Question

the balanced half - reactions for the reaction between potassium bromate and sulfur dioxide are given below.
$6\mathrm{e}^- + 6\mathrm{h}^+ + \mathrm{bro}_3^- \
ightarrow \mathrm{br}^- + 3\mathrm{h}_2\mathrm{o}$
$2\mathrm{h}_2\mathrm{o} + \mathrm{so}_2 \
ightarrow \mathrm{so}_4^{-2} + 4\mathrm{h}^+ + 2\mathrm{e}^-$
adjust and combine the reactions. what is the coefficient of the $\mathrm{h}_2\mathrm{o}$ in the combined net ionic equation?
$? \mathrm{h}_2\mathrm{o}$

Explanation:

Step1: Balance electrons in both half - reactions

The first half - reaction has 6 electrons gained: \(6e^-+6H^++BrO_3^- ightarrow Br^- + 3H_2O\)
The second half - reaction has 2 electrons lost: \(2H_2O+SO_2 ightarrow SO_4^{2 -}+4H^++2e^-\)
To balance the number of electrons, we multiply the second half - reaction by 3. So the second half - reaction becomes: \(3\times(2H_2O + SO_2 ightarrow SO_4^{2 -}+4H^++2e^-)\)
Which is \(6H_2O+3SO_2 ightarrow 3SO_4^{2 -}+12H^++6e^-\)

Step2: Add the two half - reactions together

Now we add the first half - reaction (\(6e^-+6H^++BrO_3^- ightarrow Br^- + 3H_2O\)) and the modified second half - reaction (\(6H_2O+3SO_2 ightarrow 3SO_4^{2 -}+12H^++6e^-\))

When we add them, the electrons (\(6e^-\)) cancel out.
For \(H^+\): \(6H^+\) (from first reaction) and \(12H^+\) (from second reaction) on the left and right? Wait, no, in the first reaction \(H^+\) is on the left, in the second on the right. So total \(H^+\) on left: \(6H^+\), on right: \(12H^+\). The net \(H^+\) on right is \(12H^+- 6H^+=6H^+\)
For \(H_2O\): In the first reaction, \(3H_2O\) on the right; in the second reaction, \(6H_2O\) on the left. So net \(H_2O\) on left: \(6H_2O\), on right: \(3H_2O\). The net \(H_2O\) is \(6H_2O-3H_2O = 3H_2O\) on the left? Wait, no, let's do it step by step.

Adding the two reactions:

Left side: \(6e^-+6H^++BrO_3^-+6H_2O + 3SO_2\)

Right side: \(Br^-+3H_2O+3SO_4^{2 -}+12H^++6e^-\)

Now, cancel \(6e^-\) from both sides.

Subtract \(6H^+\) from both sides (to simplify \(H^+\)):

Left side: \(BrO_3^-+6H_2O + 3SO_2\)

Right side: \(Br^-+3H_2O+3SO_4^{2 -}+6H^+\)

Now, subtract \(3H_2O\) from both sides:

Left side: \(BrO_3^-+3H_2O + 3SO_2\)

Right side: \(Br^-+3SO_4^{2 -}+6H^+\)

Wait, maybe I made a mistake in the first calculation. Let's re - do the addition:

First reaction: \(6e^-+6H^++BrO_3^- ightarrow Br^- + 3H_2O\)

Second reaction (after multiplying by 3): \(6H_2O+3SO_2 ightarrow 3SO_4^{2 -}+12H^++6e^-\)

Now, add left sides: \(6e^-+6H^++BrO_3^-+6H_2O + 3SO_2\)

Add right sides: \(Br^-+3H_2O+3SO_4^{2 -}+12H^++6e^-\)

Now, cancel \(6e^-\):

Left: \(6H^++BrO_3^-+6H_2O + 3SO_2\)

Right: \(Br^-+3H_2O+3SO_4^{2 -}+12H^+\)

Now, for \(H^+\): Subtract \(6H^+\) from both sides. Left: \(BrO_3^-+6H_2O + 3SO_2\); Right: \(Br^-+3H_2O+3SO_4^{2 -}+6H^+\)

For \(H_2O\): Subtract \(3H_2O\) from both sides. Left: \(BrO_3^-+3H_2O + 3SO_2\); Right: \(Br^-+3SO_4^{2 -}+6H^+\)

Wait, no, the coefficient of \(H_2O\) in the combined equation: In the left side (after canceling electrons and simplifying \(H^+\) and \(H_2O\)):

The \(H_2O\) from the first reaction is 3 (on right), from the second is 6 (on left). So when we add, the total \(H_2O\) on left is 6, on right is 3. So the net \(H_2O\) is \(6 - 3=3\) on the left? Wait, no, the combined equation is:

\(BrO_3^-+3SO_2 + 3H_2O ightarrow Br^-+3SO_4^{2 -}+6H^+\)

Wait, let's check the \(H_2O\) again.

First reaction: produces \(3H_2O\) (right)

Second reaction (after multiplying by 3): consumes \(6H_2O\) (left)

So total \(H_2O\) in the net reaction: \(6H_2O\) (used) - \(3H_2O\) (produced) = \(3H_2O\) used (left side)

Wait, maybe a better way:

The two half - reactions:

1. Reduction: \(BrO_3^-+6H^++6e^- ightarrow Br^- + 3H_2O\)

2. Oxidation: \(SO_2+2H_2O ightarrow SO_4^{2 -}+4H^++2e^-\) (original oxidation half - reaction, before multiplying by 3)

Multiply oxidation by 3: \(3SO_2 + 6H_2O ightarrow 3SO_4^{2 -}+12H^++6e^-\)

Now add reduction and oxidized (multiplied) reactions:

\(BrO_3^-+6H^++6e^-+3SO_2 + 6H_2O ightarrow Br^- + 3H_2O+3SO_4^{2 -}+12H^++6e^-\)

Now, cancel \(6e^-\) from both sides.

Subtract \(6H^+\) from both sides: \(BrO_3^-+3SO_2 + 6H_2O ightarrow Br^- + 3H_2O+3SO_4^{2 -}+6H^+\)

Subtract \(3H_2O\) from both sides: \(BrO_3^-+3SO_2 + 3H_2O ightarrow Br^- + 3SO_4^{2 -}+6H^+\)

So the coefficient of \(H_2O\) is 3.

Answer:

3