Define coulombic efficiency in electrochemical synthesis.

• A. The fraction of theoretical product formed per charge passed, calculated as (moles of product × nF) / total charge.
• B. The energy efficiency of converting electrical energy into chemical energy.
• C. The rate at which charge is transferred per unit area.
• D. The fraction of charge wasted as heat.

Answer: (B)

Coulombic efficiency is about how effectively the electrical charge you pass is used to form the target product in an electrochemical synthesis. It’s the fraction of total charge that actually goes into making the desired chemical rather than driving side reactions or other pathways.

In practice, you quantify it by comparing the charge that would be needed to produce the observed amount of product with the total charge passed. You multiply the amount of product formed (in moles) by the number of electrons transferred per mole of product (n) and by Faraday’s constant (F), giving the charge that went toward the desired product. Then you divide by the total charge you passed (Q) and typically express it as a percentage:

Coulombic efficiency = (moles of product × n × F) / Q × 100%.

If there are significant side reactions—like hydrogen or oxygen evolution, or other unwanted redox processes—these consume charge without producing the desired product, lowering the coulombic efficiency. Conversely, if almost every electron goes toward forming the target product, the efficiency is high.

It’s important to distinguish this from energy efficiency. Energy efficiency looks at how much electrical energy is stored in the chemical product relative to the electrical energy you put in, which depends on the cell voltage and overpotentials. You can have high coulombic efficiency but not-so-great energy efficiency if the process runs at a high voltage, and the reverse can also occur.

So, the key idea is: coulombic efficiency = fraction of total charge that ends up forming the desired product, not the overall energy exchanged or the rate of charge transfer.