Question

the two half-reactions are given below.
$6\mathrm{e}^- + 3\mathrm{h}_2\mathrm{o} + \mathrm{bro}_3^- \
ightarrow \mathrm{br}^- + 6\mathrm{oh}^-$
$\mathrm{zn} \
ightarrow \mathrm{zn}^{+2} + 2\mathrm{e}^-$
what is the coefficient for $\mathrm{h}_2\mathrm{o}$ in the combined balanced net ionic equation?

Explanation:

Step1: Balance electrons in half-reactions

The first half-reaction has 6 electrons gained, the second has 2 electrons lost. Multiply the second reaction by 3 to balance electrons: \(3\text{Zn}
ightarrow 3\text{Zn}^{+2} + 6\text{e}^-\)

Step2: Add the two half-reactions

Add the modified second half-reaction to the first:
\(6\text{e}^- + 3\text{H}_2\text{O} + \text{BrO}_3^- + 3\text{Zn}
ightarrow \text{Br}^- + 6\text{OH}^- + 3\text{Zn}^{+2} + 6\text{e}^-\)
Cancel electrons: \(3\text{H}_2\text{O} + \text{BrO}_3^- + 3\text{Zn}
ightarrow \text{Br}^- + 6\text{OH}^- + 3\text{Zn}^{+2}\)

Step3: Identify \( \text{H}_2\text{O} \) coefficient

From the balanced equation, the coefficient of \( \text{H}_2\text{O} \) is 3.

Answer:

3