Why is the value of `E_(Cu^(2+)|Cu)^(0)(+0.34V)` positive ?

(CU^(+) ((aq)) is unstable in solution and undergoes simultaneous oxidation and reduction, according to the reaction 2Cu_((aq))^(+) Cu_((aq))^(2+) +Cu_((s)) choose E^(@) for the above reaction if E_(Cu^(2+)//Cu^(@))= 0.34V and E_(Cu^(2+)//Cu^(+) )= 0.15V

(CU^(+) ((aq)) is unstable in solution and undergoes simultaneous oxidation and reduction, according to the reaction 2Cu_((aq))^(+) Cu_((aq))^(2+) +Cu_((s)) choose E^(@) for the above reaction if E_(Cu^(2+)//Cu^(@))= 0.34V and E_(Cu^(2+)//Cu^(+) )= 0.15V

The electrode potential for Cu^(2+)(aq)+e rarr Cu^(+)(aq) and Cu^(+)(aq)+e rarr Cu(s) are 0.15V and 0.50V respectively. The value of E_(Cu^2+)|Cu^(@) will be -

The electrode potential for Cu^(2+) ( aq) + e^(-) rarr Cu^(+) (aq) Cu^(+) ( aq) + e^(-) rarr Cu(s) are + 0.15V and + 0.50 V respectively. The value of E_(Cu^(2+) // Cu)^(@) will be

Which of the following statements is correct ? If E_(Cu^(2+)|Cu)^(@) = 0.34 V and E_(Sn^(2+)|Sn)^(@) = -0.136 V , E_(H^(+)|H_2)^(@) = -0.0V

E.M.F of the cell reaction , 2Ag^(+) + Cu to 2Ag + Cu^(2+) is 0.46 V . If E_(Cu^(2+) // Cu)^(0) is | 0.34 V , E_(Ag^(+)//Ag)^(0) is