Q.$9.2\times10^{12}$ litres of water is available in a lake. A power reactor using the electrolysis of water in the lake produces electricity at the rate of $2\times10^{6}\ \text{C s}^{-1}$ at an appropriate voltage. How many years would it take to completely electrolyse the water in the lake? Assume there is no loss of water except due to electrolysis.

Step 1. Mass of water (assuming density 1 kg/L): $9.2\times10^{12}\ \text{L} = 9.2\times10^{12}\ \text{kg} = 9.2\times10^{15}$ g.

Step 2. Moles of water: $n(\text{H}_2\text{O}) = \dfrac{9.2\times10^{15}}{18} = 5.111\times10^{14}$ mol.

Step 3. Electrolysis of water, $2\text{H}_2\text{O} \rightarrow 2\text{H}_2+\text{O}_2$, requires 2 moles of electrons per mole of water decomposed (4 electrons per 2 $\text{H}_2\text{O}$). So electrons needed $= 2\times5.111\times10^{14} = 1.0222\times10^{15}$ mol.

Step 4. Charge required: $Q = 1.0222\times10^{15}\times96500 \approx 9.864\times10^{19}$ C. …

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