Which of the following is the correctly balanced redox reaction for Cr2O7^2-(aq) + I^-(aq) → Cr^3+(aq) + IO3^-(aq) in acidic solution?

• A. Cr2O7^2- + 6I^- + 12H^+ → 2Cr^3+ + 3IO3^- + 6H2O
• B. Cr2O7^2- + 6I^- + 8H^+ → 2Cr^3+ + 3IO3^- + 4H2O
• C. Cr2O7^2- + 6I^- + 14H^+ → 2Cr^3+ + 6IO3^- + 7H2O
• D. Cr2O7^2- + 6I^- + 10H^+ → 2Cr^3+ + 3IO3^- + 5H2O

Answer: (A)

Verified step by step guidance

1. Identify the oxidation and reduction half-reactions. In this case, Cr2O7^2- is reduced to Cr^3+, and I^- is oxidized to IO3^-.
2. Balance the atoms in each half-reaction. For the reduction half-reaction, balance Cr atoms first, then balance oxygen atoms by adding H2O, and finally balance hydrogen atoms by adding H^+. For the oxidation half-reaction, balance iodine atoms first, then balance oxygen atoms by adding H2O, and finally balance hydrogen atoms by adding H^+.
3. Balance the charges in each half-reaction by adding electrons. For the reduction half-reaction, add electrons to the left side to balance the positive charge. For the oxidation half-reaction, add electrons to the right side to balance the negative charge.
4. Equalize the number of electrons transferred in both half-reactions by multiplying the half-reactions by appropriate coefficients.
5. Add the balanced half-reactions together, ensuring that electrons cancel out, and simplify the equation by combining like terms. Verify that the number of atoms and charges are balanced on both sides of the equation.