I am particularly interested in the reaction described in High‐Selectivity Electrochemical Conversion of CO2 to Ethanol using a Copper Nanoparticle/N‐Doped Graphene Electrode.

The reaction is

$$ \ce{2 CO + 9 H2O + 12 e- -> C2H5OH + 12 OH- E0 = 0.084 V vs SHE} $$

and has a recommended voltage of 1.2V.

My naive attempt is:

  • 1kWh @ 1.2V yields 30 moles electrons
  • 30 moles electrons has theoretical yield 2.5 moles of $\ce{C2H5OH}$

Do I also need to factor in the 63% Faradaic efficiency mentioned, so it'd be 1.575 moles of $\ce{C2H5OH}$ from 1 kWh input?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.