Electrolysis Stoichiometry

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How many faradays required to deposit 1.0 mol of Al (Al³⁺ + 3e⁻ → Al)?

RankGuru · Accepted Answer

1. **Half-Reaction**: $Al^{3+} + 3e^- 	o Al(s)$.
2. **Mole Interpretation**: To produce $1$ mole of Aluminum atoms, $3$ moles of electrons are required.
3. **Faraday definition**: $1$ Faraday ($F$) is the charge carried by $1$ mole of electrons.
4. **Calculation**: Total charge needed $= 3 	ext{ moles of electrons} = 3 	ext{ Faradays}$.
5. **Summary**: To deposit $1$ mole of $M^{n+}$, $n$ Faradays of charge are needed.

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Electrolysis Stoichiometry

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