Question

Balance elements being oxidized or reduced for these half-reactions:

Cr₂O₇⁻² = Cr⁺³
C₂O₄⁻² = CO₂

Answer 1

To balance the redox half-reactions for chromium and oxalate, one must balance atoms, oxygen with water (H2O), hydrogen with protons (H+), and charge with electrons to achieve correctly balanced reduction and oxidation reactions.

Step-by-step explanation:

The question concerns the balancing of redox half-reactions, which is an important concept in Chemistry, particularly when dealing with oxidation and reduction processes. To balance the given half-reactions for chromium and oxalate, we follow these steps: balancing the atoms, balancing oxygen with H2O, balancing hydrogen with H+, and finally balancing the charge by adding electrons.

Chromium Half-Reaction

Cr2O72- → Cr3+

Balance Cr atoms:
Cr2O72- → 2Cr3+

Balance oxygen by adding H2O:

Cr2O72- + 14H+ → 2Cr3+ + 7H2O

Balance charge by adding electrons:

14H+ + Cr2O72- + 6e- → 2Cr3+ + 7H2O

Oxalate Half-Reaction

C2O42- → CO2

Balance C atoms:
C2O42- → 2CO2

Balance oxygen by adding H2O if needed (not necessary in this case).

Balance charge by adding electrons:

C2O42- + 2e- → 2CO2

Once both half-reactions are balanced separately, ensure that the number of electrons gained in the reduction equals the number of electrons lost in the oxidation, and combine the half-reactions appropriately to obtain the overall balanced reaction.

Answer 2

To balance the given half-reactions, you must balance all elements except hydrogen and oxygen, balance oxygen by adding water, balance hydrogen by adding hydrogen ions, and balance charge by adding electrons. The balanced equation for the given half-reactions is 6Cr2O7^2- + 3C2O4^2- + 14H+ -> 12Cr^3+ + 6CO2 + 7H2O.

Step-by-step explanation:

To balance the given half-reactions:

1. Balance all elements except hydrogen and oxygen
2. Balance oxygen by adding water
3. Balance hydrogen by adding hydrogen ions (H+)
4. Balance charge by adding electrons

For the half-reactions:

• Oxidation: Cr2O7^2- -> 2Cr^3+ + 7H2O + 6e-
• Reduction: C2O4^2- -> 2CO2 + 2e-

The oxidation half-reaction requires 6 electrons, while the reduction half-reaction releases 2 electrons. Therefore, we need to multiply the reduction half-reaction by a factor of 3 to balance the number of electrons. The balanced equation is:

6Cr2O7^2- + 3C2O4^2- + 14H+ -> 12Cr^3+ + 6CO2 + 7H2O