SCIENCE · INVESTIGATE AND EXPLAIN

Separate target product from total charge

Partition electrode charge, compare current efficiencies and diagnose unreachable product targets.

Learning outcome: Calculate product amount and required duration using an assigned current efficiency, including zero-charge and zero-efficiency cases.

Starting recap

Total charge is current times duration for constant current. Only charge assigned to the target half-reaction determines its electron-based product amount.

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Try

Two runs pass 600 C but use 100% and 50% of that charge for copper deposition. Do they deposit the same amount?

Open the experiment

Predict

Predict deposited copper and other-reaction charge at 600 s with 100%, 50% and 0% assigned current efficiency.

Your prediction and reason

Experiment

Capture 600 s at 100% and 50%. Compare 300 s at 100% with 600 s at 50%. Then inspect 0% efficiency and zero duration separately. Find which sampled duration first exceeds a 1 mmol target at 50%, and explain the continuous required-time value.

Account for charge assigned to metal deposition

Worked teaching model · synthetic values · Constant external current magnitude 1.00 A. Cu²⁺ + 2e− → Cu is the target cathodic reaction; other cathodic reactions receive the remaining charge and do not involve copper. Assigned efficiency stays constant. Initially 10 mmol Cu²⁺, no copper replenishment or chemical loss of deposited copper. All supported settings leave Cu²⁺ available. A compensating anode reaction and adequate power supply are assumed, not simulated. Voltage, energy consumption, side-product identity, morphology and transport are not predicted. No physical procedure is provided.

Electrolysis duration: 300 s Assigned copper current efficiency: 100 %

Reset keeps your prediction, captures and practice history.

External supply → cathode: 1.00 C each second. Target reduction: Cu²⁺ + 2e− → Cu. The anode supports a compensating oxidation; this model does not specify its chemical products.

The bar length shows total charge; its two labeled segments partition that total. “Other” charge still participates in electrode reactions. It has not vanished or automatically become heat.

Target reaction amount and duration inference
Quantity	Value	Meaning
Target electron amount	3.109292 mmol	Target charge / 96485 C/mol, converted to mmol
Copper deposited	1.554646 mmol	Half the target electron amount
Copper(II) ions remaining	8.445354 mmol	10 mmol initially minus deposited copper
Efficiency inferred from this charge record	100%	Target charge / total charge; assigned control remains 100%
Time required for 1 mmol copper	192.97 s	Within the allowed duration range

Efficiency here is a stipulated charge fraction for simultaneous cathodic reactions, not energy efficiency or a prediction from electrode conditions. Required time is a continuous calculation; the duration control samples only every 60 seconds.

Calculated values on the same fixed scale
Row	Quantity	Value (C)
1	Total cathodic charge passed	300
2	Charge assigned to copper deposition	300
3	Charge assigned to other cathodic reactions	0

1.554646 mmol copper deposited; 8.445354 mmol Cu²⁺ remain.

Total charge = target charge + other-reaction charge. Do not add total to its two parts. At this assigned efficiency the 1 mmol copper target requires 192.97 s, within the 600 s control range.

Observation to attach to a capture

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Separate the desired reaction from the total current

A cathode may support more than one reduction. Passing 600 C therefore does not by itself establish that all 600 C reduced Cu²⁺ to copper. We assign a copper current efficiency: the fraction of total cathodic current allocated to that target reaction. In this model the fraction stays constant, so it also equals the target fraction of charge over any nonzero duration.

Let ε denote the assigned fraction, obtained by dividing the percentage control by 100. Then Q_target = εIt and n(Cu) = εIt/(2F). I is fixed at 1.00 A, t is duration in seconds, F is 96485 C/mol electrons and n(Cu) is moles of copper deposited. Every symbol describes an amount or conversion already visible in the ledger.

IUPAC’s definition of current efficiency relates an electrode reaction’s partial current to total current. Our model uses only simultaneous cathodic reactions, with no subsequent chemical production or consumption of copper; within these assumptions ε ranges from zero to one. This is a deliberately restricted case, not a universal assertion about every mixed-current electrode system.

