The positive value of E deg(Cu^2+/Cu), unlike the other 3d metals of the same row, is best explained by

the high enthalpy of atomisation and high second ionisation enthalpy of Cu, which the hydration enthalpy of Cu^2+ cannot compensate

the very high hydration enthalpy of Cu^2+ compared with all other M^2+ ions

the low enthalpy of atomisation of copper

the fact that copper has no 4s electrons

MolSynapse / Question bank / d- and f-Block Elements / Oxidation states and standard electrode potentials

JEE Main levelSingle correct optionDifficulty 2 of 3df066

The positive value of $E^{\circ} (Cu X^{2 +} / Cu)$ , unlike the other 3d metals of the same row, is best explained by

Athe high enthalpy of atomisation and high second ionisation enthalpy of Cu, which the hydration enthalpy of $Cu X^{2 +}$ cannot compensate
Bthe very high hydration enthalpy of $Cu X^{2 +}$ compared with all other $M X^{2 +}$ ions
Cthe low enthalpy of atomisation of copper
Dthe fact that copper has no $4 s$ electrons

Answer: A

E^{\circ} (M X^{2 +} / M)

reflects the sum of enthalpies of atomisation, ionisation and hydration. For Cu the first two terms are large (the second ionisation removes an electron from the stable

3 d^{10}

Cu X^{+}

), and the hydration enthalpy is not large enough to offset them, so Cu is hard to oxidise and

E^{\circ}

is positive.

Practice d- and f-Block Elements More on Oxidation states and standard electrode potentials

Previous questionNext question

The positive value of E∘(CuX2+/Cu), unlike the other 3d metals of the same row, is best explained by

Answer: A

More questions on Oxidation states and standard electrode potentials

The positive value of $E^{\circ} (Cu X^{2 +} / Cu)$ , unlike the other 3d metals of the same row, is best explained by