Question
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. 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)
Answers
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
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?
How in thw world are we supposed to be able to use part B u do not have any answers
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)
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)
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