Question

When the equation below is balanced in acidic solution using the smallest whole numbers, the coefficient in front of the H 2O( l) is ________.Fe2 (aq) BrO3-(aq) → Fe3 (aq) Br-(aq)

Answer 1

Answer: In this equation, the coefficient in front of the
`H_2O(l)` is 3.

Explanation: Looking at the given reaction, Iron is oxidizing as its oxidation number is changing from +2 to +3. reduction of Br is taking place as its oxidation number is decreasing from +5 to -1.

We write the half equations and balance them for everything. Oxygen is balanced by adding
`H_2O` and hydrogen is balanced by adding
`H^+` and the charge is balanced by adding electrons.

Oxidation half reaction:
`Fe^+^2(aq)\rightarrow Fe^+^3(aq)`

It has equal Fe on both sides and there is no other atom. Only need to balance the charge since it is +2 on left side and +3 on right side. It is balanced by adding one electron to the right side.

`Fe^+^2(aq)\rightarrow Fe^+^3(aq)+1e^-`

reduction half equation:
`BrO_3^-(aq)\rightarrow Br^-(aq)`

Br is already balanced. Add three water molecules to the right side to balance oxygen.

`BrO_3^-(aq)\rightarrow Br^-(aq)+3H_2O(l)`

Add six hydrogen ions to the left side to balance hydrogen.

`BrO_3^-(aq)+6H^+(aq)\rightarrow Br^-(aq)+3H_2O(l)`

Need to add six electrons to the left side to balance the charge.

`BrO_3^-(aq)+6H^+(aq)+6e^-\rightarrow Br^-(aq)+3H_2O(l)`

next step is to make the electrons equal and for this we need to multiply the oxidation half-reaction by six.

`6Fe^+^2(aq)\rightarrow 6Fe^+^3(aq)+6e^-`

Add the two equations to get the overall equation, If anything is common then cancel this.

The over all equation is:

`6Fe^+^2(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 6Fe^+^3(aq)+Br^-(aq)+3H_2O(l)`

In this equation, the coefficient in front of the
`H_2O(l)` is 3.