Why is copper the odd one out?

E°(M²⁺/M) built from three bars: atomisation + (IE₁ + IE₂) up, hydration down; the net decides the sign, so Cu ends positive and Mn, Ni, Zn sit below the trend.

Find the only first-row metal whose net energy comes out uphill (positive E°), and the M³⁺/M²⁺ couple with the highest E°.

Mn³⁺ and Cr²⁺ behave alike because both are d⁴.

• E°(M²⁺/M) sums up three steps: atomising the metal, removing two electrons (ΔiH₁ + ΔiH₂), and hydrating the M²⁺ ion. First-row values: Ti −1.63, V −1.18, Cr −0.90, Mn −1.18, Fe −0.44, Co −0.28, Ni −0.25, Cu +0.34 and Zn −0.76 V.
• Copper alone has a positive E°(M²⁺/M), so it cannot release H₂ from acids; only oxidising acids (nitric acid and hot concentrated sulphuric acid) attack it, and the acid itself is reduced. Turning Cu(s) into Cu²⁺(aq) costs more energy than the hydration of Cu²⁺ returns.
• The general drift to less negative values across the row follows the rising sum of the first and second ionisation enthalpies.
• Mn, Ni and Zn are more negative than the trend predicts: Mn²⁺ (d⁵) and Zn²⁺ (d¹⁰) have stable configurations, and Ni²⁺ has the most negative hydration enthalpy.
• E°(M³⁺/M²⁺) values: Ti −0.37, V −0.26, Cr −0.41, Mn +1.57, Fe +0.77, Co +1.97 V. Sc³⁺ is very stable (noble gas core), so Sc's value is low.
• Mn's value is high because Mn²⁺ (d⁵) is especially stable, and it is much higher than Cr's or Fe's because the third ionisation enthalpy of Mn (d⁵ → d⁴) is so large; this is why Mn(III) matters little. Fe's value is comparatively low because Fe³⁺ is d⁵. V's is low because V²⁺ has a half-filled t₂g set (Unit 5). Zn would have the highest value, since it would mean breaking into d¹⁰.
• Cr²⁺ and Mn³⁺ are both d⁴, yet Cr²⁺ is a reducing agent (becoming d³, a half-filled t₂g set) while Mn³⁺ is an oxidising agent (becoming d⁵).

Teaching this idea? Put the simulation on your own page or LMS with this code.