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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Reaction 1: I^- + ClO_3^- I_3^- + Cl^- (acidic medium) Step 1: Identify oxidation and reduction half-reactions. Oxidation: 3I^- I_3^- Reduction: ClO_3^- Cl^- Step 2: Balance oxidation half-reaction for I. 3I^- I_3^- (already balanced) Step 3: Balance charge in oxidation half-reaction. Left: charge -3, right: charge -1 Add 2e^- to right: 3I^- I_3^- + 2e^- Step 4: Balance reduction half-reaction for Cl. ClO_3^- Cl^- Step 5: Balance O atoms with H_2O. ClO_3^- Cl^- + 3H_2O Step 6: Balance H atoms with H^+. ClO_3^- + 6H^+ Cl^- + 3H_2O Step 7: Balance charge in reduction half-reaction. Left: charge ( -1 + 6 ) = +5, right: charge -1 Add 6e^- to left: ClO_3^- + 6H^+ + 6e^- Cl^- + 3H_2O Step 8: Equalize electrons (LCM of 2 and 6 is 6). Multiply oxidation by 3: 3(3I^- I_3^- + 2e^-) 9I^- 3I_3^- + 6e^- Reduction: ClO_3^- + 6H^+ + 6e^- Cl^- + 3H_2O Step 9: Add half-reactions. 9I^- + ClO_3^- + 6H^+ + 6e^- 3I_3^- + Cl^- + 3H_2O + 6e^- Step 10: Simplify (cancel 6e^-). 9I^- + ClO_3^- + 6H^+ 3I_3^- + Cl^- + 3H_2O 9I^- + ClO_3^- + 6H^+ 3I_3^- + Cl^- + 3H_2O Reaction 2: SO_3^2- + Lu(OH)_3 Lu + SO_4^2- (basic medium) Step 1: Identify oxidation and reduction half-reactions. Oxidation: SO_3^2- SO_4^2- Reduction: Lu(OH)_3 Lu Step 2: Balance oxidation half-reaction for S. SO_3^2- SO_4^2- Step 3: Balance O atoms with OH^-. SO_3^2- + 2OH^- SO_4^2- + H_2O Step 4: Balance charge in oxidation half-reaction. Left: charge ( -2 -2 ) = -4, right: charge ( -2 -0 ) = -2 Add 2e^- to right: SO_3^2- + 2OH^- SO_4^2- + H_2O + 2e^- Step 5: Balance reduction half-reaction for Lu. Lu(OH)_3 Lu Step 6: Balance O atoms with OH^-. Lu(OH)_3 Lu + 3OH^- Step 7: Balance charge in reduction half-reaction. Left: charge -3, right: charge ( 0 -3 ) = -3 (balanced for atoms, now electrons) Lu from +3 to 0, add 3e^- to left: Lu(OH)_3 + 3e^- Lu + 3OH^- Step 8: Equalize electrons (LCM of 2 and 3 is 6). Multiply oxidation by 3: 3(SO_3^2- + 2OH^- SO_4^2- + H_2O + 2e^-) 3SO_3^2- + 6OH^- 3SO_4^2- + 3H_2O + 6e^- Multiply reduction by 2: 2(Lu(OH)_3 + 3e^- Lu + 3OH^-) 2Lu(OH)_3 + 6e^- 2Lu + 6OH^- Step 9: Add half-reactions. 3SO_3^2- + 6OH^- + 2Lu(OH)_3 + 6e^- 3SO_4^2- + 3H_2O + 2Lu + 6OH^- + 6e^- Step 10: Simplify (cancel 6OH^- and 6e^-). 3SO_3^2- + 2Lu(OH)_3 3SO_4^2- + 3H_2O + 2Lu 3SO_3^2- + 2Lu(OH)_3 3SO_4^2- + 3H_2O + 2Lu