I thought it might be useful to put up some figures relating to the production of hydroxy from an electrolyser in order to have some base figures to work from.

The maximum gas output (theoretical) produced by standard Faraday electrolysis during the electrolysis of water tends to be the base figure against which all electrolysers and all other hydroxy processes are measured. Everyone compares their gas output with Faraday's maximum in order to evaluate the efficiency of their own electrolyser systems. And rightly so, because Faraday's laws of electrolysis are clear, concise and infallible.

So let's break this down into bite-size pieces.

Now, at the cathode an electrode potential of -0.41 V is required to produce hydrogen.

At the anode an electrode potential of +0.82 V is required to produce oxygen.

This gives us an overall potential of 1.23 volts.

This is the ideal voltage potential required to initiate electrolysis at the electrodes, and this is the voltage that Faraday's maximum gas output is based upon.

Now it takes 3.952 kW of electrical power to produce 1 mole of hydrogen and half a mole of oxygen per minute, which is 2.016 grams of hydrogen per minute and 16 grams of oxygen per minute.

If we now convert these grams into volumes of gas we get 24.46 litres of hydrogen and 12.23 litres of oxygen. So a total hydroxy gas output of 36.69 litres per minute.

W/V = A

3952 watts/1.23 volts = 3213 amps! This is 100% Faraday efficiency

But here's the thing, at 1.23 volts, minimal current flows through the water and the reality is that electrolysis takes place at a very, very slow pace. And though over time 36.69 litres of gas will be produced from 3.952kW it will take forever to get there as we simply cannot under the given circumstances generate the high current necessary to achieve this in a reasonable time.

In reality we employ voltages between around 1.5 volts and 2.0 volts, but this immediately reduces the Faraday efficiency. So too does adding an electrolyte. An electrolyte will increase the current flow for any given voltage, but it reduces the Faraday efficiency because as a consequence a slightly higher voltage potential is required now to initiate electrolysis.

You also have to consider that the electrodes themselves can play an important part in reducing Faraday efficiency as some will require greater potentials than -0.41 and +0.82. Platinum is generally regarded as the most efficient material to use as electrodes, but is obviously quite expensive.

Anything greater than Faraday's ideal 1.23 volts is known as an over-potential, and as such reduces the Faraday efficiency. And once you start to employ more voltage than you need to reach the potential to initiate electrolysis, you then start to lose efficiency due to energy being lost as heat. You will also lose efficiency due to side reactions - that brown scum you get means that side reactions have been taking place and any reaction that is unwanted will be detrimental to your gas output.

It seems very common within the HHO community to employ millilitres per minute per watt (usually denoted as mmw) as opposed to litres per minute (or hour) per kilowatt. I often see people talking about the mmw they need to exceed in order to achieve over-Faraday results.

So what mmw figure is 100% Faraday?

OK, if 100% Faraday produces 36.69 litres of hydroxy per minute at 3.952kW, we can convert this to mmw simply by division.

36.69 litres = 36690 ml

36690/3952 = 9.28 ml/m/w

But the reality of it is that you will be lucky to achieve 70% Faraday efficiency, which would be 6.49 ml/m/w with straight DC in a standard everyday Faraday electrolyser.

You also have to consider that your gas will surely contain some water vapour. Also be aware that the gas figures shown here are based on 1 atm and 25 deg C. The volume of the gases you collect will be directly dependent on atmospheric pressure and temperature.

If you see anything wrong with any of my calculations, please speak out.

The maximum gas output (theoretical) produced by standard Faraday electrolysis during the electrolysis of water tends to be the base figure against which all electrolysers and all other hydroxy processes are measured. Everyone compares their gas output with Faraday's maximum in order to evaluate the efficiency of their own electrolyser systems. And rightly so, because Faraday's laws of electrolysis are clear, concise and infallible.

So let's break this down into bite-size pieces.

Now, at the cathode an electrode potential of -0.41 V is required to produce hydrogen.

At the anode an electrode potential of +0.82 V is required to produce oxygen.

This gives us an overall potential of 1.23 volts.

This is the ideal voltage potential required to initiate electrolysis at the electrodes, and this is the voltage that Faraday's maximum gas output is based upon.

Now it takes 3.952 kW of electrical power to produce 1 mole of hydrogen and half a mole of oxygen per minute, which is 2.016 grams of hydrogen per minute and 16 grams of oxygen per minute.

If we now convert these grams into volumes of gas we get 24.46 litres of hydrogen and 12.23 litres of oxygen. So a total hydroxy gas output of 36.69 litres per minute.

W/V = A

3952 watts/1.23 volts = 3213 amps! This is 100% Faraday efficiency

But here's the thing, at 1.23 volts, minimal current flows through the water and the reality is that electrolysis takes place at a very, very slow pace. And though over time 36.69 litres of gas will be produced from 3.952kW it will take forever to get there as we simply cannot under the given circumstances generate the high current necessary to achieve this in a reasonable time.

In reality we employ voltages between around 1.5 volts and 2.0 volts, but this immediately reduces the Faraday efficiency. So too does adding an electrolyte. An electrolyte will increase the current flow for any given voltage, but it reduces the Faraday efficiency because as a consequence a slightly higher voltage potential is required now to initiate electrolysis.

You also have to consider that the electrodes themselves can play an important part in reducing Faraday efficiency as some will require greater potentials than -0.41 and +0.82. Platinum is generally regarded as the most efficient material to use as electrodes, but is obviously quite expensive.

Anything greater than Faraday's ideal 1.23 volts is known as an over-potential, and as such reduces the Faraday efficiency. And once you start to employ more voltage than you need to reach the potential to initiate electrolysis, you then start to lose efficiency due to energy being lost as heat. You will also lose efficiency due to side reactions - that brown scum you get means that side reactions have been taking place and any reaction that is unwanted will be detrimental to your gas output.

It seems very common within the HHO community to employ millilitres per minute per watt (usually denoted as mmw) as opposed to litres per minute (or hour) per kilowatt. I often see people talking about the mmw they need to exceed in order to achieve over-Faraday results.

So what mmw figure is 100% Faraday?

OK, if 100% Faraday produces 36.69 litres of hydroxy per minute at 3.952kW, we can convert this to mmw simply by division.

36.69 litres = 36690 ml

36690/3952 = 9.28 ml/m/w

But the reality of it is that you will be lucky to achieve 70% Faraday efficiency, which would be 6.49 ml/m/w with straight DC in a standard everyday Faraday electrolyser.

You also have to consider that your gas will surely contain some water vapour. Also be aware that the gas figures shown here are based on 1 atm and 25 deg C. The volume of the gases you collect will be directly dependent on atmospheric pressure and temperature.

If you see anything wrong with any of my calculations, please speak out.