[einstein]

Standard redox potentials for the systems Cu/Cu⁺ and Cu/Cu²⁺ have these values​​:
Cu_(s)=Cu⁺+e^-          E^o=-0.52V
Cu_(s)=Cu²⁺+2e^-       E^o= -0.34V
It is correct to say that negative values of potentials shows that elemental cooper
exceeds hard on its ion solution and from these values to conclude that Cu²⁺ ion is easier to be form than Cu⁺ ion and why is like this?

[DevaDevil]

The higher the oxidation potential, the easier the oxidation is.

So yes, you are indeed correct that it is easier to form Copper (II) than copper (I),
and copper (II) is more stable (see for example Cu⁺ --> Cu²⁺ + e^-; E⁰ = -0.16V, so it is very easy to further oxidize copper (I) )

You will have to explain me your question:

It is correct to say that negative values of potentials shows that elemental cooper
exceeds hard on its ion solution

because I am sorry to say I have no clue what you are asking here.

[einstein]

It was google translate fault ,  however I wanted to ask that does negative potential indicate that elemental cooper goes hard in its ion solution,and why is like that,I can't understand 

[DevaDevil]

in a clean solution of its own ion you will have Cu (s), Cu²⁺, H₂O, dissolved air and let's assume an inert anion.

The reactions of interest are the following:

Cu (s) --> Cu²⁺ (aq) + 2 e^-; E⁰ = -0.34 V
and
2 H₂O + O₂ + 4e^- → 4OH^-; E⁰ = 0.4V

so, will this happen or not?
This depends on the copper concentration and the oxygen concentrations. The equilibrium potentials are very close. But in a Cu-ion free solution, the copper will indeed go into solution, as indicated by the equilibrium potential