Question

How many coulombs are required to plate a layer of chromium metal 0.28 mm thick on an auto bumper with a total area of 0.40 m2 from a solution containing CrO2−4? The density of chromium metal is 7.20 g/cm3.

Answer 1

Explanation:

The given data is as follows.

Thickness = 0.28 mm = $0.28 \times \frac{1}{10}$ cm = 0.028 cm

Area = 0.40 $m^{2}$ = $0.40 \times 10^{4} cm^{2}$ = 4000 $cm^{2}$

As, it is known that volume = area × thickness

So,             Volume = $4000 cm^{2} \times 0.028 cm$      

                                = 112 $cm^{3}$

As density is mass divided by volume. So, mass of chromium will be calculated as follows.

                   Density = $\frac{mass}{volume}$  

            7.20 $g/cm^{3}$ = $\frac{mass}{112 cm^{3}}$      

                      mass = 806.4 g

As, mass of 1 mole of chromium is 52 g. So, number of moles in 806.4 g of chromium will be as follows.

              No. of moles = $\frac{mass}{molar mass}$                            

                                     = $\frac{806.4 g}{52 g}$

                                     = 15.50 mol

In chromate ion, ($CrO^{2-}_{4}$) charge on Cr is +6. It means that 6 electrons are needed to reduce $Cr^{+6}$ into Cr.

As, 1 mole of $Cr^{+6}$ ions require 6 moles of electrons. Therefore, moles of electrons for 15.50 mol will be calculated as follows.

                              6 × 15.50 mol = 93.04 mol

To calculate number of electrons we multiply number of moles by Avogadro's number as follows.

               $93.04 mol \times 6.02 \times 10^{23}$

                      = $560.13 \times 10^{23}$

                     = $5.6 \times 10^{25}$ electrons

There is magnitude of $6.241 \times 10^{18}$ times the charge on an electron is equal to 1 coulomb.

Hence, number of coulombs will be as follows.

              No. of coulombs = $\frac{5.6 \times 10^{25}}{6.241 \times 10^{18}}$

                                           = $0.897 \times 10^{7}$ C

or,                                        = $8.97 \times 10^{6}$ C

Thus, we can conclude that $8.97 \times 10^{6}$ C are required to plate a layer of chromium metal with given data.