The e value for copper is positive (+ 0.34V). What is possible reason for this?

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Solution

The total enthalpy change for the process M(s) $\to$ M⁺(aq) is the sum of enthalpy of atomisation, ionisation energy and hydration energy. The electrode potential are a measure of total enthalpy change.
Copper has a high energy of atomisation but a low hydration energy due to which the E⁰ value for Cu is positive.

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E^{e}

(M^{2 +} / M)

The E^θ(M²⁺/M) value for copper is positive (+0.34V). What is possibly the reason for this? (Hint: consider its high Δ_aH^θ and low Δ_hydH^θ)

2 C u^{+} \to {C u}^{+ 2} + C u, E^{\circ}

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0.34 V

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C u S, K_{s p} = 8.5 \times 10^{- 36}, E_{C u | C u^{2 +}}^{⊖} = - 0.34 V

x

y = x (1 + x)

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