This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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To balance the redox equation, we will use the half-reaction method in acidic medium. The unbalanced equation is: Cr_2O_7^2- + Fe^2+ + H^+ → Cr^3+ + Fe^3+ + H_2O Step 1: Separate the equation into two half-reactions. Oxidation half-reaction: Fe^2+ → Fe^3+ Reduction half-reaction: Cr_2O_7^2- → Cr^3+ Step 2: Balance atoms other than oxygen and hydrogen. For the oxidation half-reaction, Fe atoms are already balanced. Fe^2+ → Fe^3+ For the reduction half-reaction, balance Cr atoms: Cr_2O_7^2- → 2Cr^3+ Step 3: Balance oxygen atoms by adding H_2O. The oxidation half-reaction has no oxygen. For the reduction half-reaction, add 7H_2O to the right side to balance the 7 oxygen atoms on the left: Cr_2O_7^2- → 2Cr^3+ + 7H_2O Step 4: Balance hydrogen atoms by adding H^+. The oxidation half-reaction has no hydrogen. For the reduction half-reaction, add 14H^+ to the left side to balance the 7 × 2 = 14 hydrogen atoms on the right: 14H^+ + Cr_2O_7^2- → 2Cr^3+ + 7H_2O Step 5: Balance the charge by adding electrons (e^-). For the oxidation half-reaction: Left side charge: +2 Right side charge: +3 Add 1 electron to the right side: Fe^2+ → Fe^3+ + e^- For the reduction half-reaction: Left side charge: 14(+1) + (-2) = +12 Right side charge: 2(+3) + 0 = +6 Add 6 electrons to the left side: 6e^- + 14H^+ + Cr_2O_7^2- → 2Cr^3+ + 7H_2O Step 6: Multiply the half-reactions by appropriate integers to make the number of electrons equal. The oxidation half-reaction has 1 electron, and the reduction half-reaction has 6 electrons. Multiply the oxidation half-reaction by 6: 6(Fe^2+ → Fe^3+ + e^-) 6Fe^2+ → 6Fe^3+ + 6e^- Step 7: Add the two balanced half-reactions and cancel out common terms (electrons). (6Fe^2+ → 6Fe^3+ + 6e^-) + (6e^- + 14H^+ + Cr_2O_7^2- → 2Cr^3+ + 7H_2O) ---------------------------------------------------------------------------------------------------- Cr_2O_7^2- + 6Fe^2+ + 14H^+ → 2Cr^3+ + 6Fe^3+ + 7H_2O Step 8: Verify the balance of atoms and charge. Atoms: Cr: Left = 2, Right = 2 (Balanced) Fe: Left = 6, Right = 6 (Balanced) O: Left = 7, Right = 7 (Balanced) H: Left = 14, Right = 14 (Balanced) Charge: Left side: (-2) + 6(+2) + 14(+1) = -2 + 12 + 14 = +24 Right side: 2(+3) + 6(+3) + 0 = +6 + 18 = +24 (Balanced) The balanced equation is: Cr_2O_7^2- + 6Fe^2+ + 14H^+ → 2Cr^3+ + 6Fe^3+ + 7H_2O The coefficients of Cr_2O_7^2- and Fe^2+ are 1 and 6, respectively. The final answer is A.