Question

In the reaction
Cu+HNO3 →Cu2+ +NO2 +H2O
Deduce the oxidation state of all the elements on both the reactant and the product side.

Answer 1

LHS:
Cu: 0

H: +1
O: -2
N: +5

RHS:
Cu: 0

N: +4
O: -2

H: +1
O: -2

Answer 2

Rules for deciding oxidation number.
Anything in elemental state is 0 (eg., Cu or Na or O2)
All alkali metals(Na,K,Li,Rb) in any compunds are +1. eg, NaCl.
Oxygen is always -2.(except O2 and O3).
Hydrogen is always +1.
All other elements areobtained by balancing oxidation number and charge
If you apply these rules Cu is in elemental state on the left hence it is 0.
H is always +1 and O is -2.
In HNO3 net charge is 0. Hence sum of all oxidation numbers is0.If N has oxidation number x
So 1-2*3+x=0 x=5.
Hence N=5.
Now Cu in Cu2+ is obviously +2.
In NO2 O is -2 and net charge is 0. So x-2*2=0 x=4.

Answer 3

which reactant was oxidised and which was reduced?

Answer 4

When a reactant is oxidized, it means that it raises in oxidation number. By this...

Cu was oxidized.

N was reduced.

Answer 5

thank you, ive never done chemistry before so it's taking awhile to all make sense

Answer 6

Be sure you realize I wrote the oxidation for Cu(2+) wrong in my post. I didn't see the plus sign in your original equation.

As shankvee pointed out, its oxidation number is actually +2.