Question

Write Balance the half-reaction for reduction: 5Cl₂​+2Mn₂⁺ + 16OH⁻→10Cl−+2MnO₄−​+8H₂​O

Answer 1

The balanced half-reaction for reduction is 5Cl₂ + 2Mn²⁺ + 16OH⁻ + 10e⁺ → 10Cl⁻ + 2MnO₄⁻ + 8H₂O, where 10 electrons are added to the reactants' side to balance the charges.

Step-by-step explanation:

To balance the half-reaction for reduction given by the equation 5Cl₂ + 2Mn²⁺ + 16OH⁻ → 10Cl⁻ + 2MnO₄⁻ + 8H₂O, we need to add electrons to balance the charge. Here is the reduction half-reaction balanced for atoms and charge:

5Cl₂ + 2Mn²⁺ + 16OH⁻ + 10e⁺ → 10Cl⁻ + 2MnO₄⁻ + 8H₂O

First the chlorine and manganese atoms are balanced, then OH⁻ ions are added to balance the oxygen atoms, and water molecules are added to balance the hydrogen atoms. Lastly, to balance for charge, 10 electrons (10e⁺) are added to the reactants' side because the total charge on the reactants' side without electrons is +10, and on the products' side, it is 0.