Complete and balance the following half-reaction in acidic solution: MnO_4^- → Mn^{2+}

• A. MnO_4^- + 4H^+ + 3e^- → Mn^{2+} + 2H_2O
• B. MnO_4^- + 6H^+ + 4e^- → Mn^{2+} + 3H_2O
• C. MnO_4^- + 8H^+ + 5e^- → Mn^{2+} + 4H_2O
• D. MnO_4^- + 2H^+ + 2e^- → Mn^{2+} + H_2O

Answer: (C)

Verified step by step guidance

1. Identify the species involved in the half-reaction: permanganate ion (MnO_4^-) is reduced to manganese ion (Mn^{2+}) in acidic solution.
2. Balance the manganese atoms on both sides. Since there is one Mn atom in MnO_4^- and one in Mn^{2+}, manganese is already balanced.
3. Balance the oxygen atoms by adding water (H_2O) molecules to the side deficient in oxygen. MnO_4^- has 4 oxygen atoms, so add 4 H_2O molecules to the product side.
4. Balance the hydrogen atoms by adding H^+ ions to the side deficient in hydrogen. Since 4 H_2O molecules contain 8 hydrogen atoms, add 8 H^+ ions to the reactant side.
5. Balance the charge by adding electrons (e^-) to the more positive side to equalize the total charge on both sides. Calculate the total charge on each side and add 5 electrons to the reactant side to balance the charges.