Question

Balance charge using e⁻ for these half-reactions (only consider the elements being ox/red):

Cr₂O₇⁻² = 2Cr⁺³
C₂O₄⁻² = 2CO₂

Cr = +6
C = +3

Cr = +3
C = +4

Answer 1

In the reduction half-reaction of Cr2O72- to 2Cr3+, six electrons are added to the reactant side to balance the charge, resulting in a balanced equation. The oxidation half-reaction for C2O42- to 2CO2 does not require additional electrons to balance the charge.

Step-by-step explanation:

According to the given half-reactions, we need to balance the charge by adding electrons. For the reduction half-reaction of converting Cr2O72- to 2Cr3+, the net charge goes from +12 on the reactants' side to +6 on the products' side. Therefore, we add six electrons to the reactant side to balance the charge:

6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O

For the oxidation half-reaction, C2O42- is oxidized to 2CO2. The net charge on both sides is zero thus no electrons are needed to balance the charge in this case.