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
MnO4- --> Mn2+ (reduction)
Cl- --> Cl2 (oxidation)
ii.
MnO4 --> Mn2+
2Cl- --> Cl2
iii.
MnO4- +8H+ --> Mn2+ + 4H2O
2Cl- --> Cl2
iv.
MnO4- +8H+ + 5e- --> Mn2+ + 4H2O
2Cl- --> Cl2 + 2e-
v.
2 x MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O = 2MnO4- + 16H+ + 10e- --> 2Mn2+ + 8H2O
5 x 2Cl- --> Cl2 + 2e- = 10Cl- ---> 5Cl2 + 10e-
vi.
2MnO4- + 16H+ + 10e- +10Cl- ---> 2Mn2+ + 8H2O + 5Cl2 + 10e-
2MnO4- + 16H+ + 10Cl- ---> 2Mn2+ + 8H2O + 5Cl2 <--FINAL BALANCED EQ
how did you calculate 5 electrons from the equation MnO4- +8H+ + 5e- --> Mn2+ + 4H2O?
The 5 electrons come from the manganese going from a +7 oxidation state in the permanganate (MnO4-) to a +2 oxidation state.
Part B
i.
MnO4- +2H2O + 3e- ---> MnO2 + 4OH-
2Cl- ---> Cl2 + 2e-
ii.
2 (MnO4- +2H2O + 3e- ---> MnO2 + 4OH-) = 2MnO4- + 4H2O + 6e- ---> 2MnO2 + 8OH-
3 (2Cl- ---> Cl2 + 2e-) = 6Cl- ---> 3Cl2 + 6e-
iii.
2MnO4- + 4H2O + 6Cl- + 6e- ---> 2MnO2 + 8OH- + 3Cl2 + 6e-
Balanced Equation: 2MnO4-(aq) + 4H2O(l) + 6Cl-(aq) ---> 2MnO2(s) + 8OH-(aq) + 3Cl2(g)
How in thw world are we supposed to be able to use part B u do not have any answers
I love you for this, so helpful!!!!
i. Reduction half-reaction: MnO4–(aq) + 5e– → Mn2+(aq)
Oxidation half-reaction: 2Cl–(aq) → Cl2(g)
ii. Balanced equation for atoms (except O and H):
MnO4–(aq) + 5e– + 8H+ → Mn2+(aq) + 4H2O
2Cl–(aq) → Cl2(g)
iii. Balanced equation for atoms O and H using H2O and H+:
MnO4–(aq) + 5e– + 8H+ → Mn2+(aq) + 4H2O
2Cl–(aq) → Cl2(g) + 2e– + 2H2O
iv. Balancing the charge in the half-reactions:
MnO4–(aq) + 5e– + 8H+ → Mn2+(aq) + 4H2O
2Cl–(aq) + 2e– + 2H2O → Cl2(g) + 4OH–
v. Multiplying each half-reaction to balance charges in the reaction:
2MnO4–(aq) + 10e– + 16H+ → 2Mn2+(aq) + 8H2O
10Cl–(aq) + 10e– + 10H2O → 5Cl2(g) + 20OH–
vi. Adding the equations and simplifying to get a balanced equation:
2MnO4–(aq) + 10Cl–(aq) + 16H+ → 2Mn2+(aq) + 10Cl2(g) + 8H2O
b. Assuming a reaction takes place in a basic solution to form the given products:
i. Balanced half-reactions for atoms and charge:
MnO4– + 4H2O + 3e– → MnO2 + 4OH–
2Cl– → Cl2 + 2e–
ii. Multiply to balance the charges in the reaction:
2MnO4– + 8H2O + 6e– → 2MnO2 + 8OH–
10Cl– → 5Cl2 + 10e–
iii. Adding the equations and simplifying to get a balanced equation:
2MnO4– + 16OH– + 10Cl– → 2MnO2 + 5Cl2 + 8H2O
i. Write the reduction and oxidation half-reactions (without electrons):
In the given reaction, MnO4–(aq) is reduced to Mn2+, and Cl–(aq) is oxidized to Cl2(g).
Reduction half-reaction: MnO4–(aq) → Mn2+
Oxidation half-reaction: Cl–(aq) → Cl2(g)
ii. Balance the equations for atoms (except O and H):
To balance the equations for atoms, count the number of each element on both sides of the half-reactions and adjust coefficients as needed.
Reduction half-reaction: MnO4–(aq) → Mn2+
Oxidation half-reaction: 2 Cl–(aq) → Cl2(g)
iii. Balance the equations for atoms O and H using H2O and H+:
To balance the equations for atoms O and H, add water (H2O) molecules to balance oxygen atoms and hydrogen ions (H+) to balance hydrogen atoms.
Reduction half-reaction: MnO4–(aq) + 4H2O(l) → Mn2+ + 4OH–(aq)
Oxidation half-reaction: 2 Cl–(aq) → Cl2(g)
iv. Balance the charge in the half-reactions:
To balance the charge, add electrons (e–) to the side of each half-reaction that has a higher charge.
Reduction half-reaction: MnO4–(aq) + 4H2O(l) + 3e– → Mn2+ + 4OH–(aq)
Oxidation half-reaction: 2 Cl–(aq) → Cl2(g) + 2e–
v. Multiply each half-reaction by the proper number to balance charges in the reaction:
To balance the charges, multiply the reduction half-reaction by 2 and the oxidation half-reaction by 3.
Reduction half-reaction: 2 MnO4–(aq) + 8H2O(l) + 6e– → 2 Mn2+ + 8OH–(aq)
Oxidation half-reaction: 6 Cl–(aq) → 3 Cl2(g) + 6e–
vi. Add the equations and simplify to get a balanced equation:
Finally, add the two half-reactions together and simplify by canceling out common species on both sides.
Overall balanced equation: 2 MnO4–(aq) + 8H2O(l) + 6 Cl–(aq) → 2 Mn2+ + 8OH–(aq) + 3 Cl2(g)
b. Assume a reaction takes place in a basic solution to form the given products:
i. Balance the given half-reactions for atoms and charge:
In this case, the reduction half-reaction remains the same, and the oxidation half-reaction is balanced for atoms and charge.
Reduction half-reaction: MnO4– + H2O → MnO2 + OH–
Oxidation half-reaction: 2 Cl– → Cl2
ii. Multiply to balance the charges in the reaction:
The reduction half-reaction is already balanced in terms of charge, so no additional multiplication is needed. However, the oxidation half-reaction needs to be multiplied by 2 to balance the charges.
Reduction half-reaction: MnO4– + H2O → MnO2 + OH–
Oxidation half-reaction: 4 Cl– → 2 Cl2
iii. Add the equations and simplify to get a balanced equation:
Add both half-reactions together and simplify by canceling out common species on both sides.
Overall balanced equation: MnO4– + 4 Cl– + H2O → MnO2 + Cl2 + OH–