Question

now lets look at the oxidation half - reaction. for the oxidation half - reaction, how many hydroxide ions go on the reactant side? ?oh⁻ + br⁻ → bro₃⁻

Explanation:

Step1: Balance Br atoms

The Br atoms are already balanced (1 Br on each side).

Step2: Balance O atoms

On the product side, there are 3 O atoms in \( \text{BrO}_3^- \). To balance O, we add \( \text{H}_2\text{O} \) on the reactant side? Wait, no, in basic solution, we use \( \text{OH}^- \) and \( \text{H}_2\text{O} \). Wait, let's do it properly.

First, let's write the unbalanced half-reaction: \( \text{Br}^- ightarrow \text{BrO}_3^- \) (oxidation, since Br goes from -1 to +5, loss of electrons).

Balance O: Add 3 \( \text{H}_2\text{O} \) on the left? No, in basic solution, we balance O by adding \( \text{OH}^- \) and \( \text{H}_2\text{O} \). Wait, the correct approach for basic solution:

1. Balance Br: \( \text{Br}^- ightarrow \text{BrO}_3^- \) (already 1 Br each)
2. Balance O: Add 3 \( \text{H}_2\text{O} \) on the left? No, wait, in basic solution, to balance O, we add \( \text{OH}^- \) to the side with less O and \( \text{H}_2\text{O} \) to the other. Wait, the product has 3 O, so we need 3 O on the left. So add 3 \( \text{H}_2\text{O} \) on the left? No, that would add H. Wait, maybe better to balance in acidic first then convert to basic.

In acidic solution:

- Balance Br: \( \text{Br}^- ightarrow \text{BrO}_3^- \)
- Balance O: Add 3 \( \text{H}_2\text{O} \) on left: \( 3\text{H}_2\text{O} + \text{Br}^- ightarrow \text{BrO}_3^- \)
- Balance H: Add 6 \( \text{H}^+ \) on right: \( 3\text{H}_2\text{O} + \text{Br}^- ightarrow \text{BrO}_3^- + 6\text{H}^+ \)
- Balance charge: Left charge: -1, Right charge: -1 + 6(+1) = +5. So add 6 \( e^- \) on right: \( 3\text{H}_2\text{O} + \text{Br}^- ightarrow \text{BrO}_3^- + 6\text{H}^+ + 6e^- \)

Now convert to basic by adding 6 \( \text{OH}^- \) to both sides:

Left: \( 3\text{H}_2\text{O} + \text{Br}^- + 6\text{OH}^- \)
Right: \( \text{BrO}_3^- + 6\text{H}^+ + 6e^- + 6\text{OH}^- \)

Combine \( \text{H}^+ \) and \( \text{OH}^- \) to form \( \text{H}_2\text{O} \):

Right: \( \text{BrO}_3^- + 6\text{H}_2\text{O} + 6e^- \)

Now simplify \( \text{H}_2\text{O} \):

Left has 3 \( \text{H}_2\text{O} \), right has 6 \( \text{H}_2\text{O} \). Subtract 3 \( \text{H}_2\text{O} \) from both sides:

Left: \( \text{Br}^- + 6\text{OH}^- \)
Right: \( \text{BrO}_3^- + 3\text{H}_2\text{O} + 6e^- \)

Wait, but the original equation given is \( \text{OH}^- + \text{Br}^- ightarrow \text{BrO}_3^- \) (missing some terms, like \( \text{H}_2\text{O} \) and electrons, but the question is about \( \text{OH}^- \) on reactant side. Wait, maybe the equation is missing \( \text{H}_2\text{O} \) and electrons, but let's check the O balance.

Wait, the given equation is \( [?]\text{OH}^- + \text{Br}^- ightarrow \text{BrO}_3^- \). Let's balance O: product has 3 O, so reactant needs 3 O from \( \text{OH}^- \)? No, \( \text{OH}^- \) has 1 O each, so 3 \( \text{OH}^- \) would give 3 O, but then we have H. Wait, maybe the correct balancing for basic solution:

The correct half-reaction (oxidation) in basic solution:

1. Balance Br: \( \text{Br}^- ightarrow \text{BrO}_3^- \)
2. Balance O: Add 3 \( \text{H}_2\text{O} \) to the left (wait, no, in basic, to balance O, we add \( \text{OH}^- \) to the side with less O. Wait, product has 3 O, so we need 3 O on left. So add 3 \( \text{OH}^- \) to left? No, that would add H. Wait, I think I made a mistake earlier. Let's do it properly for basic solution:

