Given the two half-reactions, cr2o72- + 14 h+ + 6 e− imported asset 2 cr3+ + 7 h2o and h2c2o4 imported asset 2 co2 + 2 h+ + 2 e−, the next step you should take toward balancing the final equation is:

Two half reactions:

14H⁺ + Cr₂O₇²⁻+ 6e⁻ --> 2Cr³⁺ + 7H₂O (i)

H₂C₂O₄ --> 2CO₂(g) + 2H⁺ + 2e⁻ (ii)

To balance the equations, the next step is balance the number the electrons in both the reactions, by multiplying equation (ii) by 3.

(H₂C₂O₄ --> 2CO₂(g) + 2H⁺ + 2e⁻)×3

=(3H₂C₂O₄ --> 6CO₂(g) + 6H⁺ + 6e⁻)(iii)

Now second step is the addition of both the reactions.

14H⁺ + Cr₂O₇²⁻+ 6e⁻ --> 2Cr³⁺ + 7H₂O (i)

3H₂C₂O₄ --> 6CO₂(g) + 6H⁺ + 6e⁻(iii)

Balance equation:

Cr₂O₇²⁻ + 3H₂C₂O₄ + 8H⁺ > 2Cr³⁺ + 6CO₂(g) + 7H₂O