Question

A current of $1.93A$ is passed through $500ml$ of $0.8M - CuS{O_4}$ solution for $5000\sec $ , using inert electrodes. Assuming no change in volume of electrolytic solution and constant temperature of $25^\circ C$ , the molar concentration of ${H^ + }$ ions in final solution is
(a) $0.1M$
(b) $0.05M$
(c) $0.2M$
(d) $0.4M$

Answer 1

Hint: Here in these questions we have to understand the terms like using inert electrodes and no change in volume of the electrolytic solution and then using the formulas with the help of given data to find out the answer.

Formulas used:As we know, charge flowed formula:
$Q = I \times t$
Where, $Q$ is the charge flowed, $I$ is the current passed and $t$ is the time for which current flows.
$1$ Faraday consists of $96500C$of charge.
The concentration of the solution when the moles are given:
$N = \dfrac{n}{V}$
Where, $n$ is the number of moles, $N$ is the concentration of the solution and $V$ is the volume of the solution.

Complete step-by-step answer:Now, we start by analysing the given data:
Given,
Current passed, $I = 1.93A$
Temperature, $T = 25^\circ C$
Time, $t = 5000\sec $
Volume, $V = 500ml$
Morality of the solution, $M = 0.8M$
Here, the reaction of the given data,
Reaction at cathode:
$C{u^{2 + }} + 2{e^ - } \to Cu$
Reaction at anode:
${H_2}O \to \dfrac{1}{2}{O_2} + 2{H^ + } + 2{e^ - }$
First, we can find these by finding the number of faradays,
From laws of electrolysis,
$Q = I \times t$
Substituting the data given in the question
$Q = 1.93 \times 5000 = 9650C$
$1$ Faraday consists of $96500C$of charge.
Hence, $9650C$ charge consists $\dfrac{{9650}}{{96500}} = 0.1$ (by unitary method)
Hence, $0.1$ faraday charge is flowed.
When $2$ faraday charge is flowed, it releases $2$ moles of ${H^ + }$ .
So, when $0.1$ Faraday charge will flow there are $0.1$ mole of the ${H^ + }$ released.
The concentration of the solution when the moles are given:
$N = \dfrac{n}{V}$
Where, $n$ is the number of moles, $N$ is the concentration of the solution and $V$ is the volume of the solution.
Now, putting $n = 0.1$ mole and the volume as given, i.e, $500ml$
The concentration of the solution,
$N = \dfrac{{0.1 \times 1000}}{{500}} = 0.2M$
Hence, the concentration of the resulting solution will be $0.2M$ .

Hence, the correct option is (c) $0.2M$ .

Note:Faraday’s law of electrolysis has two laws, first law states that the amount of substance deposited or liberated at any electrode is directly proportional to the amount of charge flowed and the second law states that the amount of substance deposited or liberated at any electrode is directly proportional to its equivalent weight.