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Question: A layer of chromium metal \(\,0.25mm\,\) thick is to be plated on an auto bumper with a total area o...

A layer of chromium metal  0.25mm \,0.25mm\, thick is to be plated on an auto bumper with a total area of  0.32m2 \,0.32{m^2}\, from a solution containing  CrO42− \,Cr{O_4}^{2 - }\,​. What current flow is required for this electroplating if the bumper is to be plated in  60s \,60s\,? The density of chromium metal is  7.20g/cm3 \,7.20g/c{m^3}\,.
A. 4.9×103A \,4.9 \times {10^3}A\,
B. 1.78×103A \,1.78 \times {10^3}A\,
C. 5.3×104A \,5.3 \times {10^4}A\,
D. 10.69×104A \,10.69 \times {10^4}A\,

Explanation

Solution

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 \,1F = 96500C\,
Formula used:
 n=wGMM \,n = \dfrac{w}{{GMM}}\, where  n \,n\, is the number of moles,  w \,w\, is the weight and  GMM \,GMM\, is gram molecular mass.
 Q=F×e \,Q = F \times e\,
 F= \,F = \, Faraday’s constant  =96500C \, = 96500C\, and  e= \,e = \,number of electrons
Q=I×t Q = I \times t\,
So,  I=Qt \,I = \dfrac{Q}{t}\,
Where, I= I = \, current passed,  Q= \,Q = \, charge deposited  t= \,t = \, time taken

Complete step by step answer:
Let us first calculate the total volume of the metal layer here;
 Volume=area×thickness \,Volume = area \times thickness\,
So,  0.25×103m×0.32m2=8.0×10−5m3 \,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  cm3 \,c{m^3}\,
So,  8.0×10−5×106cm3=80cm3 \,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×volume mass = density \times volume\,
 7.20g/cm3×80cm3=576g \,7.20g/c{m^3} \times 80c{m^3} = 576g\,
Let us now calculate the number of moles of chromium here;
 n=wGMM  \,n = \dfrac{w}{{GMM}}\,\, where  n \,n\, is the number of moles,  w \,w\, is the weight and  GMM \,GMM\, is gram molecular mass.
Molar mass of chromium is  52g \,52g\,, and weight we found out as  576g \,576g\,
Therefore,  n=57652=11.07mol \,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 \,1mol\,Cr = 6mol\,electrons\,;
 11.07mol=66.4mol electrons (11.07×6) \,11.07mol = 66.4mol\,electrons\,(11.07 \times 6)\,
So, from this we can calculate the charge deposited which is  Q \,Q\,
 Q=F×e \,Q = F \times e\,
 F= \,F = \,Faraday’s constant  =96500C \, = 96500C\,and  e= \,e = \,number of electrons  =66.4mol \, = 66.4mol\,
So, by substituting we get;
 =96500×66.4 =6.41×106 C \, = 96500 \times 66.4\, = 6.41 \times {10^6}\,C\,
Now we know, Q=I×t Q = I \times t\,
So,  I=Qt \,I = \dfrac{Q}{t}\,
Where, I= I = \,current passed,  Q= \,Q = \,charge deposited  t= \,t = \,time taken
Here we have,  Q=6.41×106C\,Q = 6.41 \times {10^6}Cand  t=60s \,t = 60s\,
Hence, by substituting we get;
I=6.41×10660=106833.33  I = \dfrac{{6.41 \times {{10}^6}}}{{60}} = 106833.33\,\,
Which is equal to the value of 10.69×104 A 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 \,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.