Steps for balancing oxidation half-reaction in basic solution:

a. Balance Br: \( \text{Br}^- ightarrow \text{BrO}_3^- \) (1 Br each)

b. Balance O: Add 3 \( \text{H}_2\text{O} \) to the left (to provide O): \( 3\text{H}_2\text{O} + \text{Br}^- ightarrow \text{BrO}_3^- \)

c. Balance H: Add 6 \( \text{H}^+ \) to the right (since left has 6 H from 3 \( \text{H}_2\text{O} \)): \( 3\text{H}_2\text{O} + \text{Br}^- ightarrow \text{BrO}_3^- + 6\text{H}^+ \)

d. Now, since it's basic solution, add 6 \( \text{OH}^- \) to both sides to neutralize \( \text{H}^+ \):

Left: \( 3\text{H}_2\text{O} + \text{Br}^- + 6\text{OH}^- \)

Right: \( \text{BrO}_3^- + 6\text{H}^+ + 6\text{OH}^- \)

e. Combine \( \text{H}^+ \) and \( \text{OH}^- \) to form \( \text{H}_2\text{O} \):

Right: \( \text{BrO}_3^- + 6\text{H}_2\text{O} \)

f. Simplify \( \text{H}_2\text{O} \):

Left has 3 \( \text{H}_2\text{O} \), right has 6 \( \text{H}_2\text{O} \). Subtract 3 \( \text{H}_2\text{O} \) from both sides:

Left: \( \text{Br}^- + 6\text{OH}^- \)

Right: \( \text{BrO}_3^- + 3\text{H}_2\text{O} \)

g. Balance charge: Left charge: -1 + 6(-1) = -7; Right charge: -1 + 0 (from \( \text{H}_2\text{O} \)) = -1. So we need to add 6 \( e^- \) to the right (since oxidation is loss of electrons, so electrons should be on the right):

\( \text{Br}^- + 6\text{OH}^- ightarrow \text{BrO}_3^- + 3\text{H}_2\text{O} + 6e^- \)

Wait, but the given equation is \( [?]\text{OH}^- + \text{Br}^- ightarrow \text{BrO}_3^- \) (missing \( \text{H}_2\text{O} \) and electrons), but the question is about the number of \( \text{OH}^- \) on the reactant side. From the balanced equation, we have 6 \( \text{OH}^- \) on the left (reactant) side. Wait, but maybe the original equation was missing some terms, but according to the balancing, the number of \( \text{OH}^- \) is 6. Wait, let's check again.

Wait, the given equation is \( [?]\text{OH}^- + \text{Br}^- ightarrow \text{BrO}_3^- \). Let's balance O: product has 3 O, so we need 3 O on the left. Each \( \text{OH}^- \) has 1 O, so 3 \( \text{OH}^- \)? No, that would give 3 O, but then H would be 3 H on left, so we need to balance H. If we add 3 \( \text{OH}^- \) on left, then we have 3 O and 3 H. Then we need to add \( \text{H}_2\text{O} \) on right? Wait, maybe the correct answer is 6. Wait, let's check the oxidation state of Br: in \( \text{Br}^- \), it's -1; in \( \text{BrO}_3^- \), it's +5. So the change in oxidation state is +6 (from -1 to +5), so 6 electrons lost. In basic solution, for each electron lost, we need to balance charge with \( \text{OH}^- \). Wait, the charge on left: \( \text{Br}^- \) is -1, \( \text{OH}^- \) is -1 each, so total charge left: -1 + (-1)*n (where n is number of \( \text{OH}^- \)). Charge on right: \( \text{BrO}_3^- \) is -1. So:

-1 + (-n) = -1 + 6 (since 6 electrons are lost, so charge increases by 6? Wait, no: oxidation is loss of electrons, so electrons are on the right. So the charge balance should be:

Left charge: -1 + (-n) (from \( \text{OH}^- \))

Right charge: -1 + 0 (if no electrons) but actually, electrons are on the right, so:

Left: -1 - n

Right: -1 - 6 (since 6 \( e^- \) are on right, each -1, so total charge right: -1 + (-6) = -7)

Wait, no, let's do charge balance properly. In the balanced equation (including electrons), the charge on left should equal charge on right plus electrons (since electrons are lost, so left charge = right charge + electrons charge).

Left charge: \( \text{Br}^- \) (-1) + 6 \( \text{OH}^- \) (6*(-1)) = -1 -6 = -7

Right charge: \( \text{BrO}_3^- \) (-1) + 3 \( \text{H}_2\text{O} \) (0) + 6 \( e^- \) (6*(-1)) = -1 -6 = -7

So that's balanced. So the number of \( \text{OH}^- \) is 6.

Step3: Conclusion

From the balanced oxidation half-reaction in basic solution, the number of \( \text{OH}^- \) ions on the reactant side is 6.

Answer:

6