Electrolysis, charge and chemical amount

The external source supplies the driving direction

Electrolysis uses an external electrical source to drive the specified reaction direction. Oxidation still occurs at the anode and reduction at the cathode. In ordinary driven electrolysis, the anode is positive and the cathode negative, unlike the galvanic-discharge signs in the previous model.

The necessary finite-current operating voltage can include kinetic overpotentials and ohmic loss as well as the reversible thermodynamic requirement. Reaction products also depend on medium and competing processes. A formal Al(III) + 3e⁻ → Al amount relation is not a claim that aluminium deposits from an ordinary aqueous Al³⁺ solution.

Charge counts electrons before it counts product

For constant current I over time t, electrical charge q_el = It, with ampere-seconds equal to coulombs. For varying current, use q_el = ∫I(t)dt or sum the interval contributions. The subscript distinguishes this electrical charge from the activity quotient Q.

Electron amount is n(e⁻) = q_el/F. If one product entity requires z electrons in the specified electrode reaction, its ideal amount is q_el/(zF). For Cu²⁺ + 2e⁻ → Cu, divide by 2. Negligible non-Faradaic charging, sufficient reactant supply and retained product are assumptions of this simple yield model.

Allocate the desired part of the current

An original compatible copper-deposition model uses constant I = 2.00 A for 30.0 min = 1.80 × 10³ s and supplied molar mass M_Cu = 63.546 g mol⁻¹. Total charge is 3.60 × 10³ C, giving ideal copper mass about 1.19 g. The original graph shows mass versus supplied charge on this reaction basis.

In this simple one-direction deposition model, a constant desired-product current fraction η = 0.800 allocates only 80.0% of total charge to copper formation. m = ηq_elM_Cu/(2F) then gives about 0.948 g at the same charge. This current fraction is not electrical-energy efficiency, and subsequent product dissolution would require a broader balance.

Integrate the current you actually have

For a separate hypothetical interval model, 1.00 A for 10.0 min followed by 2.00 A for 10.0 min gives q_el = 1.00 × 600 + 2.00 × 600 = 1.80 × 10³ C. Using the final current for the entire interval would incorrectly give 2.40 × 10³ C.

For a specified deposit mass with constant current and efficiency, solve t = mzF/(ηIM). A lower selected-product efficiency increases the required total charge and time under this model. It does not by itself quantify the electrical energy, which also depends on voltage along the process.

Worked examples

Try it yourself

1. A compatible Cu²⁺ deposition model passes 1.80 × 10³ C, with all current forming retained copper and negligible non-Faradaic charge. Use F = 96485.332 C mol⁻¹ and M(Cu) = 63.546 g mol⁻¹. What copper mass follows?

   Show the answer and reasoning

   Approximately 0.593 g.

   The electron amount is 1800/F ≈ 0.01866 mol. Divide by two for Cu²⁺ + 2e⁻ → Cu, then multiply by 63.546 g mol⁻¹: m = 1800 × 63.546/(2 × 96485.332) ≈ 0.5927 g.

2. Which time unit gives coulombs when multiplied by current in amperes?

   Show the answer and reasoning

   Seconds.

   One ampere-second equals one coulomb.

3. Does 80.0% desired-product current efficiency mean 80.0% electrical-energy efficiency?

   Show the answer and reasoning

   No.

   Product charge allocation and energy conversion are different quantities.

4. May the final current be multiplied by the entire time when the current changed earlier?

   Show the answer and reasoning

   No.

   Integrate the current or sum the actual interval charges.

What to carry forward

• State the driven reaction, medium and electrode roles.
• Convert charge into electron amount before product amount.
• Retain supply, product-retention and current-allocation assumptions.
• Changing current and electrical energy require their own integrals.