Question

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:

Answer 1

The next step is to balance the electrons between the reduction half-reaction and the oxidation half-reaction by multiplying the oxidation half-reaction by 3, then adding the two balanced half-reactions together.

Step-by-step explanation:

The next step toward balancing the final equation for the provided redox reactions is to balance the electrons between the two half-reactions. For the reduction half-reaction given by Cr2O72- + 14 H+ + 6 e- → 2 Cr3+ + 7 H2O, we already see that 6 electrons are added to the reactant side to balance charges. The oxidation half-reaction, H2C2O4 → 2 CO2 + 2 H+ + 2 e-, indicates 2 electrons are added to the product side. In order to combine these half-reactions into a full balanced equation, we need to equalize the number of electrons transferred in both half-reactions. This can be done by multiplying the oxidation half-reaction by 3 to match the 6 electrons in the reduction reaction.

Once the electrons are balanced, you can then add the two half-reactions together and cancel out any species that appear on both the reactant and product sides.

Answer 2

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