The d- and f-Block Elements

Chemistry · Class 12

Lesson 5 of 13 · 6 min

Electrode potentials

NCERT §4.3.5; §4.3.6

Meera drops a strip of copper and a strip of zinc into dilute acid. The zinc fizzes; the copper sits there. Electrode potentials explain why.

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The lesson in notes

In short

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).

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