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# If 9 gm ${H_2}O$ is electrolysed completely with the current of 50% efficiency then A) 96500 charge is requiredB) $2 \times 96500{\text{ C}}$ charge is requiredC) 5.6 L of ${O_2}$ at STP will be formedD) 11.2 L of ${O_2}$ at STP will be formed.

Last updated date: 16th Jun 2024
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Hint: First, you need to do the electrolysis of water. The quantity of electricity is called Faraday. Current efficiency is 50% given, so we require two times the amount of electricity. Also, volume of 1 mole of a substance at STP is equal to 22.4 L.

Complete step by step solution:
Electrolysis reaction of water (${H_2}O$) is as follows:
${H_2}O \to {H_2} + \dfrac{1}{2}{O_2}$
Molar mass of ${H_2}O$ = 18 g/mol.
Molar mass of ${O_2}$ = 16 g/mol.
- On complete electrolysis, 1 mole of ${H_2}O$ is giving $\dfrac{1}{2}$ or 0.5 mole of ${O_2}$.
Now, you must know that volume occupied by 1 mole of any substance at STP is 22.4 L.
Therefore, by unitary method, volume occupied by 0.5 moles of ${O_2}$ at STP is equal to 11.2 L.
Or, we can also say that from 18 g of ${H_2}O$, 11.2 L of ${O_2}$ at STP is formed.
Now, we are given that 9 g of water is electrolysed completely.
Since, from 18 g of ${H_2}O$, 11.2 L of ${O_2}$ at STP is formed.
From 9 g of ${H_2}O$, $\dfrac{{11.2}}{2}L$ of ${O_2}$ at STP will be formed.
Or, From 9 g of ${H_2}O$, $5.6L$ of ${O_2}$ at STP will be formed. Hence, option C is correct.
- Moles of ${H_2}O$ in 9 g = $\dfrac{{{\text{Given mass}}}}{{{\text{Molar mass}}}} = \dfrac{{9g}}{{18g}} = 0.5{\text{ moles}}$
For electrolysis of 1 mole of ${H_2}O$, 2 moles of electrons are involved.
Therefore, for 0.5 mol of ${H_2}O$, 1 mole of electrons are required.
We are given that current efficiency is 50% when 9 g of ${H_2}O$ is electrolysed. When current efficiency gets halved, we need two times the amount of electricity (faraday). So, for 50% current efficiency, we require 2F of electricity i.e. $2 \times 96500{\text{ C}}$. Thus, option B is also correct.

Therefore, both options B and C are correct.

Note: Charge on one mole of electrons is equal to around 96500 C, which is the value of faraday constant i.e., F. It should be noted that, current efficiency for 18 g would be 100% that is, 1 F. Logically, when current efficiency decreases, amount of electricity required increases.