Anonymous
Anonymous asked in Science & MathematicsChemistry · 6 years ago

chemistry balancing questions 5points?

Balance the follwing oxidation-reduction equations. Assume each reaction takes place in basic solution.

(A) SO3^(2-) + Cl2---------> SO4^(2-) + Cl^(1-)

(B) MnO2 +O2--------> MnO4^(2-) + H2O

(C) MnO4^(1-) + BrO2^(1-) --------> MnO2 + BrO4^(1-)

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  • Lexi R
    Lv 7
    6 years ago
    Best Answer

    (C) MnO4^(1-) + BrO2^(1-) --------> MnO2 + BrO4^(1-)

    split it into the two half equations

    MnO4^- is going to MnO2

    MnO4^- --> MnO2

    balance the O by adding H2O to the side deficiant in O

    MnO4^- --> MnO2 + 2H2O

    balance the H by adding H+ to the other side

    MnO4^- + 4H+ --> MnO2 + 2H2O

    convert to basic conditions by adding the same number of OH- as you have H+ to BOTH sides of the equation

    MnO4^- + 4H+ +4OH- --> MnO2 + 2H2O + 4OH-

    where you have a H+ and OH- on both sides = H2O

    MnO4^- + 4H2O --> MnO2 + 2H2O + 4OH-

    cancel out the H2O that occurs on both sides

    MnO4^- + 2H2O --> MnO2 + 4OH-

    now the atoms are balanced, balance the charge.

    Add up the total charge on each side of the arrow

    LHS = 1 x MnO4^- = -1

    RHS = 4 x OH- = 4-

    add enough electrons to the least negative side to balance the charge

    add 3 e to the LHS

    MnO4^- + 2H2O +3e --> MnO2 + 4OH-

    balanced reduction half equation

    repeat for the oxidatin half equation

    BrO2^- ---> BrO4^-

    BrO2^- + 2H2O ---> BrO4^-

    BrO2^- + 2H2O ---> BrO4^- +4H+

    BrO2^- + 2H2O + 4OH- ---> BrO4^- +4H+ + 4OH-

    BrO2^- + 2H2O + 4OH- ---> BrO4^- + 4H2O

    BrO2^- + 4OH- ---> BrO4^- + 2H2O

    LHS = 1 x BrO2^- + 4 x OH- = 5-

    RHS = 1 x BrO4^- = 1-

    add 4 e to the RHS

    BrO2^- + 4OH- ---> BrO4^- + 2H2O + 4e

    balanced oxidation half equation

    the electrons in the 2 half equations do not balance, multiply each half equation by the number of electrons in the other half equation

    3 x (BrO2^- + 4OH- ---> BrO4^- + 2H2O + 4e)

    3BrO2^- + 12OH- ---> 3BrO4^- + 6H2O + 12e

    4 x (MnO4^- + 2H2O +3e --> MnO2 + 4OH-)

    4MnO4^- + 8H2O + 12e --> 4MnO2 + 16OH-

    add them together

    3BrO2^- + 12OH- ---> 3BrO4^- + 6H2O + 12e

    4MnO4^- + 8H2O + 12e --> 4MnO2 + 16OH-

    ---------------------------------------------------------------------------------

    3BrO2^- + 4MnO4^- + 12OH- + 8H2O +12e ---> 3BrO4^- + 4MnO2 + 6H2O + 16OH- + 12e

    simplify to get the balanced equation

    3BrO2^- + 4MnO4^- + 2H2O ---> 3BrO4^- + 4MnO2 + 4OH-

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