 2 Cr+3 + 7 H2O --------(i) Oxidation half reaction C2O4-2 → 2CO2 + 2e- Multiplying by 3 , we got, 3C2O4-2 → 6CO2 + 6e --------(ii) adding equation (i) and (ii) , we got- Cr2O7-2 + 14H+ + 3C2O4-2 → 2 Cr+3 + 6CO2 + 7H2O -----(iii) |
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