Convert electrical current into reaction amount

Supply work to drive an electrode reaction

A galvanic cell can supply electrical work through a favorable reaction. In electrolysis, an external source supplies work to drive the selected process. Oxidation still occurs at the anode and reduction at the cathode. The reaction definitions do not change when the energy source changes. In an operating electrolytic cell the external source connects its negative side to the cathode and positive side to the anode; electrode names should not be memorized solely from signs observed in a galvanic-cell diagram.

Our target cathodic half-reaction is Cu²⁺ + 2e− → Cu. We assume a supply maintains a constant current magnitude of 1.00 A and that a compensating anode reaction supports the charge transfer. Its chemistry is not specified here. Therefore this target half-reaction is not a complete cell equation or an instruction for assembling a cell.

The OpenStax introduction to electrolysis provides the electrode definitions and charge–amount relations. Our fixed-current experiment is constructed bookkeeping, not a measured plating run.

Translate the current into charge

A current of one ampere transfers one coulomb each second. For constant current magnitude I and duration t, Q = It. Q is charge magnitude in coulombs, I is current in amperes and t is time in seconds. If current changes, multiplying its final value by the full duration is generally wrong; the transferred charge is the accumulated area under the current-versus-time record.

Keep assigned copper current efficiency at 100%. At 300 s, the model passes 1.00 C/s × 300 s = 300 C. At 600 s it passes 600 C. Current remains 1.00 A: a longer duration increases accumulated charge, not the rate at which charge flows. The duration slider represents an assigned elapsed interval. It does not start a real-time process or background animation.

The charge chart reports a total and its two parts. Do not add all three rows as if they were independent supplies of charge. At 100% target efficiency, target charge equals total and the other-reaction allocation is zero.

Use electron stoichiometry to obtain product

The rounded Faraday constant is F = 96485 C/mol electrons. Dividing the 300 C target charge by F gives about 0.003109 mol electrons, or 3.109 mmol. The target reaction consumes two electrons per copper ion. Thus n(Cu) = Q_target/(2F), where n(Cu) is deposited copper amount in moles and Q_target is charge allocated to this reaction.

The predicted copper amount is about 1.555 mmol, not 3.109 mmol. The latter is the electron amount. The coefficient two is essential. If a different supplied target reaction needed one electron per metal ion, the same target charge would produce twice as many moles of metal. That would not mean its deposited mass was twice as large: masses also depend on molar masses.

The model begins with 10 mmol Cu²⁺ and assumes no replenishment from the anode and no chemical loss of deposited copper. After this run, about 8.445 mmol Cu²⁺ remain. Add dissolved and newly deposited copper: the result remains 10 mmol. The original substrate or electrode body is not included in this tracked copper inventory.

Use duration as a controlled comparison

Capture 0, 300 and 600 s while holding efficiency at 100%. At zero time, no charge has passed, no copper is deposited and all 10 mmol Cu²⁺ remain. At 600 s, deposition is about 3.109 mmol, twice the 300 s amount. These settings stay well below depletion of the initial supply. Extrapolating the same rule indefinitely would eventually become invalid.

A real run need not retain constant efficiency or current as conditions change. Nor does calculated amount establish coating thickness, uniformity, adhesion or surface coverage. Those need geometry, material properties and observations. This model deliberately does not turn a charge count into a claim about product quality.

State what is missing from an energy claim

The record “1.00 A for 300 s” determines charge but not electrical energy. If a separate measurement established a constant 2.00 V across a device for that interval, the supplied electrical energy would be 2.00 J/C × 300 C = 600 J. The 2.00 V is a hypothetical additional measurement, not a prediction of this widget or a required driving voltage.

A thermodynamic cell potential alone also does not establish the practical voltage required at a chosen current. Electrode kinetics, resistance and competing reactions can matter. An applied voltage does not guarantee that all charge makes the desired product; the next lesson investigates that allocation.

Transfer: A hypothetical constant-current record gives 0.50 A for 240 s, with all target charge assigned to Cu²⁺ reduction. Calculate charge and copper amount, then name one additional quantity needed to determine electrical energy input.

Check the electrolysis transfer

Q = 0.50 × 240 = 120 C. Copper amount = 120/(2 × 96485) mol, approximately 0.622 mmol. Electrical energy requires voltage information over the interval; if voltage is constant, energy is voltage times charge. Current and time alone do not specify the energy.