N2H4 + 2Fe(OH)2 ------> N2 + 2Fe+ 4 H2O

Question

N2H4 + 2Fe(OH)2 ------> N2 + 2Fe+ 4 H2O

In the above redox reaction, use oxidation numbers to identify the element oxidized, the element reduced, the oxidizing agent and the reducing agent.

a) name of the element oxidized:

b) name of the element reduced:

c) formula of the oxidizing agent:

d) formula of the reducing agent:

Answer 1

N₂H₄ + 2Fe(OH)₂ ------> N₂ + 2Fe + 4 H₂O

H combined with non metals has oxidation number of +1

H combined with metals has oxidation number of -1

O has an oxidation number of -2, except in peroxides

Elements have oxidation numbers of zero

Sum of all oxidation numbers should add up to the charge on the molecule, if any.

Now, we can assign oxidation numbers

N₂H₄ + 2Fe(OH)₂ ------> N₂ + 2Fe + 4 H₂O

^{-2 +1.........+2...-2..+1.............0...........0..........+1...-2}

So N in N₂H₄ goes from -2 on the left to zero on the right. It has been oxidized

H in N₂H₄ goes from +1 on the left to +1 on the right. No change

Fe goes from +2 on the left to zero on the right. It has been reduced

O in Fe(OH)2 goes from -2 on the left to -2 on the right. No change

H in Fe(OH)2 goes from +1 on the left to +1 on the right. No change.

a) Element oxidized is N

b) Element reduced is Fe

c) Formula of oxidizing agent is Fe(OH)₂

d) Formula of reducing agent is N₂H₄