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The E^(0)(M^(2+)//M) value for copper is...

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

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`E^(@) ( M^(2+) //M) ` for any metal is related to the sum of the enthalpy changes taking place in the following steps `: `
`M(s) + Delta_(a) H rarr M(g)`, `(Delta_(a) H =` enthalpy of atomisation sublimation )
`M(g) + Delta+(i) H rarr M^(2+) (g) `, `( Delta_(i) H =` ionization enthalpy )
`M^(2+) (g) + aq) rarr M^(2+) ( aq) + Delta_(hyd) H` `( Delta_(hyd) H =` hydration enthalpy )
Copper has high enthalpy of atomisation (i.e., energy absorbed ) and low enthalpy of hydration , ( i.e., energy released ). The high energy required to transform Cu(s) to `Cu^(2+) (aq)` , i.e., sum of enthalpies of sublimation and ionization is not balanced by its hydration enthalpy . Hence, `E^(@) (Cu^(2+) //Cu) ` is positive.
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