Question

A layer of chromium metal $\,0.25mm\,$ thick is to be plated on an auto bumper with a total area of $\,0.32{m^2}\,$ from a solution containing $\,Cr{O_4}^{2 - }\,$​. What current flow is required for this electroplating if the bumper is to be plated in $\,60s\,$? The density of chromium metal is $\,7.20g/c{m^3}\,$.
A.$\,4.9 \times {10^3}A\,$
B.$\,1.78 \times {10^3}A\,$
C.$\,5.3 \times {10^4}A\,$
D.$\,10.69 \times {10^4}A\,$

Answer 1

Hint:
Here electroplating is carried out. In electroplating, the deposited metal becomes part of the plating / coating of the existing product. In order to do the calculations of electroplating , we need a constant termed as Faraday’s constant. This constant reflects the electric charge magnitude per mole of electrons in chemistry and physics.
$\,1F = 96500C\,$
Formula used:
$\,n = \dfrac{w}{{GMM}}\,$ where $\,n\,$ is the number of moles, $\,w\,$ is the weight and $\,GMM\,$ is gram molecular mass.
$\,Q = F \times e\,$
$\,F = \,$ Faraday’s constant $\, = 96500C\,$ and $\,e = \,$number of electrons
$Q = I \times t\,$
So, $\,I = \dfrac{Q}{t}\,$
Where, $I = \,$ current passed, $\,Q = \,$ charge deposited $\,t = \,$ time taken

Complete step by step answer:
Let us first calculate the total volume of the metal layer here;
$\,Volume = area \times thickness\,$
So, $\,0.25 \times {10^3}m \times 0.32{m^2} = 8.0 \times {10^{ - 5}}{m^3}\,$
Now, we have to convert this to $\,c{m^3}\,$
So, $\,8.0 \times {10^{ - 5}} \times {10^6}c{m^3} = 80c{m^3}\,$
Now, we have to calculate the mass of chromium metal here
So, $mass = density \times volume\,$
$\,7.20g/c{m^3} \times 80c{m^3} = 576g\,$
Let us now calculate the number of moles of chromium here;
$\,n = \dfrac{w}{{GMM}}\,\,$ where $\,n\,$ is the number of moles, $\,w\,$ is the weight and $\,GMM\,$ is gram molecular mass.
Molar mass of chromium is $\,52g\,$, and weight we found out as $\,576g\,$
Therefore, $\,n = \dfrac{{576}}{{52}} = 11.07mol\,$
Now, let us calculate the number of moles of electrons in this number of moles of chromium;
Since, $\,1mol\,Cr = 6mol\,electrons\,$;
$\,11.07mol = 66.4mol\,electrons\,(11.07 \times 6)\,$
So, from this we can calculate the charge deposited which is $\,Q\,$
$\,Q = F \times e\,$
$\,F = \,$Faraday’s constant $\, = 96500C\,$and $\,e = \,$number of electrons $\, = 66.4mol\,$
So, by substituting we get;
$\, = 96500 \times 66.4\, = 6.41 \times {10^6}\,C\,$
Now we know, $Q = I \times t\,$
So, $\,I = \dfrac{Q}{t}\,$
Where, $I = \,$current passed, $\,Q = \,$charge deposited $\,t = \,$time taken
Here we have, $\,Q = 6.41 \times {10^6}C$and $\,t = 60s\,$
Hence, by substituting we get;
$I = \dfrac{{6.41 \times {{10}^6}}}{{60}} = 106833.33\,\,$
Which is equal to the value of $10.69 \times {10^4}\,A\,$
Hence, option D is the correct answer for this question.

Note:One Faraday is one mole of electrons. It is again equal to $\,96500C\,$.The use of faraday helps one to calculate how many moles of a material are formed during electrolysis, given that we know the transferred total charge or we can use this to find the charge also from number of moles as we saw in above solution. In anyways, faraday constant is an important factor in electrolytic calculations.