Balance the following redox reaction in acidic solution and determine how many water molecules are in the equation.
MnO₄^- + ClO^-
ClO₃^- + Mn²⁺ + H₂O

Question

Balance the following redox reaction in acidic solution and determine how many water molecules are in the equation. MnO₄^- + ClO^- ClO₃^- + Mn²⁺ + H₂O

Quizplus · Accepted Answer

First, we assign oxidation states to all the elements: Mn(+II) + (S in ₄^-)(-II) + Cl(+VII) + (S in ClO^-)(-II) → Mn(+VII) + (S in Mn²⁺)(+II) + Cl(-I) + (S in H₂O)(-II) Next, we identify which elements are oxidized and reduced, and determine the number of electrons transferred: Mn(+II) → Mn(+VII) (oxidized, loses 5 electrons) Cl(+VII) → Cl(-I) (reduced, gains 8 electrons) We then balance the electrons by multiplying the manganese species by 5 and the chloride species by 8: 5Mn(+II) + 8Cl(+VII) + (S in ₄^-)(-II) + (S in ClO^-)(-II) → 5Mn(+VII) + 8Cl(-I) + (S in Mn²⁺)(+II) + (S in H₂O)(-II) Next, we balance the sulfur atoms by adding 14H+ to the left side and 12H+ to the right side: 5Mn(+II) + 8Cl(+VII) + 14H+ + (S in ₄^-)(-II) + (S in ClO^-)(-II) → 5Mn(+VII) + 8Cl(-I) + 12H+ + (S in Mn²⁺)(+II) + (S in H₂O)(-II) We can then balance the hydrogen atoms by adding water molecules: 5Mn(+II) + 8Cl(+VII) + 14H+ + (S in ₄^-)(-II) + (S in ClO^-)(-II) → 5Mn(+VII) + 8Cl(-I) + 12H2O + (S in Mn²⁺)(+II) + (S in H₂O)(-II) Therefore, there are 12 water molecules in the balanced equation, and the answer is C.

Balance the Following Redox Reaction in Acidic Solution and Determine