Redox ReactionsMedium

Standard potential and spontaneity

Determine if Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is spontaneous.
E°(Zn²⁺/Zn) = -0.76 V, E°(Cu²⁺/Cu) = +0.34 V.

Given data

E°(Zn²⁺/Zn) = -0.76 VE°(Cu²⁺/Cu) = +0.34 V

Steps

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Identify oxidation half-reaction (anode): Zn → Zn²⁺. E°_ox = ?
Write your working — the numerical result must appear in the text V
Calculate E°_cell = E°_cat + E°_an.
If E°_cell > 0, is the reaction spontaneous?

Full worked solution

Identify oxidation half-reaction (anode): Zn → Zn²⁺. E°_ox = ?
$-(-0.76)$
$-(-0.76) = +0.76 = \mathbf{0.76\,V}$ . Oxidation potential.
Calculate E°_cell = E°_cat + E°_an.
$0.34 + 0.76$
$0.34 + 0.76 = 1.10 = \mathbf{1.10\,V}$ . Cell potential.
If E°_cell > 0, is the reaction spontaneous?
$1.10\,\mathrm{V} > 0 = \mathbf{1.10\,V}$ . ΔG° < 0 → spontaneous.

Result:E°_cell = 1.10 V > 0 → spontaneous reaction.

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