Peter asked in Science & MathematicsChemistry · 6 years ago

# 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

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-

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-

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