Question

potassium bromate reacts with sulfur dioxide in acidic solution. balance the reduction half - reaction below.
bro₃⁻ → br⁻
how many electrons are transferred in this half - reaction?
? e⁻

Explanation:

Step1: Determine oxidation states

In \( \text{BrO}_3^- \), let the oxidation state of Br be \( x \). Oxygen has an oxidation state of -2 (except in peroxides, etc.). So, \( x + 3(-2) = -1 \) (since the overall charge of \( \text{BrO}_3^- \) is -1). Solving: \( x - 6 = -1 \) → \( x = +5 \). In \( \text{Br}^- \), the oxidation state of Br is -1.

Step2: Calculate electron transfer

The change in oxidation state of Br is from +5 to -1. The number of electrons gained (since it's a reduction half - reaction, gain of electrons) is \( 5 - (-1)=6 \). So, 6 electrons are transferred.

Answer:

6