In the reaction
Cu+HNO3 →Cu2+ +NO2 +H2O
Deduce the oxidation state of all the elements on both the reactant and the product side.
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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.
Now Cu in Cu2+ is obviously +2.
In NO2 O is -2 and net charge is 0. So x-2*2=0 x=4.
which reactant was oxidised and which was reduced?