Question

Part A) Balance the following using the ½ reaction method

Permanganate ion plus bromide ion yields manganese oxide and BrO3 -      (basic solution)

Part B) Given the following data and using Hess’ Law, Calculate change in Enthalpy for the reaction

Overall reaction :     NO(g)   + O (g)   ------→ NO2(g)

                               2O3(g)    ------→ 3O2(g)                                   ∆H = - 427 kJ

                           O2(g)     ------→ 2O(g)                                  ∆H = 495 kJ

                            NO(g)   + O3 (g)   ------→ NO2 (g) + O2 (g)      ∆H = - 199 kJ

Answer 1

Step 1: Write the “skeleton” half-reactions

ox: Br⁻ −→ BrO3 ^–

red: MnO₄⁻ −→ MnO₂

Step 2: Balance each half-reaction “atomically”

ox: Br⁻ + 3 H₂O −→ BrO₃^–

ox: Br⁻ + 3 H₂O −→ BrO₃⁻ + 6 H⁺

red: MnO₄⁻ −→ MnO₂ + 2 H₂O

red: MnO₄⁻ + 4 H⁺ −→ MnO₂ + 2 H₂O

Step 3: Balance the electric charges by adding electrons

ox: Br⁻ + 3 H₂O −→ BrO₃⁻ + 6 H⁺ + 6 e⁻

red: MnO₄⁻ + 4 H⁺ + 3 e⁻ −→ MnO₂ + 2 H₂O

Step 4: Prepare the two half-equations for summation by making the number of electrons the same in both, i.e, find the least common multiple

ox: Br⁻ + 3 H₂O −→ BrO₃⁻ + 6 H⁺ + 6 e⁻

2 × red: 2 MnO₄⁻ + 8 H⁺ + 6 e⁻ −→ 2 MnO₂ + 4 H₂O

Step 5: Combine the two half-reactions

Br⁻ + 3 H₂O + 2 MnO₄⁻ + 8 H⁺ + 6 e⁻ −→ BrO₃⁻ + 6 H⁺ + 6 e⁻ + 2 MnO₂ + 4 H₂O

Step 6: Simplify

Br⁻ + 2 MnO₄⁻ + 2 H⁺ −→ BrO₃⁻ + 2 MnO₂ + H₂O

Step 6a: Change to basic solution by adding as many OH− to both sides as there are H+

Br− + 2 MnO₄⁻ + 2 H⁺ + 2 OH⁻ −→ BrO₃⁻ + 2 MnO₂ + H₂O + 2 OH⁻

“neutralization”: Combine the H+ and the OH− to form H2O

Br− + 2 MnO₄⁻ + 2 H₂O −→ BrO₃⁻ + 2 MnO₂ + H₂O + 2 OH⁻

simplify     Br⁻ + 2 MnO₄⁻ + H₂O −→ BrO₃⁻ + 2 MnO₂ + 2 OH⁻

Step 7: Indicate the state of each species

Br⁻(aq) + 2 MnO₄⁻(aq) + H₂O(l) −→ BrO₃⁻(aq) + +2 MnO₂(s) + 2 OH⁻(aq)