ElectrochemistryHard

Galvanic series — emf calculation

Calculate the standard emf of: Al(s) | Al³⁺(aq) || Cu²⁺(aq) | Cu(s).
E°(Al³⁺/Al) = -1.66 V, E°(Cu²⁺/Cu) = +0.34 V.

Given data

E°(Al³⁺/Al) = -1.66 V (reduction)E°(Cu²⁺/Cu) = +0.34 V (reduction)

Steps

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Identify anode and cathode. Al oxidizes (more negative E°). E°_ox = ?
Write your working — the numerical result must appear in the text V
Cathode: Cu²⁺ reduces. E°_cat = +0.34 V. Calculate E°_cell.
Balance the reaction: Al + Cu²⁺ → Al³⁺ + Cu. Total e⁻ transferred?

Full worked solution

Identify anode and cathode. Al oxidizes (more negative E°). E°_ox = ?
$-(-1.66)$
E°_ox(Al) = -(-1.66) = +1.66 V.
Cathode: Cu²⁺ reduces. E°_cat = +0.34 V. Calculate E°_cell.
$0.34 + 1.66$
E°_cell = 0.34 + 1.66 = 2.00 V.
Balance the reaction: Al + Cu²⁺ → Al³⁺ + Cu. Total e⁻ transferred?
$3 \cdot 2$
2Al + 3Cu²⁺ → 2Al³⁺ + 3Cu, 6e⁻ transferred.

Result:E°_cell = 2.00 V.

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