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.

The charge carried by one mole of electrons (one Faraday) is 96500 C; dividing the given 9650 C by this Faraday constant gives the number of moles of electrons, which is then converted to a number of electrons using Avogadro's number.

Step 1 — moles of electrons: one Faraday, $F = 96500\ C$, is the charge of exactly one mole ($6.022\times10^{23}$) of electrons. The number of moles of electrons corresponding to 9650 C is: $n = \dfrac{9650\ C}{96500\ C\,mol^{-1}} = 0.1\ mol$

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The Faraday constant, $F=N_A\times e\approx 96500\ C\,mol^{-1}$, is the total charge of Avogadro's number of electrons — i.e. the charge carried by one mole of electrons.

$F=N_A\times e$, where $N_A=6.022\times10^{23}\ mol^{-1}$ (Avogadro's number) and $e=1.602\times10^{-19}\ C$ (charge on a single electron); multiplying gives $F\approx96500\ C\,mol^{-1}$. Option (b) describes the charge of a single electron, not a mole of them. Option (c) understates the number of electrons ($6.022\times10^{23}$, not $6.22\times10^{10}$). Option (a) describes the charge needed to deposit one equivalent of substance during electrolysis (which numerically equals 1 Fara …