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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?

Text Solution

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`E^(@)(M^(2+)//M)` for any metal is based upon three factors which have been discussed in the text part.
`M(s) +Delta_(a)H to M(g), (Delta_(a)H=" Enthalpy of atomisation ")`
`M(g) + Delta_(i)H to M^(2+)(g), ( Delta_(i)H=" Ionisation enthalpy")`
`M^(2+)(g)+aq to M^(2+)(aq) , ( Delta_("hyd")H= " Hydration enthalpy")`
Copper has very high enthalpy of atomisation ( energy required ) and low enthalpy of hydration ( energy released ). In nut shell the `Delta_(i) H ` i.e. ionisation enthalpy needed is not compensated by the energy released. Therefore `E^(@)(Cu^(2+)//Cu)` is positive.
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