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

Step-by-step explanation:

The given data is as follows.

Thickness = 0.28 mm =
`0.28 * (1)/(10)` cm = 0.028 cm

Area = 0.40
`m^(2)` =
`0.40 * 10^(4) cm^(2)` = 4000
`cm^(2)`

As, it is known that volume = area × thickness

So, Volume =
`4000 cm^(2) * 0.028 cm`

= 112
`cm^(3)`

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

Density =
`(mass)/(volume)`

7.20
`g/cm^(3)` =
`(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 =
`(mass)/(molar mass)`

=
`(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 * 6.02 * 10^(23)`

=
`560.13 * 10^(23)`

=
`5.6 * 10^(25)` electrons

There is magnitude of
`6.241 * 10^(18)` times the charge on an electron is equal to 1 coulomb.

Hence, number of coulombs will be as follows.

No. of coulombs =
`(5.6 * 10^(25))/(6.241 * 10^(18))`

=
`0.897 * 10^(7)` C

or, =
`8.97 * 10^(6)` C

Thus, we can conclude that
`8.97 * 10^(6)` C are required to plate a layer of chromium metal with given data.