MnO4–(aq) + Cl–(aq) Mn2+ + Cl2(g) (unbalanced)
i. Write the reduction and oxidation half-reactions (without electrons). (.5 point)
ii. Balance the equations for atoms (except O and H). (.5 point)
iii. Balance the equations for atoms O and H using H2O and H+. (.5 point)
iv. Balance the charge in the half-reactions. (.5 point)
v. Multiply each half-reaction by the proper number to balance charges in the reaction. (.5 point)
vi. Add the equations and simplify to get a balanced equation. (.5 point)
b. Assume a reaction takes place in a basic solution to form the given products:
MnO4–(aq) + Cl–(aq) MnO2(s) + Cl2(g) (unbalanced)
i. Balance the given half-reactions for atoms and charge. (.5 point)
MnO4– + H2O MnO2 + OH–
Cl– Cl2
ii. Multiply to balance the charges in the reaction. (.5 point)
iii. Add the equations and simplify to get a balanced equation. (.5 point)
i. The reduction half-reaction: MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)
The oxidation half-reaction: Cl–(aq) → Cl2(g)
ii. Balanced for atoms (except O and H):
MnO4–(aq) + H+(aq) + e– → Mn2+(aq) + H2O(l)
2Cl–(aq) → Cl2(g)
iii. Balanced for atoms O and H using H2O and H+:
MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)
2Cl–(aq) → Cl2(g) + 2e–
iv. Balanced charge in the half-reactions:
MnO4–(aq) + 8H+(aq) + 5e– → Mn2+(aq) + 4H2O(l)
10Cl–(aq) → 5Cl2(g) + 10e–
v. Multiplied each half-reaction to balance charges in the reaction:
2MnO4–(aq) + 16H+(aq) + 10e– → 2Mn2+(aq) + 8H2O(l)
10Cl–(aq) → 5Cl2(g) + 10e–
vi. Added the equations and simplified to get a balanced equation:
2MnO4–(aq) + 16H+(aq) + 10Cl–(aq) → 2Mn2+(aq) + 5Cl2(g) + 8H2O(l)
b. Assuming a reaction takes place in a basic solution:
i. Balanced half-reactions for atoms and charge:
MnO4– + H2O → MnO2 + OH–
2Cl– → Cl2
ii. Multiplied to balance the charges in the reaction:
2MnO4– + 2H2O → 2MnO2 + 4OH–
3Cl– → Cl2 + 6e–
iii. Added the equations and simplified to get a balanced equation:
2MnO4– + 2H2O + 3Cl– → 2MnO2 + Cl2 + 4OH–
i. To write the reduction and oxidation half-reactions, we first need to identify the species being reduced and the species being oxidized. In this case, MnO4- is being reduced to Mn2+ and Cl- is being oxidized to Cl2.
Reduction half-reaction: MnO4-(aq) + 8H+(aq) + 5e- -> Mn2+(aq) + 4H2O(l)
Oxidation half-reaction: 2Cl-(aq) -> Cl2(g)
ii. To balance the equations for atoms (except O and H), we count the number of each atom on both sides of the half-reaction and adjust coefficients as needed.
Reduction half-reaction: MnO4-(aq) + 8H+(aq) + 5e- -> Mn2+(aq) + 4H2O(l)
Oxidation half-reaction: 2Cl-(aq) -> Cl2(g)
iii. To balance the equations for atoms O and H, we can add H2O on the appropriate side of the reactions and H+ ions to balance the hydrogen atoms.
Reduction half-reaction: MnO4-(aq) + 8H+(aq) + 5e- -> Mn2+(aq) + 4H2O(l)
Oxidation half-reaction: 2Cl-(aq) -> Cl2(g) + 2e-
iv. To balance the charge in the half-reactions, we add electrons (e-) to one side of the reaction as needed.
Reduction half-reaction: MnO4-(aq) + 8H+(aq) + 5e- -> Mn2+(aq) + 4H2O(l)
Oxidation half-reaction: 2Cl-(aq) -> Cl2(g) + 2e-
v. To balance the charges in the reaction, we need to multiply each half-reaction by a proper number.
Reduction half-reaction: 2MnO4-(aq) + 16H+(aq) + 10e- -> 2Mn2+(aq) + 8H2O(l)
Oxidation half-reaction: 10Cl-(aq) -> 5Cl2(g) + 10e-
vi. Finally, we add the equations and simplify to get a balanced equation.
2MnO4-(aq) + 16H+(aq) + 10Cl-(aq) -> 2Mn2+(aq) + 5Cl2(g) + 8H2O(l)
b.
i. MnO4-(aq) + H2O --> MnO2 + OH-
Cl- --> Cl2
ii. Since OH- is produced in the reduction half-reaction, we need to multiply it by 2 to balance the charges.
MnO4-(aq) + 2H2O + 4e- --> MnO2 + 4OH-
Cl- --> Cl2 + 2e-
iii. Adding the equations and simplifying, we get the balanced equation:
MnO4-(aq) + 2H2O + Cl- --> MnO2 + Cl2 + 4OH-