Question

Balance the following equation in basic conditions. Phases are optional.

N2H4+Cu(OH)2-->N2+Cu

Answer 1

the given reaction is

N2H4 + Cu(OH)2 ---> N2 + Cu

consider the half reactions

Cu(OH)2 ----> Cu

N2H4 ---> N2

1) first balance the atoms other than 0 and H

so

Cu(OH)2 ---> Cu

N2H4 ---> N2

2) now

balance oxygen by adding H20

so

Cu(OH)2 ---> Cu + 2H20

N2H4 ---> N2

now

3) balance hydrogen by adding protons

so

Cu(OH)2 + 2H+ ---> Cu + 2H20

N2H4 ---> N2 + 4H+

4) now

balance charge with e-

so

Cu(OH)2 + 2H+ + 2e- ----> Cu + 2H20

N2H4 ---> N2 + 4H+ + 4e-

5)

use multiples to get equal electrons in both reactions

so

2Cu(OH)2 + 4H+ + 4e- ---> 2Cu + 4H20

N2h4 ---> N2 + 4H+ + 4e-

6)

now add the reaction and cancel out the common terms

we get

2Cu(OH)2 + N2h4 ---> 2Cu + N2 + 4H20

So

the balanced reaction is

2 Cu(OH)2 + N2H4 ---> 2 Cu + N2 + 4H20