The `E^(0)(M^(2+)//M)` value for copper is positive `(+0.34V)`. What is possibly the reason for this?

Answer

`E^(0)(M^(2+)//M)` for any metals is related to the sum of the enthaply changes taking place in the following steps:` M(s)+Delta_(a)H to M(g), (Delta_(a) H="enthalpy of atomization")`
`M(g)+Delta_(i)H to M^(2+)(g), (Delta_(i)H="ionization enthalpy" I+II)`
`M^(2+)(g)+ag to m^(2+)(aq)+Delta_(hyd)H (Delta_(i)H="hydration enthalpy") `
Copper has high enthalpy of atomization (i.e.,energy absorbed ) and low enthalpy of hydration,(i.e.,energy released). The high energy energy required to transform `Cu(S)` to `Cu^(2+) (aq)` is not balanced by its hydration enthalpy. Hence, `E^(0)(Cu^(2+)//Cu)` is positive