Question

the electrons are on the reactant side. how many electrons are needed to balance the charge in the half - reaction? ?e⁻ + 6h⁺ + bro₃⁻ → br⁻ + 3h₂o

Explanation:

Step1: Calculate charge on left side

The left side has \(6\text{H}^+\) (each \(+1\), so total \(+6\)), \(\text{BrO}_3^-\) (charge \(-1\)), and \(x\) electrons (each \(-1\), so total \(-x\)). So total left charge: \(+6 - 1 - x=+5 - x\).

Step2: Calculate charge on right side

The right side has \(\text{Br}^-\) (charge \(-1\)) and \(3\text{H}_2\text{O}\) (neutral, charge \(0\)). So total right charge: \(-1\).

Step3: Set charges equal and solve for x

For charge balance, left charge = right charge:
\(+5 - x=-1\)
Subtract \(5\) from both sides: \(-x=-6\)
Multiply by \(-1\): \(x = 6\).

Answer:

\(6\)