Question

H2C2O4 + K2Cr2O7 + H2SO4 = Cr2(SO4)3 + K2SO4 + CO2 + H2O balance using ion electron method​

Answer 1

Answer:

Explanation:

3 H2C2O4 + K2Cr2O7 + 4 H2SO4 → Cr2(SO4)3 + K2SO4 + 6 CO2 + 7 H2O

This is an oxidation-reduction (redox) reaction:

6 CIII - 6 e- → 6 CIV (oxidation)

2 CrVI + 6 e- → 2 CrIII (reduction)

H2C2O4 is a reducing agent, K2Cr2O7 is an oxidizing agent.

Answer 2

The balanced equation to the following reaction is as follow -

3H₂C₂O₄ + K₂Cr₂O₇ + 4H₂SO₄ = Cr₂(SO₄)₃ + K₂SO₄ + 6CO₂ + 7H₂O

• A balanced equation is the equation of reactant and products in which the law of conservation of masses is being followed, that is the number of moles of reactant should be equal to the number of moles of product.

• Balancing using the ion-electron method is the method that involves separating the whole equation into two parts. The first is oxidation and the other is reduction.
• Each of these is balanced separately and then combined to form a balanced redox reaction.
• We strip the equation that involves all common ions like 0H⁻, H⁺, H₂0
• Equalize all the electrons by adding sufficient hydrogen ions, and electrons to balance the electron on both halves.
• In the process of equalizing the electrons, we get half of a balance reaction, and then we sum the whole of the reaction to get the desired reaction.
• It is also called an ion-electron method and half-reaction method.