Question

the redox reaction given below occurs in basic solution.
zn + bro₃⁻ → zn⁺² + br⁻
balance the half - reaction:
bro₃⁻ → br⁻
how many electrons are transferred?
?e⁻

Explanation:

Step1: Balance atoms other than O and H

The bromine atoms are already balanced: \( \text{BrO}_3^- ightarrow \text{Br}^- \) has 1 Br on each side.

Step2: Balance O by adding \( \text{H}_2\text{O} \)

There are 3 O on the left, so add \( 3\text{H}_2\text{O} \) to the right:
\( \text{BrO}_3^- ightarrow \text{Br}^- + 3\text{H}_2\text{O} \)

Step3: Balance H by adding \( \text{H}^+ \)

Now there are 6 H on the right (from \( 3\text{H}_2\text{O} \)), so add \( 6\text{H}^+ \) to the left:
\( \text{BrO}_3^- + 6\text{H}^+ ightarrow \text{Br}^- + 3\text{H}_2\text{O} \)

Step4: Since it's basic solution, add \( \text{OH}^- \) to neutralize \( \text{H}^+ \)

Add \( 6\text{OH}^- \) to both sides:
\( \text{BrO}_3^- + 6\text{H}^+ + 6\text{OH}^- ightarrow \text{Br}^- + 3\text{H}_2\text{O} + 6\text{OH}^- \)
Simplify \( \text{H}^+ + \text{OH}^- = \text{H}_2\text{O} \):
\( \text{BrO}_3^- + 3\text{H}_2\text{O} ightarrow \text{Br}^- + 3\text{H}_2\text{O} + 6\text{OH}^- \)
Cancel \( 3\text{H}_2\text{O} \) from both sides:
\( \text{BrO}_3^- ightarrow \text{Br}^- + 6\text{OH}^- \) Wait, no, correction: Wait, after adding \( 6\text{OH}^- \) to both sides, the left side becomes \( \text{BrO}_3^- + 6\text{H}_2\text{O} \) (since \( 6\text{H}^+ + 6\text{OH}^- = 6\text{H}_2\text{O} \)), and the right side is \( \text{Br}^- + 3\text{H}_2\text{O} + 6\text{OH}^- \). Wait, I made a mistake earlier. Let's redo step 4 properly.

Correct step 4:
After step 3: \( \text{BrO}_3^- + 6\text{H}^+ ightarrow \text{Br}^- + 3\text{H}_2\text{O} \)
In basic solution, add \( 6\text{OH}^- \) to both sides:
Left: \( \text{BrO}_3^- + 6\text{H}^+ + 6\text{OH}^- = \text{BrO}_3^- + 6\text{H}_2\text{O} \)
Right: \( \text{Br}^- + 3\text{H}_2\text{O} + 6\text{OH}^- \)
So the equation becomes:
\( \text{BrO}_3^- + 6\text{H}_2\text{O} ightarrow \text{Br}^- + 3\text{H}_2\text{O} + 6\text{OH}^- \)
Cancel \( 3\text{H}_2\text{O} \):
\( \text{BrO}_3^- + 3\text{H}_2\text{O} ightarrow \text{Br}^- + 6\text{OH}^- \)

Now, balance charge by adding electrons. Let's find the charge on each side.

Left side: \( \text{BrO}_3^- \) has charge -1, \( 3\text{H}_2\text{O} \) is neutral, so total charge -1.

Right side: \( \text{Br}^- \) has charge -1, \( 6\text{OH}^- \) has total charge -6, so total charge -7.

Wait, no, wait: Wait, in the corrected step 4, after canceling \( 3\text{H}_2\text{O} \), the equation is \( \text{BrO}_3^- + 3\text{H}_2\text{O} ightarrow \text{Br}^- + 6\text{OH}^- \)? No, that can't be. Wait, I messed up the balancing. Let's start over for charge balance.

Alternative approach: After step 3 (acidic), the equation is \( \text{BrO}_3^- + 6\text{H}^+ ightarrow \text{Br}^- + 3\text{H}_2\text{O} \)

Charge on left: \( -1 + 6(+1) = +5 \)

Charge on right: \( -1 + 0 = -1 \)

To balance charge, we need to add electrons to the left (since it's more positive) to make charges equal.

Let \( x \) be the number of electrons. So:

\( +5 - x = -1 \)

Solving for \( x \): \( x = 6 \)

So add 6 electrons to the left:

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

Now, for basic solution, we convert \( \text{H}^+ \) to \( \text{H}_2\text{O} \) by adding \( \text{OH}^- \), but the electron transfer is determined from the charge balance in the acidic step (or we can check the oxidation state).

Oxidation state of Br in \( \text{BrO}_3^- \): Let Br be \( x \), so \( x + 3(-2) = -1 \) → \( x -6 = -1 \) → \( x = +5 \)

Oxidation state of Br in \( \text{Br}^- \) is -1.

So the change in oxidation state is \( +5 \) to \( -1 \), which is a gain of 6 electrons (since \( -1 - (+5) = -6 \), so 6 electrons are gained).

Answer:

6