How to balance this redox equation?

How do you balance this basic redox solution?

nh4^+ + h2o2 --> n2o + h2

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  • Lexi R
    Lv 7
    6 years ago
    Favorite Answer

    NH4^+ ----> N2O

    balance the N's

    2NH4^+ ----> N2O

    balance the O by adding H2O to the side that is deficient in O

    2NH4^+ + H2O ----> N2O

    balance the H by adding H+ to the side deficient in H

    2NH4^+ + H2O ----> N2O + 10H+

    to change over to basic solution add 1 OH- to each side of the arrow for each H+ that was added

    2NH4^+ + H2O + 10OH- ----> N2O + 10H+ + 10OH-

    H+ + OH- makes H2O

    2NH4^+ + H2O + 10OH- ----> N2O + 10H2O

    cancel out H2O that occurs on both sides

    2NH4^+ + 10OH- ----> N2O + 9H2O

    now work out the electrons. Add up the total charge on each side of the arrow

    LHS: 2 x NH4+ + 10OH- = 8-

    RHS: 0

    now, add as many electrons to the least negative side as needed to balance the charge.

    ie, add 8e to the RHS

    2NH4^+ + 10OH- ----> N2O + 9H2O +8e

    this is the balanced oxidation half equation

    H2O2 -----> H2

    H2O2 -----> H2 + 2H2O

    H2O2 + 4H+ ---> H2 + 2H2O

    H2O2 + 4H+ + 4OH- ---> H2 + 2H2O + 4OH-

    H2O2 + 4H2O ---> H2 + 2H2O + 4OH-

    H2O2 + 2H2O ---> H2 + 4OH-

    LHS = 0

    RHS = 4 x OH- = 4-

    add 4e to the LHS

    H2O2 + 2H2O +4e ---> H2 + 4OH-

    this is the balanced reduction half equaion

    notice that the electrons are not equal. You need 2 of the reduction half equation to provide enough electrons to balance the oxidation half equation

    2 x (H2O2 + 2H2O +4e ---> H2 + 4OH-)

    2H2O2 + 4H2O +8e ---> 2H2 + 8OH-

    Now add them together,

    2H2O2 + 4H2O +8e ---> 2H2 + 8OH-

    2NH4^+ + 10OH- ----> N2O + 9H2O +8e

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

    2H2O2 + 2NH4^+ + 10OH- +4H2O + 8e ----> 2H2 + N2O + 9H2O + 8OH- +8e

    cancel out things that occur on both sides

    2H2O2 + 2NH4^+ + 2OH- ----> 2H2 + N2O + 5H2O

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