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

Explanation

`E^(ɵ)((M^(2+))/(M))` for any metal is related to the sum of the enthalpy changes taking place in the following steps:
`M_(s)overset(triangle_aH^(ɵ))toM_(g)(triangle_aH^(ɵ)=` enthalpy of atomisation)
`M_(g)overset(triangle_iH^(ɵ))toM_(g)^(2+)(triangle_iH^(ɵ)=`ionisation enthalpy)
`M_(g)^(2+)+aq.overset(triangle_(hyd)H^(ɵ))toM_(aq)^(2+)(triangle_(hyd)H^(ɵ)=`hydration enthalpy)
Copper has high (positive) enthalpy of atomisation and low (negative) enthalpy of hydration.
Hence `E^(ɵ)((Cu^(2+))/(Cu))` is positive.