Question

How many minutes will it takes to plate out 5.2 g of Cr from a $\text{ C}{{\text{r}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\text{ }$ solution using a current of 9.65 A? (Atomic weight of Cr = 52.0)
A) $\text{ }200\text{ }$
B) $\text{ 5}0\text{ }$
C) $\text{ 1}00\text{ }$
D) $\text{ 1}03\text{ }$

Answer 1

This question is based on the concept of Faraday’s law applicable in chemistry. The faraday's equation states the relationship between the current and time. The relation is given as follows,
$\text{ w = }\dfrac{\text{E}}{96500}\times \text{i}\times \text{t }$
Where w is the weight of the species, E is the equivalent weight of the species, ‘i’ is the current flowing through the solution , ‘t’ is the time required, and $\text{ }96500\text{ }$ is a faraday constant.

Complete Solution :
Now, we have given the following data in the question:
Current flowing through the solution is $\text{ i = 9}\text{. 65 A }$
The molecular mass of chromium is,$\text{ 52 g/mol }$
Weight of the $\text{ C}{{\text{r}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\text{ }$ given as, $\text{ w = 5}\text{.2 g }$
- First, we will discuss this question step by step. The compound given is $\text{ C}{{\text{r}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\text{ }$ when dissolve, it breaks into the chromium ion, and the sulphate ion, i.e.
$\text{ C}{{\text{r}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\to \text{C}{{\text{r}}^{3+}}\text{+S}{{\text{O}}_{4}}^{2-}\text{ }$

- At the cathode, the chromium undergoes the reduction reaction by accepting the 3 electrons. The reaction is as shown below,
$\text{ C}{{\text{r}}^{\text{3+}}}\text{ + 3}{{\text{e}}^{-}}\to \text{ Cr }$
Thus, the number of electrons involved in the reaction is equal to 3. So, here the equivalent weight of the $\text{ C}{{\text{r}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\text{ }$ is:

$\text{ Eq}\text{.Weight = }\dfrac{\text{Molar mass of C}{{\text{r}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}}{\text{No}\text{.of }{{\text{e}}^{-}}\text{ }}\text{ = }\dfrac{52}{3}\text{ }$

- We are interested to determine the time required for the deposition of $\text{ w = 5}\text{.2 g }$ weight. Modified the equation for time 't’ and Substitute values in the faraday's equation we have,
$\text{ }\begin{matrix} \text{t} & \text{=} & \dfrac{\text{96500 }\\!\\!\times\\!\\!\text{ w}}{\text{E }\\!\\!\times\\!\\!\text{ i}} \\\ \text{t} & \text{=} & \dfrac{\text{96500 }\\!\\!\times\\!\\!\text{ 5}\text{.2}}{\text{17}\text{.33 }\\!\\!\times\\!\\!\text{ 9}\text{.65}} \\\ \text{t} & \text{=} & \text{3000sec} \\\ \end{matrix}\text{ }$

- But we want to find the time in terms of minutes .we know that the,
$\text{ 1 min = 60 sec }$ .Thus divide the time in seconds by the 60 to convert into the minutes. We have,
$\text{ t = }\dfrac{3000\text{ se }}{60\text{ se}}\text{ }\times \text{ 1 min = 50 min }$
Thus, the 50 minutes will require takes to plate out 5.2 g of Cr from a $\text{ C}{{\text{r}}_{2}}{{\text{(S}{{\text{O}}_{4}}\text{)}}_{3}}\text{ }$ solution using a current of 9.65 A.
So, the correct answer is “Option B”.

Note: Note that, in solving such a question always consider the units in which the time to be found. Faraday's law always gives time in terms of the second. Thus, convert the time as per the requirements. Consider the following table for conversions:

Time	In second
$\text{ 1 min }$	$\text{60 sec }$
$\text{ 1 hr }$	$\text{ 60 min = 60 }\times \text{ 60 sec = 3600 sec }$
$\text{ 1 Day }$	$\text{ 24 hr = 24 }\times \text{ 3600 sec = 86400 sec }$.