Question

Dal lake has water $8.2\times {{10}^{12}}$ litre approximately. A power reactor produces electricity at the rate of $1.5\times {{10}^{6}}$ coulomb per second an appropriate voltage. How many years would it take to electrolyse the lake?
(A)- 3 million year
(B)- 4.1 million year
(C)- 1.9 million year
(D)- None of these

Answer 1

Hint: The amount of electricity passed is related to the number of moles of electrons in the reaction and also the current passed through the cell. This will provide the relation to obtain the time taken for the electrolysis from Faraday’s law.

Complete step by step answer:
Given the water at Dal lake is electrolysed by the power reactor. In order to determine the time taken to electrolyse the water of volume $8.2\times {{10}^{12}}$ 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: $2{{H}_{2}}O\to 4{{H}^{+}}+{{O}_{2}}+4{{e}^{-}}$
At the cathode: $2{{H}_{2}}O+2{{e}^{-}}\to {{H}_{2}}+2O{{H}^{-}}$

On multiplying the reaction at cathode by 2, we get the overall reaction as:
$2{{H}_{2}}O\to 2{{H}_{2}}+{{O}_{2}}$

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:
$Q=I\times t={{n}_{e}}\times F$ where F is the Faraday’s constant = 96500 C/mol ------- (a)

So, as we have 4 moles of electrons electrolyse 2 moles of water, from stoichiometry.
The molecular weight of water being 18g/mol, then 2 moles of water will be $(18\times 2)=36g$.

Also, the density of water being 1 g/mL or ${{10}^{3}}g/L$. The volume of water will be $\dfrac{36}{{{10}^{3}}}L$.
Then, form unitary method, the moles of electrons for the given volume $=8.2\times {{10}^{12}}L$ is$=\dfrac{4\times {{10}^{3}}}{36}\times 8.2\times {{10}^{12}}\,moles$ ---------- (b)

Now substituting the value of given the amount of current passed, that is, $1.5\times {{10}^{6}}$ coulomb/ second, equation (b) and F in equation (a), we will obtain the time taken as follows:
$t=\dfrac{{{n}_{e}}F}{I}$
$=\dfrac{4\times {{10}^{3}}\times 8.2\times {{10}^{12}}}{36}\times 96500\times \dfrac{1}{1.5\times {{10}^{6}}}=5.861\times {{10}^{13}}\text{seconds}$$=\dfrac{5.861\times {{10}^{13}}}{3.15\times {{10}^{7}}}=1.86\times {{10}^{6}}\approx 1.9\times {{10}^{6}}years$
So, the correct answer is “Option C”.

Note: The conversion of the moles of water to volume in the lake, through density is used. As, $Density=\dfrac{mass}{volume}$ .
Also, the time in seconds is divided by the number to convert into years, since $\text{1year = 3}\text{.15 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{7}}}\text{seconds}$.