Question

A current of 1.70A is passed through 300mL of 0.160M solution of $ZnS{{O}_{4}}$ for 230s with a current efficiency of 90%. The molarity (M) of $Z{{n}^{2+}}$ after deposition of Zn is $1.54\times {{10}^{-x}}$, then what is the value of x? Assume the volume of solution to remain constant during the electrolysis.

Answer 1

Hint: Finding the molarity of the zinc ions left in the solution will give us the answer. To calculate it, first we have to find the number of zinc ions left in the solution which can be calculated by subtracting the initial number of equivalents by the number of ions lost. To find the number of equivalents of the ions lost, we can use the formula $\dfrac{i\times t}{96500}$.

Complete answer:
In the question it is given to us that a current of 1.70A is passed with 90% efficiency through a zinc sulphate solution.
As the efficiency of current passed is 90%, therefore we can write that the current passed through the zinc sulphate solution is $1.70\times 90%$.
Or, we can write that- $i=\dfrac{1.70\times 90}{100}=1.53A$. i.e. current that was passed through the solution was 1.53A.
Initially, number of equivalents of $Z{{n}^{2+}}$= volume $\times $ normality
We know, normality is the product of molarity and the n-factor. Here, the n-factor is 2 for $Z{{n}^{2+}}$.
Therefore, the number of equivalents of $Z{{n}^{2+}}$= $300\times 0.16\times 2=96$.
Now, we know that on passing current, the $Z{{n}^{2+}}$ ions will be lost and will be deposited as Zn. So, to calculate the number of equivalence of zinc ions lost, we can simply multiply the current passed by the solution by the time for which it was passed and we know that faraday constant gives us the magnitude of electric charge per mole of electrons.
Therefore,
Number of equivalent of lost $Z{{n}^{2+}}$=$\dfrac{i\times t}{96500}=\dfrac{1.53\times 230}{96500}=3.646\times {{10}^{-3}}milliequivalent\text{ of Z}{{\text{n}}^{2+}}$
As we know that milliequivalent is one thousandth of equivalent, therefore,
 Number of equivalents lost = 3.646 equivalents
Now, to find out the number of equivalents left in the solution, we can subtract the initial value by the lost value of equivalents.
So, the number of equivalents of $Z{{n}^{2+}}$ left in the solution = (96 – 3.646) = 92.354.
Therefore, molarity of the $Z{{n}^{2+}}$ left in the solution = $\dfrac{number\text{ }of\text{ }equivalents}{volume\times n-factor}=\dfrac{92.354}{300\times 2}=0.15349=1.54\times {{10}^{-1}}M$
As we can see from the above calculation that the molarity is $1.54\times {{10}^{-1}}M$.
Therefore, the value of ‘x’ is 1, which is the required answer.

Note: We should not be confused between the normality and the molarity of a solution. Molarity is the number of moles of solute per litres of solvent and normality is the gram equivalent or number of equivalents per litre of solvent.