Question

reduction half - reaction
3e⁻ + 2h₂o + mno₄⁻ → mno₂ + 4oh⁻
oxidation half - reaction
6oh⁻ + br⁻ → bro₃⁻ + 3h₂o + 6e⁻
what do you need to multiply the reduction half - reaction by in order to balance the electron charge?

Explanation:

Step1: Identify electrons in each half-reaction

Reduction half - reaction has \(3e^-\), oxidation half - reaction has \(6e^-\).

Step2: Find the multiplier

To balance electrons, we need the number of electrons in reduction and oxidation half - reactions to be equal. Let the multiplier for reduction half - reaction be \(x\). We want \(3x = 6\). Solving for \(x\), we get \(x=\frac{6}{3}=2\).

Answer:

2