Question

the products oh⁻ and h⁺ combine to make six h₂o molecules, allowing for cancellation of the three h₂o molecules on the reactant side.
6oh⁻ + 3h₂o + br⁻ → bro₃⁻ + 6h⁺ + 6oh⁻
6oh⁻ + 3h₂o + br⁻ → bro₃⁻ + 6h₂o
6oh⁻ + br⁻ → bro₃⁻ + 3h₂o
balance the charges, how many electrons must be added to the product side?
6oh⁻ + br⁻ → bro₃⁻ + 3h₂o + ?e⁻

Explanation:

Step1: Calculate charge on reactant side

Reactants: \(6\ce{OH^-}\) (charge: \(6\times(-1) = -6\)) and \(\ce{Br^-}\) (charge: \(-1\)). Total charge: \(-6 + (-1) = -7\).

Step2: Calculate charge on product side (excluding electrons)

Products: \(\ce{BrO_3^-}\) (charge: \(-1\)) and \(3\ce{H_2O}\) (charge: \(0\)). Total charge: \(-1 + 0 = -1\).

Step3: Determine electrons needed

Let \(x\) be the number of electrons (each electron has charge \(-1\)). We need: \(-1 + (-x) = -7\) (since reactant charge = product charge + electron charge). Solving: \(-x = -7 + 1 = -6\) → \(x = 6\).

Answer:

6