When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for   h2o, and on which side of the balanced equation should it appear?

mno4–(aq)+br–(aq)→mn2+(aq)+br2(l)
select one:

a. 1, reactant side

b. 2, product side

c. 8, product side

d. 16, reactant side

e. 4, product side

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)