chemistry

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)

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  1. 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

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  2. how did you calculate 5 electrons from the equation MnO4- +8H+ + 5e- --> Mn2+ + 4H2O?

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

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