Calculating Efficiency of an Electrochemical cell

[deleted-819671]

I did the water to fuel to water project
https://www.sciencebuddies.org/science- ... /fuel-cell
but all of the voltages that I got from the cell were less than 1.23 volts. According to the website, this would be over 100% efficiency, which is not possible.

After the experiment, I connected the multimeter probes to either side of the 10K Ohm resistor, to find out how much voltage it drops. I found that it drops 18.64 volts from four 9V batteries whose sum of volts given was 36.1V. Using Ohm's law, the current was 18.64/10000 = 1.864 mA.

How did the creators of the project get 1.23 V as the ideal voltage? How is the ideal voltage found?

Thank you.

[norman40]

Hello snbhamidi,

The ideal voltage you asked about is the standard cell potential for splitting water into oxygen and hydrogen. It means that at least 1.23 volts are required to make the reaction proceed. More information is available at the following link.

https://en.wikipedia.org/wiki/Electrolysis_of_water

The voltage measurement you described suggests that that the voltage output from your batteries is low.
Please review Technical Note #1 in the project procedure for an explanation of the expected voltage and current from the batteries in the circuit. You might try replacing all four batteries and then repeat your voltage measurements.

I hope this helps. Please post again if you have more questions.

A. Norman