# Eletrolysis of water

The electrolyser used in the process at 25°C and 10.0 bar exhibits a global yield of 82% with a potential of 1.481 V. Determine the electrical energy that can be electrolyzed by year in order to get molar flow of hydrogen: $F(H_2) = 12.016 \times 10^6\ mol/h$. So, can anyone help with the calculation of energy required for this electrolysis ?

Update:
Yes, it's homework. I actually calculated the needed energy but i don't know if it's true.

Electrolysis: $\ce{2H_2O + Electricity -> 2H_2 + O_2}$.

The half-reaction is: $\ce{4e^- + 4H_2O(l) \to 2H_2(g) + 4OH^-(aq)}$ .

the molar flow of electrons is: $F(e) = 2F(\ce{H_2})/0.92 = 12.016 \times \frac{10^6}{0.92} = 14.653\cdot 10^6 \pu{mol/h}$.

Then, $1F(faraday\ constant) = q(e)\ Na = 1.602 \times 10^{-19}\times 6.023\times10^{23} = 96488.46 C/mol$.

$F(e) \ F = 14.653 \times 10^6 mol/h \times 96488.46 C/mol = 1413838.6 \times 10^6 C/h$

$I(current) = q/t = 1413838.6 \times 10^6 C/h \times 1h/3600s = 392.74 \times 10^6 A$

$E(energy) = UIt= 1.481 \times 392.74\times 10^6 \times 31536000s (1 year) = 1.834\times 10^{16} J$

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• This seems like homework, so show what you have done or attempted so far. We are not a free homework completion service. – Solar Mike Apr 13 '18 at 17:35