Respuesta :
Answer: In this equation, the coefficient in front of the [tex]H_2O(l)[/tex] 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 [tex]H_2O[/tex] and hydrogen is balanced by adding [tex]H^+[/tex] and the charge is balanced by adding electrons.
Oxidation half reaction: [tex]Fe^+^2(aq)\rightarrow Fe^+^3(aq)[/tex]
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.
[tex]Fe^+^2(aq)\rightarrow Fe^+^3(aq)+1e^-[/tex]
reduction half equation: [tex]BrO_3^-(aq)\rightarrow Br^-(aq)[/tex]
Br is already balanced. Add three water molecules to the right side to balance oxygen.
[tex]BrO_3^-(aq)\rightarrow Br^-(aq)+3H_2O(l)[/tex]
Add six hydrogen ions to the left side to balance hydrogen.
[tex]BrO_3^-(aq)+6H^+(aq)\rightarrow Br^-(aq)+3H_2O(l)[/tex]
Need to add six electrons to the left side to balance the charge.
[tex]BrO_3^-(aq)+6H^+(aq)+6e^-\rightarrow Br^-(aq)+3H_2O(l)[/tex]
next step is to make the electrons equal and for this we need to multiply the oxidation half-reaction by six.
[tex]6Fe^+^2(aq)\rightarrow 6Fe^+^3(aq)+6e^-[/tex]
Add the two equations to get the overall equation, If anything is common then cancel this.
The over all equation is:
[tex]6Fe^+^2(aq)+BrO_3^-(aq)+6H^+(aq)\rightarrow 6Fe^+^3(aq)+Br^-(aq)+3H_2O(l)[/tex]
In this equation, the coefficient in front of the [tex]H_2O(l)[/tex] is 3.
