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Electrolysis

Apply charge accounting to the original stipulated electrolysis gas models with accepted revisions; keep water/solute competition and gas conditions explicit.

Supporting original reading. Follow the AP topic pathway for reviewed scope, worked reasoning and targeted practice.

Chemistry II · original Chapter 9 · combined Chapter 17

Section Breakdown

Electrolysis
Electrolysis of Water
Electrolysis of Brine

Section 3

Electrolysis

So far, we’ve only used Faraday’s Laws as they apply to galvanic cell corrosion. What about the reverse? If you apply a current to water, what amount of the water would be electrolyzed?

Faraday’s laws don’t only apply to galvanic cells. In fact, their original derivation came from studies on the electrolysis of water:

$2H_2O_{(l)}\rightarrow 2H_{2(g)}+O_{2(g)}$

It is comprised of the half reactions:

$2H^+_{(aq)}+2e^-\rightarrow H_{2(g)}$ $E^\circ = 0.000V$

$O_{2(g)}+4H^+_{(aq)}+4e^-\rightarrow 2H_2O_{(l)}$ $E^\circ =1.229V$

As the electrolysis of water results in Hydrogen being reduced and oxygen being oxidized, that is against the $E^\circ$ -based conventions. This means that we are looking at an electrolytic cell.

Even though we’ve only looked at galvanic cells with Faraday’s law, the $Q$ can represent charge added to the cell, instead of given off by the cell. This allows us to calculate the amount of current needed to electrolyze a certain amount of substance. As we already have spoken much about water, we can begin there.

Example Problem 9.3.1

Reviewed explicit gas convention: ideal water electrolysis passes 4.00 A for 6.00 s with 100% Faradaic efficiency for H₂ formation. Find the ideal H₂ volume at 273.15 K and 100.0 kPa. Use F=96485.3 C mol⁻¹ and R=8.314462618 L kPa mol⁻¹ K⁻¹.

Q=It=4.00×6.00=24.0 C. Two moles of electrons form one mole of H₂, so n(H₂)=24.0/(2F)=1.24371×10⁻⁴ mol.

V=nRT/P=[24.0/(2×96485.3)]×8.314462618×273.15/100.0=0.00282459 L≈2.82 mL.

The temperature and pressure are stated explicitly; a different STP pressure convention gives a different molar volume. The source gas calculation is preserved in provenance.

Q=It=4.00×6.00=24.0 C. Two moles of electrons form one mole of H₂, so n(H₂)=24.0/(2F)=1.24371×10⁻⁴ mol.

V=nRT/P=[24.0/(2×96485.3)]×8.314462618×273.15/100.0=0.00282459 L≈2.82 mL.

The temperature and pressure are stated explicitly; a different STP pressure convention gives a different molar volume. The source gas calculation is preserved in provenance.

Let’s look at an unfamiliar electrolytic cell, the electrolysis of table salt, $NaCl$ . When it is electrolyzed as an aqueous solution, its half-reactions are:

For the stipulated aqueous hydrogen-forming cathode: 2 H₂O(l)+2 e⁻→H₂(g)+2 OH⁻(aq). The coefficient 2 on hydroxide is required to conserve both atoms and charge. Chlorine formation at the anode uses 2 Cl⁻→Cl₂+2 e⁻; its competition with oxygen evolution depends on conditions.

Ordinary aqueous Na⁺ is not plated as stable sodium metal by merely applying enough voltage: water reduction competes, and sodium reacts with water. Sodium production uses an appropriate molten or nonaqueous process. For aqueous electrolysis, identify the possible water and solute half-reactions and the stated conditions first.

$Na^+_{(aq)}+e^-\rightarrow Na_{(s)}$ $E^\circ=-2.71V$

Now, elemental sodium is a nasty feller. It will violently react with water, resulting in Hydrogen gas and lye:

$2Na_{(s)}+2H_2O_{(l)}\rightarrow 2NaOH_{(aq)}+H_{2(g)}$

Molten-salt sodium production and aqueous chloride electrolysis use different chemical models. In ordinary aqueous solution water reduction competes, and elemental sodium reacts with water. Standard electrode potentials alone do not set a practical aqueous-cell operating-voltage window; account for composition, pH, electrodes and overpotentials.

Example Problem 9.3.2

Reviewed chloride-electrolysis model: stipulate that an aqueous chloride cell produces H₂ at the cathode and Cl₂ at the anode, each with 100% Faradaic efficiency. Current is constant at 5.00 A for 1.00 h. Find the ideal volume of each separately collected gas at 273.15 K and 100.0 kPa. Use F=96485.3 C mol⁻¹ and R=8.314462618 L kPa mol⁻¹ K⁻¹. This task does not infer an operating-voltage window from saturated NaCl.

The stipulated half-reactions are 2 H₂O+2 e⁻→H₂+2 OH⁻ and 2 Cl⁻→Cl₂+2 e⁻. They give 2 H₂O+2 Cl⁻→H₂+Cl₂+2 OH⁻, conserving atoms and charge.

Q=It=5.00×3600=18000 C. Each gas requires two moles of electrons per mole, so n(H₂)=n(Cl₂)=18000/(2F)=0.0932785 mol.

For either gas, V=nRT/P=[18000/(2×96485.3)]×8.314462618×273.15/100.0=2.11844 L≈2.12 L.

This stipulated ideal amount calculation does not establish real electrode selectivity or operating voltage. A real aqueous-cell model also needs activities, pH, electrodes, overpotentials and competing water oxidation/reduction. Sodium is not simply plated as stable metal from ordinary aqueous NaCl.