Complete and balance the following redox reaction in acidic solution. Be sure to include the proper phases for all species within the reaction.Mn²⁺(aq)+BiO₃⁻(aq)→MnO₄⁻(aq)+Bi³⁺(aq)☐⁴⁻☐³⁻☐²⁻☐⁻☐⁺☐²⁺☐³⁺☐⁴⁺1234567890☐₁☐₂☐₃☐₄☐₅☐₆☐₇☐₈☐₉☐₀+()→⇌(s)(l)(g)(aq)HH₂OOH⁻H₃O⁺Bie⁻MnOH⁺Reset

The first step in balancing a redox reaction is to identify the oxidation and reduction half-reactions.

In this case, Mn²⁺ is being oxidized to MnO₄⁻ and BiO₃⁻ is being reduced to Bi³⁺.

The half-reactions are:

Mn²⁺ → MnO₄⁻

BiO₃⁻ → Bi³⁺

Next, we balance the atoms other than O and H in each half-reaction. The Mn and Bi are already balanced.

Then, we balance the O atoms by adding H₂O:

Mn²⁺ → MnO₄⁻ + 4H₂O

BiO₃⁻ + 2H₂O → Bi³⁺

Next, we balance the H atoms by adding H⁺:

Mn²⁺ + 8H⁺ → MnO₄⁻ + 4H₂O

BiO₃⁻ + 2H₂O → Bi³⁺ + 6H⁺

Then, we balance the charge by adding electrons:

Mn²⁺ + 8H⁺ + 5e⁻ → MnO₄⁻ + 4H₂O

BiO₃⁻ + 2H₂O + 2e⁻ → Bi³⁺ + 6H⁺

Finally, we multiply the half-reactions by factors that will make the number of electrons in the two half-reactions equal, and then add the half-reactions together:

2(Mn²⁺ + 8H⁺ + 5e⁻ → MnO₄⁻ + 4H₂O)

5(BiO₃⁻ + 2H₂O + 2e⁺ → Bi³⁺ + 6H⁺)

This gives:

2Mn²⁺ + 16H⁺ + 10e⁻ + 5BiO₃⁻ + 10H₂O → 2MnO₄⁻ + 8H₂O + 5Bi³⁺ + 30H⁺

Simplifying, we get:

2Mn²⁺(aq) + 5BiO₃⁻(aq) + 14H⁺(aq) → 2MnO₄⁻(aq) + 5Bi³⁺(aq) + 7H₂O(l)

So, the balanced redox reaction in acidic solution is:

2Mn²⁺(aq) + 5BiO₃⁻(aq) + 14H⁺(aq) → 2MnO₄⁻(aq) + 5Bi³⁺(aq) + 7H₂O(l)