Balance the equation in basic conditions. Phases are optional. equation: N_{2}H_{4} + Cu(OH)_{2} -> N_{2} + Cu N2H4+Cu(OH)2⟶N2+Cu

Answer 1

N₂H₄ + Cu(OH)₂ ==> N₂ + Cu ... unbalanced

N₂H₄ ==> N₂ ... oxidation half reaction

N₂H₄ +4OH- ==> N₂ + 4H2O ... balanced for N, O and H using base (OH-) and H2O

N₂H₄ +4OH- ==> N₂ + 4H2O + 4e- ... balanced for mass and charge

Cu(OH)₂ ==> Cu ... reduction half reaction

Cu(OH)₂ ==> Cu + 2H2O ... balanced for Cu and O

Cu(OH)₂ + 2H2O ==> Cu + 2H2O + 2OH- ... balanced for Cu, O and H using base (OH)- and H2O

Cu(OH)₂ + 2H2O + 2e- ==> Cu + 2H2O + 2OH- ... balanced for mass and charge

Since the reduction half reaction has only 2e-, we must multiply it by 2 to equal the electrons in the oxidation half reaction. This gives us a reduction half reaction of 2Cu(OH)₂ + 4H2O + 4e- ==> 2Cu + 4H2O + 4OH-

Add this to the oxidation half reaction and we have...

N₂H₄ +4OH- + 2Cu(OH)₂ + 4H2O + 4e- ==> N₂ + 4H2O + 4e- + 2Cu + 4H2O + 4OH-

Combining and cancelling like terms, we end up with our final balanced redox equation:

N₂H₄ + 2Cu(OH)₂ + ==> N₂ + 2Cu + 4H2O ... BALANCED REDOX EQUATION