[Lex Moen]

what is wrong at this page, is there anuthing wrong? http://www.atakama.org/littlemath.php

[Arkcon]

Well, just looking quickly at an anonymous page, its sloppily written, and uses inappropriate jargon.  But I can cut and paste the text block for everyone here's benefit:

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how efficient is the technique?

wikipedia states the efficiency of an electrolysiscell to be at a minimum of 60%, though higher percentages are claimed.
Let us assume 60%.
The efficiency of a hydrogenfuel cell is about 60% at minimum see wikipedia
So, when a solarpanel produces X Joules per unit of time, that means 0,6 X Joules is fixated into hydrogen and oxygen in the electrolysisvessel.
In the hydrogenfuelcell the hydrogen and oxygen are combined again to water generating electricity with an efficiency of 60%.

This means, that of the 0,6 X Joules per unit time, 0,36 X Joules of electricity is generated in the hydrogencell.

this 0,36 X Joule is used again to electrolyse an extra amount of water with an efficiency of 60%.

So, this adds another 0,216 X joules fixated in hydrogen in the electrolysisvessel.

this 0,216 X joule in hydrogen generates (60% efficiency of the hydrogencell) another 0,190 X joule of electricity in the hydrogencell,
which with an efficiency of 60% (in the electrolysisvessel)leads to 0,114 X Joule of fixated hydrogen.

so, this adds up to 0,6 + 0,216 + 0,114 + 0,04 = 97% efficiency

Does the similar name mean that this is your web page?  You might have just said so from the outset.  There are many layers of assumptions here, it is hard to know where exactly it might be wrong.  But yes, it isn't likely that consumption of the hydrogen produced by electrolysis produces nearly as much energy as the electrolysis itself.  Generally, there are losses in the engineering to take in to account.

[Enthalpy]

You can't add the production of the two electrolysis steps, because you have already consumed the first produced gas in the fuel cell.

[Jack Bauer]

Not sure what you are trying to achieve here but basically you are losing 40% to heat during every conversion (this is very rough btw - depends on the electrochemical device).

I don't follow this at at all - so, this adds up to 0,6 + 0,216 + 0,114 + 0,04 = 97% efficiency