Current Efficiency and Side Reactions

Introduction

The Faraday calculation gives an ideal product amount when every electron is used for the chosen electrode reaction. A real cell can send current through competing reactions, so less desired product forms. Current efficiency measures the fraction of charge that actually makes that product.

Core explanation

For a defined desired electrode reaction, current efficiency η = charge used for that reaction / total charge passed. If the desired product amount is proportional to charge by a fixed half-reaction, η can be calculated as actual electrochemical product amount divided by charge-based theoretical amount, provided the measured amount reflects formation accurately. A 0.80 efficiency means 80% of charge produced the desired species and 20% went to other faradaic processes or otherwise failed the simple assignment.

Suppose a charge predicts 10.0 g Cu from Cu²⁺ + 2e⁻ → Cu, but 8.0 g is deposited because some cathode current evolves hydrogen. Then η = 8.0/10.0 = 0.80, or 80%. For planning, actual expected mass = η × theoretical mass if the efficiency is known under comparable operating conditions. A 90% efficiency applied to a 5.00 g ideal prediction gives 4.50 g expected deposit.

Care is needed with “actual mass.” If 10.0 g copper was electrochemically formed but 2.0 g flaked off before weighing, the part gains only 8.0 g. The measured coating recovery is 80% of ideal, but that does not prove the electrical current efficiency was 80%; the loss was mechanical. Similarly, dissolved gas may escape collection, so collected volume can understate formed gas. Separate reaction selectivity from collection yield.

Side reactions depend on electrolyte, potential, electrode surface and concentrations. A cathode plating copper may also reduce water or hydrogen ions to hydrogen. An anode intended to evolve chlorine may also evolve oxygen. At very short times, charge can temporarily alter the electrode double layer without proportionate chemical product. A simple school efficiency problem normally specifies which interpretation to use.

The total current in one external circuit is the same charge flow through both electrode connections, but desired-product efficiencies at anode and cathode can differ. One side can have a clean reaction while the other has competing reactions. Product amount comparisons require the half-reaction on the particular side.

Step-by-step reasoning

1. Calculate ideal product from total charge and the desired half-reaction. 2. Identify the actual amount formed or recovered as specified. 3. Divide actual by theoretical and multiply by 100 for percent. 4. Explain likely side reactions or losses, distinguishing their types. 5. Check efficiency is between zero and 100% for the stated simple measure.

Visual explanation

Draw a 100 C charge bar split into 80 C for copper deposition and 20 C for hydrogen evolution. Next to it show a separate arrow for physical coating loss after formation, which is not the same charge split.

Real-world analogy

A factory may direct 80% of machine time to the desired product and the rest to other tasks. Losing finished items during shipping is a different problem. Current efficiency concerns how charge is used; recovery concerns what is collected.

Real-world example

In electroplating, hydrogen bubbles at a cathode may indicate a competing reduction. They can reduce the fraction of current available to deposit metal and may affect coating texture.

Why?

Why is measured product sometimes less than Faraday's ideal prediction? Some electrons can support other reactions, and formed product can also be lost before measurement. These causes should not be conflated.

Common misconception

“Any shortfall proves the Faraday constant is wrong.” The constant converts charge to electron amount; side reactions and collection losses explain many shortfalls.

Worked example

A run passes charge equivalent to 0.200 mol electrons. Cu²⁺ reduction could produce 0.100 mol Cu ideally. The recovered, well-adhered deposit is 0.0750 mol, and assume no physical loss after deposition. Efficiency is 0.0750/0.100 = 0.750, or 75.0%. The electron amount assigned to copper is 2 × 0.0750 = 0.150 mol, leaving 0.050 mol electron-equivalent charge for other processes.

Quick check

1. If 4.0 g is formed where 5.0 g is theoretically possible, what is simple current efficiency if no collection loss occurs? Answer: 4.0/5.0 × 100 = 80% of charge produced the desired product under the stated assumptions.

Exam focus

State what “actual” measures and whether losses are electrochemical or physical. Use the half-reaction-based theoretical amount in the denominator.

Advanced insight

Faradaic efficiency can change during a run as local ion concentration, surface condition and potential change. One reported overall percentage averages over those changing conditions.

Summary

Current efficiency compares charge assigned to a desired reaction with total charge. It can be inferred from product amounts when measurement reflects formation, but physical recovery loss is a separate issue. Side reactions lower desired yield.

Practice questions

1. A charge predicts 2.00 g deposit; 1.60 g forms with no physical loss. Find efficiency. Answer: 1.60/2.00 × 100 = 80.0%. 2. Does metal flaking after deposition necessarily lower faradaic efficiency? Answer: No. It lowers recovered mass, but the charge may still have formed the metal before it detached. 3. Name one side reaction during aqueous metal plating. Answer: Reduction of water or hydrogen ions to H₂ can consume cathode current that would otherwise deposit metal.