Dal Lake has about 8.2 × 1012 L of water. A power reactor produces electricity at the rate of 1.5 × 106 C s−1 at a suitable voltage. How many years would it take to electrolyse all the water of the lake into H2 and O2? (Density of water = 1 g mL−1, 1 F = 96500 C mol−1, 1 year = 3.15 × 107 s)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1.9 million years
By Faraday’s first law, the amount decomposed is fixed by the charge passed. Splitting one molecule of water takes 2 electrons, so 1 mol of \( \ce{H2O} \) needs 2 F:
\( \displaystyle \ce{H2O(l) -> H2(g) + 1/2O2(g)} \)
Mass of water (1 L = 1000 mL = 1000 g):
\( \displaystyle m=8.2\times10^{12}\,\mathrm{L}\times1000\,\mathrm{g\,L^{-1}}=8.2\times10^{15}\,\mathrm{g} \)
Moles of water:
\( \displaystyle n=\frac{8.2\times10^{15}\,\mathrm{g}}{18\,\mathrm{g\,mol^{-1}}}=4.56\times10^{14}\,\mathrm{mol} \)
Charge needed:
\( \displaystyle Q=2\times4.56\times10^{14}\,\mathrm{mol}\times96500\,\mathrm{C\,mol^{-1}}=8.80\times10^{19}\,\mathrm{C} \)
Time at 1.5 × 106 C s−1:
\( \displaystyle t=\frac{Q}{I}=\frac{8.80\times10^{19}\,\mathrm{C}}{1.5\times10^{6}\,\mathrm{C\,s^{-1}}}=5.87\times10^{13}\,\mathrm{s} \)
In years:
\( \displaystyle t=\frac{5.87\times10^{13}\,\mathrm{s}}{3.15\times10^{7}\,\mathrm{s\,year^{-1}}}=1.86\times10^{6}\,\text{years} \)
This is about 1.9 million years.