Compare two runs with the same charge

At 600 s and 100% assigned copper efficiency, total and target charge are both 600 C. Copper deposition is about 3.109 mmol. At 600 s and 50%, total charge is still 600 C, but copper receives only 300 C. Its deposited amount falls to about 1.555 mmol. The remaining 300 C is assigned to other cathodic reactions.

The stacked bar has one fixed 0–600 C scale. Its first segment shows copper charge; its second shows the remainder. The parts add to total charge. Other-reaction charge does not disappear, and it is not automatically heat. The model assumes those other reactions do not involve copper, so they do not alter the tracked copper inventory. It does not identify their products because no competing half-reactions or experimental conditions have been supplied.

Do not confuse current efficiency with energy efficiency. A 50% charge allocation does not establish that 50% of the input energy is stored in the desired product. Energy accounting additionally requires voltage and thermodynamic information. A long run with high current efficiency could still use substantial electrical energy.

Make an equal-product comparison

Compare 300 s at 100% with 600 s at 50%. Both allocate 300 C to copper and deposit the same calculated amount. The second run passes twice as much total charge and allocates 300 C elsewhere. Equal target product is not evidence of equal resource use or equal by-products.

Hold efficiency fixed and double duration. This model doubles both target charge and deposition because it assumes a constant current and constant efficiency. In a physical system, efficiency can change with concentration, electrode condition and transport. Here it is an input, not something predicted from those mechanisms. Captures record the consequences of an assumption; they do not validate that assumption experimentally.

Solve for a target duration

Depositing 1 mmol copper requires 0.001 × 2 = 0.002 mol target electrons, corresponding to 192.97 C. At 1 A and efficiency ε, the target current is ε amperes. Required duration is t_target = 192.97/ε seconds for this fixed-current case.

At 100%, the duration is 192.97 s. At 50%, it is 385.94 s. The control uses 60-second steps, so 360 s is below that target and 420 s is the first available setting above it. The continuous estimate is not the same as the first sampled duration. Do not round the estimate before deciding which sampled setting reaches the target.

At 30%, the required duration is approximately 643.23 s. That is beyond the permitted 600 s experiment window. This means the target is outside the supported run duration at that efficiency, not that copper deposition is impossible. At 0%, however, the target receives no charge for any duration under the stated model. Dividing by zero is not allowed; the interface reports no finite time rather than a fabricated numerical answer.

Interpret the empty record carefully

Set duration to zero at 50% assigned efficiency. The control still says 50%, but both total and target charge are zero. An efficiency inferred from this record would require 0/0 and is undefined. An assumed parameter and evidence estimating that parameter are different things.

Now use positive duration with 0% assigned efficiency. Total charge is positive, target charge is zero, and the charge record gives zero target efficiency. This is a valid zero ratio. It is different from the empty-record case. Explain both in your reflection.

Transfer: A hypothetical run passes 400 C and deposits 1.000 mmol copper, with no later chemical changes. Infer the copper current efficiency. Then state whether this result alone identifies the other electrode products or the energy efficiency.

Check the current-efficiency transfer

The deposited copper corresponds to 2 × 0.001 × 96485 = 192.97 C target charge. Its fraction of 400 C is 0.482425, or about 48.24%. The other 207.03 C is not assigned to copper in this accounting. Product identity and energy efficiency cannot be inferred from this charge fraction alone.

Apply

Public practice checks a specific claim. It does not establish independent mastery.

Explain why unproductive-for-copper charge is not missing charge, why charge efficiency differs from energy efficiency, and why no-charge evidence cannot establish a measured efficiency.

Your explanation and limitations

Continue

Completion requires a saved prediction, at least one capture and a reflection. Presence checks do not grade your reasoning.

Skipping retains experiments and awards no mastery.

Previous: Convert electrical current into reaction amount Next: Diagnose corrosion and protection at a damaged surface

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