Question

\[ Cr _{2} O _{7}^{2-}+ C _{2} O _{4}^{2-} \stackrel{ H ^{+}}{\longrightarrow} Cr ^{3+}+ CO _{2} \] Balance by onidation method

Answer 1

Cr₂O₇^2- + C₂O₄^-2 \(\xrightarrow{H^{+}}\) Cr⁺³ + Co₂

hall cell reaction

Cr₂ O₇^2- + 6e → 2cr⁺³ - (reduction)

C₂O4^-2 → 2 Co₂ + 2e^- (oxidation)

Reduction half reaction.

balancing oxygen atom - Cr₂O₇^-2 + 6e → 2Cr⁺³ + 7H₂O

balancing Hydrogenation

Cr₂O₇^2- + 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)