Question

Which of the following is correctly balanced redox half reaction? Group of answer choices A. 14H+ + 9e- + Cr2O72- ⟶ Cr3+ + 7H2O B. 14H+ + 6e- + Cr2O72- ⟶ 2Cr3+ + 7H2O C. 14H+ + Cr2O72- ⟶ 2Cr3+ + 7H2O + 6e- D. 14H+ + Cr2O72- ⟶ Cr3+ + 7H2O + 9e-

Answer 1

Answer:

The correct option is: B. 14 H⁺ + 6 e⁻ + Cr₂O₇²⁻ ⟶ 2 Cr³⁺ + 7 H₂O

Explanation:

Redox reactions is an reaction in which the oxidation and reduction reactions occur simultaneously due to the simultaneous movement of electrons from one chemical species to another.

The reduction of a chemical species is represented in a reduction half- reaction and the oxidation of a chemical species is represented in a oxidation half- reaction.

To balance the reduction half-reaction for the reduction of Cr₂O₇²⁻ to Cr³⁺:

Cr₂O₇²⁻ ⟶ Cr³⁺

First the number of Cr atoms on the reactant and product side is balanced

Cr₂O₇²⁻ ⟶ 2 Cr³⁺

Now, Cr is preset in +6 oxidation state in Cr₂O₇²⁻ and +3 oxidation state in Cr³⁺. So each Cr gains 3 electrons to get reduced.

Therefore, 6 electrons are gained by 2 Cr atoms of Cr₂O₇²⁻ to get reduced.

Cr₂O₇²⁻ + 6 e⁻ ⟶ 2 Cr³⁺

Now the total charge on the reactant side is (-8) and the total charge on the product side is (+6).

From the given options it is evident that the reaction must be balanced in acidic conditions.

Therefore, to balance the total charge on the reactant and product side, 14 H⁺ is added on the reactant side.

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺

Now to balance the number of hydrogen and oxygen atoms, 7 H₂O is added on the product side.

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O

Therefore, the correct balanced reduction half-reaction is:

Cr₂O₇²⁻ + 6 e⁻ + 14 H⁺ ⟶ 2 Cr³⁺ + 7 H₂O