Dal Lake has water 8.2 × 10¹² L approximately. A power reactor produces electricity at the rate of 1.5 × 10⁶ coulomb per second at an appropriate voltage. How many years would it take to electrolyze the lake?

1.9 million years

The correct answer is option 3. 1.9 million years.

Given the water at Dal lake is electrolyzed by the power reactor. In order to determine the time taken to electrolyze the water of volume \(8.2 × 10^{12}\) litres, we will make use of Faraday's first law of electrolysis.

The reaction taking place during the process is as follows:

At the anode: \(2H_2O \longrightarrow 4H^+ + O_2 + 4e^−\)

At the cathode: \(2H_2O + 2e^− \longrightarrow H_2 + 2OH^−\)

On multiplying the reaction at cathode by 2, we get the overall reaction as:

\(2H_2O \longrightarrow 2H_2 + O_2\)

So, during the electrolysis, 4 moles of electrons are used to electrolyse the water molecule in the lake. Then, from Faraday’s law, the quantity of substance formed at the electrode is proportional to the amount of electricity, \(Q\) passed, where

\(Q = I \times t\)

Here, \(Q\) is stoichiometrically equivalent to moles of electrons supplied during the reaction. That is 4 moles of electrons. So, we have the relation between the current, time and the number of moles of electrons \((n_e)\) as follows:

\(Q = I × t = n_e × F\)

where \(F\) is the Faraday’s constant \( = 96500 C/mol\) ------- (i)

So, as we have 4 moles of electrons electrolyze \(2\) moles of water, from stoichiometry.

The molecular weight of water being \(18g/mol\), then \(2\) moles of water will be \((18 × 2) = 36g\).

Also, the density of water being \(1 g/mL\) or \(10^3\ , \ g/L\). The volume of water will be \(\frac{36}{10^3}\) L.

Then, form unitary method, the moles of electrons for the given volume \(= 8.2 × 10^{12} L\) is \(= \frac{4 × 10^3}{36} × 8.2 × 10^{12}\ , \ moles\) ---------- (ii)

Now substituting the value of given the amount of current passed, that is, \(1.5 × 10^6\)

coulomb/ second, equation (ii) and \(F\) in equation (i), we will obtain the time taken as follows:

\(t = \frac{n_eF}{I}\)

\(t = \frac{4 \times 10^3 \times 8.2 \times 10^2}{36} \times 96500 \times \frac{1}{1.5 \times 10^6} = 5.861 \times 10^{13}\, \ seconds\)

\(t = \frac{5.861 \times 10^{13}}{3.15 \times 10^7} = 1.86 \times 10^6 \approx 1.9 \times 10^6\, \ years\) [Since, \(1\, \ year = 3.15 \times 10^7\)]